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## Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity

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### Samacheer Kalvi 12th Physics Current Electricity Textual Evaluation Solved

**Samacheer Kalvi 12th Physics Current Electricity Multiple Choice Questions **

**Current Electricity Class 12 Problems With Solutions State Board Question 1.**

The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor?

(a) 2 ohm

(b) 4 ohm

(c) 8 ohm

(d) 1 ohm

Answer:

(a) 2 ohm

**12th Physics Chapter 2 Book Back Answers Question 2.**

A wire of resistance 2 ohms per meter is bent to form a circle of radius 1 m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is-

(a) π Ω

(b) \(\frac { π }{ 2 }\) Ω

(c) 2π Ω

(d) \(\frac { π }{ 4 }\) Ω

Answer:

(b) \(\frac { π }{ 2 }\) Ω

**Samacheer Kalvi Guru 12th Physics Question 3.**

A toaster operating at 240 V has a resistance of 120 Ω. The power is

(a) 400 W

(b) 2 W

(c) 480 W

(d) 240 W

Answer:

(c) 480 W

**Current Electricity Class 12 State Board Question 4.**

A carbon resistor of (47 ± 4.7) k Ω to be marked with rings of different colours for its identification. The colour code sequence will be

(a) Yellow – Green – Violet – Gold

(b) Yellow – Violet – Orange – Silver

(c) Violet – Yellow – Orange – Silver

(d) Green – Orange – Violet – Gold

Answer:

(b) Yellow – Violet – Orange – Silver

**12th Physics Lesson 2 Book Back Answers Question 5.**

What is the value of resistance of the following resistor?

(a) 100 k Ω

(b) 10 k Ω M

(c) 1 k Ω

(d) 1000 k Ω

Answer:

(a) 100 k Ω

**12th Physics Samacheer Kalvi Question 6.**

Two wires of A and B with circular cross section made up of the same material with equal lengths. Suppose R_{A} = 3 R_{B}, then what is the ratio of radius of wire A to that of B?

(a) 3

(b) √3

(c) \(\frac { 1 }{ √3 }\)

(d) \(\frac { 1 }{ 3 }\)

Answer:

(c) \(\frac { 1 }{ √3 }\)

**12th Physics Solutions Samacheer Kalvi Question 7.**

A wire connected to a power supply of 230 V has power dissipation P_{1} Suppose the wire is cut into two equal pieces and connected parallel to the same power supply. In this case power dissipation is P_{2}. The ratio \(\frac {{ p }_{2}}{{ p }_{1}}\) is.

(a) 1

(b) 2

(c) 3

(d) 4

Answer:

(d) 4

**Samacheer Kalvi 12th Physics Question 8.**

In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60W bulb for use in India is R, the resistance of a 60W bulb for use in USA will be

(a) R

(b) 2R

(c) \(\frac { R }{ 4 }\)

(d) \(\frac { R }{ 2 }\)

Answer:

(c) \(\frac { R }{ 4 }\)

**Current Electricity Class 12 Pdf Question 9.**

In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1kW are connected. The voltage of electric mains is 220 V. The minimum capacity of the main fuse of the building will be (IIT-JEE 2014)

(a) 14 A

(b) 8 A

(c) 10 A

(d) 12 A

Answer:

(d) 12 A

**Current Electricity Short Notes Question 10.**

There is a current of 1.0 A in the circuit shown below. What is the resistance of P ?

(a) 1.5 Ω

(b) 2.5 Ω

(c) 3.5 Ω

(d) 4.5 Ω

Answer:

(c) 3.5 Ω

**Samacheerkalvi.Guru 12th Physics Question 11.**

What is the current out of the battery?

(а) 1 A

(b) 2 A

(c) 3 A

(d) 4 A

Answer:

(а) 1 A

**Samacheer Kalvi 12th Physics Solutions Question 12.**

The temperature coefficient of resistance of a wire is 0.00125 per °C. At 300 K, its resistance is 1 Ω. The resistance of the wire will be 2 Ω. at

(a) 1154 K

(ft) 1100 K

(c) 1400 K

(d) 1127 K

Answer:

(d) 1127 K

**Samacheer Kalvi.Guru 12th Physics Question 13.**

The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10Ω is

(a) 0.2 Ω

(b) 0.5 Ω

(c) 0.8 Ω

(d) 1.0 Ω

Answer:

(b) 0.5 Ω

**Samacheer Kalvi 12th Physics Solution Book Question 14.**

A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of

(a) each of them increases

(b) each of them decreases

(c) copper increases and germanium decreases

(d) copper decreases and germanium increases

Answer:

(d) copper decreases and germanium increases

**12 Physics Samacheer Kalvi Question 15.**

In Joule’s heating law, when I and t are constant, if the H is taken along the y axis and I^{2} along the x axis, the graph is

(a) straight line

(b) parabola

(c) circle

(d) ellipse

Answer:

(a) straight line

**Samacheer Kalvi 12th Physics Current Electricity Short Answer Questions**

**Samacheer Kalvi Class 12 Physics Solutions Question 1.**

Why current is a scalar?

Answer:

The current I is defined as the scalar product of current density and area vector in which the charges cross.

I = \(\vec { j } \) . \(\vec { A } \)

The dot product of two vector quantity is a scalar form. Hence, current is called as a scalar quantity.

**Samacheer Kalvi Guru Physics Question 2.**

Distinguish between drift velocity and mobility.

Answer:

**Physics Solution Class 12 Samacheer Kalvi Question 3.**

State microscopic form of Ohm’s law.

Answer:

Current density J at a point in a conductor is the amount of current flowing per unit area of the conductor around that point provided the area is held in a direction normal to the current.

**Physics Class 12 Samacheer Kalvi Solutions Question 4.**

State macroscopic form of Ohm’s law.

Answer:

The macroscopic form of Ohm’s Law relates voltage, current and resistance. Ohm’s Law states that the current through an object is proportional to the voltage across it and inversely proportional to the object’s resistance.

V = IR.

**Samacheer 12th Physics Solutions Question 5.**

What are ohmic and non-ohmic devices?

Answer:

Materials for which the current against voltage graph is a straight line through the origin, are said to obey Ohm’s law and their behaviour is said to be ohmic. Materials or devices that do not follow Ohm’s law are said to be non-ohmic.

**Samacheer Kalvi Physics Question 6.**

Define electrical resistivity.

Answer:

Electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having unit area of cross section.

**12th Physics Samacheer Kalvi Solutions Question 7.**

Define temperature coefficient of resistance.

Answer:

It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity at T_{0}

**Current Electricity Class 12 Notes Question 8.**

What is superconductivity?

Answer:

The ability of certain metals, their compounds and alloys to conduct electricity with zero resistance at very low temperatures is called superconductivity.

**Class 12 Samacheer Physics Solutions Question 9.**

What is electric power and electric energy?

Answer:

1. Electric power:

It is the rate at which an electric appliance converts electric energy into other forms of energy. Or, it is the rate at which work is done by a source of emf in maintaining an electric current through a circuit.

P = \(\frac { W }{ t }\) = VI = I^{2}R = \(\frac {{ V }_{2}}{ R }\)

2. Electric energy:

It is the total workdone in maintaining an electric current in an electric circuit for a given time.

W = Pt = VIt joule = I^{2}Rt joule.

**12th Samacheer Physics Solutions Question 10.**

Define current density.

Answer:

The current density (J) is defined as the current per unit area of cross section of the conductor

J = \(\frac { 1 }{ A }\)

The S.I unit of current density.

\(\frac { A }{{ m }^{2}}\)

Or

Am^{2}

**Current Electricity Class 12 Question 11.**

Derive the expression for power P = VI in electrical circuit.

Answer:

The electrical power P is the rate at which the electrical potential energy is delivered,

P = \(\frac { dU }{ dt }\) = \(\frac { d }{ dt }\) (V.dQ) = V\(\frac { dQ }{ dt }\)

Since the electric current I = \(\frac { dQ }{ dt }\)

So the equation can be rewritten as P = VI.

**Current Electricity Class 12 Notes Pdf Download Question 12.**

Write down the various forms of expression for power in electrical circuit.

Answer:

The electric power P is the rate at which the electrical potential energy is delivered,

P = \(\frac { dU }{ dt }\) = \(\frac { 1 }{ dt }\) (V.dQ) = V.\(\frac { dQ }{ dt }\)

[dU = V.dQ]

The electric power delivered by the battery to any electrical system.

P = VI

The electric power delivered to the resistance R is expressed in other forms.

P = VI = I(IR) = I^{2}R

P = IV = \((\frac { V }{ R })\) V = \(\frac {{ V }^{2}}{ R }\).

**Physics Class 12 Samacheer Kalvi Question 13.**

State Kirchhoff’s current rule.

Answer:

It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a statement of conservation of electric charge.

**12th Std Samacheer Physics Solutions Question 14.**

State Kirchhoff’s voltage rule.

Answer:

It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system.

**Current Electricity Class 12 Problems With Solutions Pdf Question 15.**

State the principle of potentiometer.

Answer:

The basic principle of a potentiometer is that when a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.

**Current Electricity Class 12 Solutions Question 16.**

What do you mean by internal resistance of a cell?

Answer:

The resistance offered by the electrolyte of a cell to the flow of current between its electrodes is called internal resistance of the cell. An ideal battery has zero internal resistance and the potential difference across the battery equals to its emf. But a real battery is made of electrodes and electrolyte, there is resistance to the flow of charges within the battery. A freshly prepared cell has low internal resistance and it increases with ageing.

Question 17.

State Joule’s law of heating.

Answer:

It states that the heat developed in an electrical circuit due to the flow of current varies directly as:

- the square of the current
- the resistance of the circuit and
- the time of flow.

H = I^{2}R?

Question 18.

What is Seebeck effect?

Answer:

Seebeck discovered that in a closed circuit consisting of two dissimilar metals, when the junctions are maintained at different temperatures, an emf (potential difference) is developed.

Question 19.

What is Thomson effect?

Answer:

Thomson showed that if two points in a conductor are at different temperatures, the density of electrons at these points will differ and as a result the potential difference is created between these points. Thomson effect is also reversible.

Question 20.

What is Peltier effect?

Answer:

When an electric current is passed through a circuit of a thermocouple, heat is evolved at one junction and absorbed at the other junction. This is known as Peltier effect.

Question 21.

State the applications of Seebeck effect.

Answer:

Applications of Seebeck effect:

- Seebeck effect is used in thermoelectric generators (Seebeck generators). These thermoelectric generators are used in power plants to convert waste heat into electricity.
- This effect is utilised in automobiles as automotive thermoelectric generators for increasing fuel efficiency.
- Seebeck effect is used in thermocouples and thermopiles to measure the temperature difference between the two objects.

**Samacheer Kalvi 12th Physics Current Electricity Long Answer Questions**

Question 1.

Describe the microscopic model of current and obtain genera! form of Ohm’s Law.

Answer:

Microscopic model of current: Consider a conductor with area of cross-section A and an electric field E applied from right to left. Suppose there are n electrons per unit volume in the conductor and assume that all the electrons move with the same drift velocity \(\vec { v } \)_{d}.

The drift velocity of the electrons = v_{d}

The electrons move through a distance dx within a small interval of dt

v_{d} = \(\frac { dx }{ dt }\); dx = v_{d}dt ….. (1)

Since A is the area of cross section of the conductor, the electrons available in the volume or length dx is

= volume x number per unit volume

= A dx × n …… (2)

Substituting for dx from equation (1) in (2)

= (A v_{d} dt)n

Total charge in volume element dQ = (charge) x (number of electrons in the volume element)

dQ= (e)(A v_{d} dt)n

Hence the current, I = \(\frac { dQ }{ dt }\) = \(\frac{n e A v_{d} d t}{d t}\)

I = ne A v_{d} …….. (3)

Current denshy (J):

The current density (J) is defined as the current per unit area of cross section of the conductor

J = \(\frac { I }{ A }\)

The S.I. unit of current density,\(\frac { A }{{ m }^{2}}\) (or) Am^{-2}

J = \(\frac {{ neA v }_{d}}{ A }\) (from equation 3)

J = nev_{d} …….. (4)

The above expression is valid only when the direction of the current is perpendicular to the area A. In general, the current density is a vector quantity and it is given by

\(\vec { J } \) = ne\(\vec { v } \)_{d}

Substituting i from equation \(\vec { v } \)_{d} = \(\frac { -eτ }{ m }\) \(\vec { E } \)

\(\vec { J } \) = –\(\frac{n \cdot e^{2} \tau}{m}\)\(\vec { E } \) …… (5)

\(\vec { J } \) = -σ\(\vec { E } \)

But conventionally, we take the direction of (conventional) current density as the direction of electric field. So the above equation becomes

\(\vec { J } \) = σ\(\vec { E } \) …….. (6)

where σ = \(\frac{n \cdot e^{2} \tau}{m}\) is called conductivity.

The equation 6 is called microscopic form of ohm’s law.

Question 2.

Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.

Answer:

Ohm’s law: The Ohm’s law can be derived from the equation J = σE. Consider a segment of wire of length l and cross sectional area A.

When a potential difference V is applied across the wire, a net electric field is created in the wire which constitutes the current. For simplicity, we assume that the electric field is uniform in the entire length of the wire, the potential difference (voltage V) can be written as V = El

As we know, the magnitude of current density

J = σE = σ\(\frac { V }{ l }\) ……. (1)

But J = \(\frac { I }{ A }\), so we write the equation as

\(\frac { I }{ A }\) σ\(\frac { V }{ l }\)

By rearranging the above equation, we get

V = I \(\left( \frac { l }{ \sigma A } \right) \) ……… (2)

The quantity \(\frac { l }{ \sigma A } \)is called resistance of the conductor and it is denoted as R. Note that the oA resistance is directly proportional to the length of the conductor and inversely proportional to area of cross section.

Therefore, the macroscopic form of Ohm’s law can be stated as

V = IR.

Question 3.

Explain the equivalent resistance of a series and parallel resistor network.

Answer:

1. Resistors in series:

When two or more resistors are connected end to end, they are said to be in series. The resistors could be simple resistors or bulbs or heating elements or other devices. Fig. (a) shows three resistors R_{1}, R_{2} and R_{3} connected in series.

The amount of charge passing through resistor R_{1} must also pass through resistors R_{2} and R_{3} since the charges cannot accumulate anywhere in the circuit. Due to this reason, the current I passing through all the three resistors is the same. According to Ohm’s law, if same current pass through different resistors of different values, then the potential difference across each resistor must be different. Let V_{1}, V_{2} and V_{3} be the potential difference (voltage) across each of the resistors R_{1}, R_{2} and R_{3} respectively, then we can write V_{1 }= IR_{1,} V_{2} = IR_{2} and V_{3}= IR_{3}. But the total voltage V is equal to the sum of voltages across each resistor.

V = V_{1} + V_{2} + V_{3}

= IR_{1} + IR_{2} + IR_{3} ….. (1)

V = I(R_{1} + R_{2} +R_{3})

V = I.R_{S} …… (2)

where R_{s} = R_{1} + R_{2} R_{3} ……. (3)

When several resistances are connected in series, the total or equivalent resistance is the sum of the individual resistances as shown in fig. (b).

Note:

The value of equivalent resistance in series connection will be greater than each individual resistance.

2. Resistors in parallel:

Resistors are in parallel when they are connected across the same potential difference as shown in figer.

In this case, the total current I that leaves the battery in split into three separate paths. Let I_{1}, I_{2} and I_{3} be the current through the resistors R_{1}, R_{2} and R_{3} respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.

I = I_{1} + I_{2} + I_{3} ……. (1)

Since the voltage across each resistor is the same, applying Ohm’s law to each resistor, we have

I_{1} = \(\frac { V }{{ R }_{1}}\) I_{2} = \(\frac { V }{{ R }_{2}}\),I_{1} = \(\frac { V }{{ R }_{3}}\)

Substituting these values in equation (1), we get

I = \(\frac { V }{{ R }_{1}}\) + \(\frac { V }{{ R }_{2}}\) +\(\frac { V }{{ R }_{3}}\) = V\(\left[ \frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } \right] \)

I = \(\frac { V }{{ R }_{p}}\)

\(\frac { 1 }{{ R }_{p}}\) = \(\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } \)

Here R_{P} is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination as shown in the fig. (b).

Note:

The value of equivalent resistance in parallel connection will be lesser than each individual resistance.

Question 4.

Explain the determination of the internal resistance of a cell using voltmeter.

Answer:

Determination of internal resistance:

The emf of cell ξ is measured by connecting a high resistance voltmeter across it without connecting the external resistance R. Since the voltmeter draws very little current for deflection, the circuit may be considered as open. Hence, the voltmeter reading gives the emf of the cell. Then, external resistance R is included in the circuit and current I is established in the circuit. The potential difference across R is equal to the potential difference across the cell (V).

The potential drop across the resistor R is

V= IR …… (1)

Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell ξ. It is because, certain amount of voltage (Ir) has dropped across the internal resistance r.

V = ξ – Ir

Ir = ξ – V …… (2)

Dividing equation (2) by equation (1), we get

\(\frac { Ir }{IR}\) = \(\frac { ξ – V }{V}\)

r = |\(\frac { ξ – V }{V}\)| R …… (3)

Since ξ, V and R are known, internal resistance r can be determined.

Question 5.

State and explain Kirchhoff’s rules.

Answer:

Kirchhoff’s first rule (current rule or junction rule):

Statement: It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a statement of conservation of electric charge.

Explanation:

All charges that enter a given junction in a circuit must leave that junction since charge cannot build up or disappear at a junction. Current entering the junction is taken as positive and current leaving the junction is taken as negative.

Applying this law to the junction A,

I_{1} + I_{2} – I_{3} – I_{4} – I_{5} = o

Or

I_{1} + I_{2} = + I_{3} I_{4} + I_{5}

Kirchhoff’s second rule (voltage rule or loop rule):

Statement: It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system. (The energy supplied by the emf sources is equal to the sum of the energy delivered to all resistors).

Explanation:

The product of current and resistance is taken as positive when the direction of the current is followed. Suppose if the direction of current is opposite to the direction of the loop, then product of current and voltage across the resistor is negative. It is shown in Fig. (a) and (b). The emf is considered positive when proceeding from the negative to the positive terminal of the cell. It is shown in Fig. (c) and (d).

Kirchhoff voltage rule has to be applied only when all currents in the circuit reach a steady state condition (the current in various branches are constant).

Question 6.

Obtain the condition for bridge balance in Wheatstone’s bridge.

Answer:

An important application of Kirchhoff’s rules is the Wheatstone’s bridge. It is used to compare resistances and also helps in determining the unknown resistance in electrical network. The – bridge consists of four resistances P, Q, R and S connected. A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is I_{G} and its resistance is G.

Applying KirchhofFs current rule to junction B,

I_{1} – I_{G} – I_{3} = 0 …….. (1)

Applying Kirchhoff’s current rule to junction D,

I_{2} – I_{G} – I_{4} = 0 …….. (2)

Applying Kirchhoff’s voltage rule to loop ABDA,

I_{1}P + I_{G}G – I_{2}R = 0 …….. (3)

Applying Kirchhoff’s voltage rule to loop ABCDA,

I_{1}P + I_{3}Q – I_{4}S – I_{2}R = 0 …….. (4)

When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B and D, no current flows through galvanometer (I_{G} = 0). Substituting I_{G} = 0 in equation, (1), (2) and (3), we get

I_{1} = I_{3} …….. (5)

I_{2} = I_{4} …….. (6)

I_{1}P = I_{2}R …….. (7)

Substituting the equation (5) and (6) in equation (4)

I_{1}P + I_{1}Q – I_{2}R = 0

I_{1}(P + Q) = I_{2} (R + S) …….. (8)

Dividing equation (8) by equation (7), we get

\(\frac { P + Q }{ P }\) = \(\frac { R + S }{ R }\)

1 + \(\frac { Q }{ P }\) = 1 + \(\frac { S }{ R }\)

⇒ \(\frac { Q }{ P }\) = \(\frac { S }{ R }\)

\(\frac { P }{ Q }\) = \(\frac { R }{ S }\) …….. (9)

This is the bridge balance condition. Only under this condition, galvanometer shows null deflection. Suppose we know the values of two adjacent resistances, the other two resistances can be compared. If three of the resistances are known, the value of unknown resistance (fourth one) can be determined.

Question 7.

Explain the determination of unknown resistance using meter bridge.

Answer:

The meter bridge is another form of Wheatstone’s bridge. It consists of a uniform manganin wire AB of one meter length. This wire is stretched along a meter scale on a wooden board between two copper strips C and D. Between these two copper strips another copper strip E is mounted to enclose two gaps G_{1} and G_{2} An unknown resistance P is connected in G_{1} and a standard resistance Q is connected in G_{2}.

A jockey (conducting wire) is connected to the terminal E on the central copper strip through a galvanometer (G) and a high resistance (HR). The exact position of jockey on the wire can be read on the scale. A Lechlanche cell and a key (K) are connected across the ends of the bridge wire.

The position of the jockey on the wire is adjusted so that the galvanometer shows zero deflection. Let the point be J. The lengths AJ and JB of the bridge wire now replace the resistance R and S of the Wheatstone’s bridge. Then

\(\frac { P }{ Q }\) = \(\frac { R }{ S }\) = \(\frac { R’.AJ }{ R’.JB }\) …….. (1)

where R’ is the resistance per unit length of wire

\(\frac { P }{ Q }\) = \(\frac { AJ }{ JB }\) = \(\frac {{ l }_{1}}{ { l }_{2} }\) …….. (2)

P = Q \(\frac {{ l }_{1}}{ { l }_{2} }\) ……… (3)

The bridge wire is soldered at the ends of the copper strips. Due to imperfect contact, some resistance might be introduced at the contact. These are called end resistances. This error can be eliminated, if another set of readings are taken with P and Q interchanged and the average value of P is found.

To find the specific resistance of the material of the wire in the coil P, the radius r and length l of the wire is measured. The specific resistance or resistivity r can be calculated using the relation

Resistance = ρ-\(\frac { l }{ A }\)

By rearranging the above equation, we get

ρ = Resistance x \(\frac { A }{ l }\) …….. (4)

If P is the unknown resistance, equation (4) becomes

ρ = P\(\frac {{ πr }^{2}}{ l }\).

Question 8.

How the emf of two cells are compared using potentiometer?

Answer:

Comparison of emf of two cells with a potentiometer:

To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer wire CD is connected to a battery Bt and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emf ξ_{1} and ξ_{2} to be compared are connected to the terminals M_{1,} N_{1} and M_{2}, N_{2} of the DPDT switch.

The positive terminals of Bt, ξ_{1} and ξ_{2} should be connected to the same end C. The DPDT switch is pressed towards M_{1}, N_{1} so that cell ξ_{1} is included in the secondary circuit and the balancing length l_{1} is found by adjusting the jockey for zero deflection, Then the second cells ξ_{2} is included in the circuit and the balancing length l_{2} is determined. Let r be the resistance per unit length of the potentiometer wire and I be the current flowing through the wire.

we have.

ξ_{1} = Irl_{1} …… (1)

ξ_{2} = Irl_{2} ……. (2)

By dividing (1) by (2)

\(\frac {{ ξ }^{1}}{ { ξ }^{2} }\) = \(\frac {{ l }^{1}}{ { l }^{2} }\) ……. (3)

By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.

**Samacheer Kalvi 12th Physics Current Electricity Numerical Problems**

Question 1.

The following graphs represent the current versus voltage and voltage versus current for the six conductors A,B,C,D,E and F. Which conductor has least resistance and which has maximum resistance?

Solution:

According to ohm’s law, V = IR

Resistance of conductor, R = \(\frac { V }{ I }\)

Graph-I:

Conductor A, I = 4 A and V = 2 V

R = \(\frac { V }{ I }\) = \(\frac { 2 }{ 4 }\) = 0.5 Ω

Conductor B, I = 3 A and V = 4 V

R = \(\frac { V }{ I }\) = \(\frac { 4 }{ 3 }\) = 1.33 Ω

Conductor C, I = 2 A and V = 5 V

R= \(\frac { V }{ I }\) = \(\frac { 5 }{ 2 }\) = 2.5 Q Ω.

Graph-II:

Conductor D, I = 2 A and V = 4 V

R = \(\frac { V }{ I }\) = \(\frac { 4 }{ 2 }\) = 2 Ω d

Conductor E, I = 4A and V = 3 V

R = \(\frac { V }{ I }\) = \(\frac { 3 }{ 4 }\) = 1.75 Ω

Conductor F, I = 5A and V = 2 V

R = \(\frac { V }{ I }\) = \(\frac { 2 }{ 5 }\) = 0.4 Ω

Conductor F has least resistance, R_{F} = 0.4 Ω,

Conductor C has maximum resistance, R_{C} = 2.5 G.

Question 2.

Lightning is very good example of natural current. In typical lightning, there is 10^{9} J energy transfer across the potential difference of 5 x 10^{7} V during a time interval of 0.2 s.

Using this information, estimate the following quantities (a) total amount of charge transferred between cloud and ground (b) the current in the lightning bolt (c) the power delivered in 0.2 s.

Solution:

During the lightning energy, E = 10^{9} J

Potential energy, V = 5 x 10^{7} V

Time interval, t = 0.2 s

(a) Amount of charge transferred between cloud and ground,

q = It

(b) Current in the lighting bolt, E = VIt

I = \(\frac { E }{ Vt }\) = \(\frac{10^{9}}{5 \times 10^{7} \times 0.2}\) = 1 × 10^{9} × 1^{-7}

I = 1 × 10^{2}

I = 100 A

∴ q = It = 100 × 0.2

q = 20 C.

(c) Power delivered, E = VIt

P = VI = 5 × 10^{7} × 100 = 500 × 10^{7}

I = 5 × 10^{9} W

P = 5 GW.

Question 3.

A copper wire of 10^{6} m^{2} area of cross section, carries a current of 2 A. If the number of electrons per cubic meter is 8 x 10^{28}, calculate the current density and average drift velocity.

Solution:

Cross-sections area of copper wire, A = 10^{6} m^{2}

I = 2 A

Number of electron, n = 8 x 10^{28}

Current density, J = \(\frac { 1 }{ A }\) = \(\frac { 2 }{{ 10 }^{-6}}\)

J = 2 × 10^{6} Am^{-2}

Average drift velocity, V_{d} = \(\frac { 1 }{ neA }\)

e is the charge of electron = 1.6 × 10^{-9} C

V_{d} = \(\frac{2}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}}\) = \(\frac { 1 }{{ 64. × 10 }{3}}\)

V_{d} = 0.15625 × 10^{-3}

V_{d} = 15.6 × 10^{-5} ms^{-1}

Question 4.

The resistance of a nichrome wire at 0 °C is 10 Ω. If its temperature coefficient of resistance is 0.004/°C, find its resistance at boiling point of water. Comment on the result.

Solution:

Resistance of a nichrome wire at 0°C, R_{0} = 10 Ω

Temperature co-efficient of resistance, α = 0.004/°C

Resistance at boiling point of water, RT = ?

Temperature of boiling point of water, T = 100 °C

R_{T}=R_{0} ( 1 + αT) = 10[1 + (0.004 x 100)]

R_{T}= 10(1 +0.4) = 10 x 1.4

R_{T} = 14 Ω

As the temperature increases the resistance of the wire also increases.

Question 5.

The rod given in the figure is made up of two different materials.

Both have square cross sections of 3 mm side. The resistivity of the first material is 4 x 10^{-3} Ω.m and it is 25 cm long while second material has resistivity of 5 x 10^{-3} Ω.m and is of 70 cm long. What is the resistivity of rod between its ends?

Solution:

Square cross section of side, a = 3 mm = 3 x 10^{-3} m

Cross section of side, A = a^{2} = 9 x 10^{6} m

First material:

Resistivity of the material, ρ_{1} = 4 x 10^{-3} Ωm

length, l_{1} = 25 cm = 25 x 10^{-2} m

Resistance of the lord, R_{1} = \(\frac{\rho_{l} l_{l}}{A}\) = \(\frac{4 \times 10^{-3} \times 25 \times 10^{-2}}{9 \times 10^{-6}}\) = \(\frac{100 \times 10^{-5} \times 10^{6}}{9}\)

R_{1} = 11.11 x 10^{1} Ω

Second material:

Resistivity of the material, ρ_{2} = 5 x 10^{-3} Ωm

length, l_{2} = 70 cm = 70 x 10^{-2} m

Resistance of the rod, R_{2} = \(\frac{\rho_{2} l_{2}}{A}\) = \(\frac{5 \times 10^{-3} \times 70 \times 10^{-2}}{9 \times 10^{-6}}\) = \(\frac{350 \times 10^{-5} \times 10^{6}}{9}\)

R_{2} = 38.88 x 10^{1}

R_{2} = 389 Ω

Total resistance between the ends f the rods

R = R_{1} + R_{2} = 111 + 389

= 500 Ω

Question 6.

Three identical lamps each having a resistance R are connected to the battery of emf as shown in the figure.

Suddenly the switch S is closed, (a) Calculate the current in the circuit when S is open and closed (b)

What happens to the intensities of the bulbs A,B and C. (c) Calculate the voltage across the three bulbs when S is open and closed (d) Calculate the power delivered to the circuit when S is opened and closed (e) Does the power delivered to the circuit decreases, increases or remain same?

Solution:

Resistance of the identical lamp = R

Emf of the battery = ξ

According to Ohm’s Law, ξ = IR

(a) Current:

When Switch is open— The current in the circuit. Total resistance of the bulb,

R_{s} = R_{1} + R_{2} + R_{3}

R_{1} = R_{2} = R_{3} = R

R_{s} = R + R + R = 3R

∴ Current, I = \(\frac { ξ }{{ R }_{s}}\)

⇒ I_{0} = \(\frac { ξ }{ 3R }\)

Switch is closed— The current in the circuit. Total resistance of the bulb,

R_{s} = R + R = 2R

Current I = \(\frac { ξ }{{ R }_{s}}\)

I_{c} = \(\frac { ξ }{ 2R }\).

(b) Intensity:

When switch is open — All the bulbs glow with equal intensity.

When switch is closed — The intensities of the bulbs A and B equally increase. Bulb C will not glow since no current pass through it.

(c) Voltage across three bulbs:

When switch is open — Voltage across bulb A, V_{A} = I_{0} R = \(\frac { ξ }{ 3R }\) x R = \(\frac { ξ }{ 3 }\)

similarly:

Voltage across bulb B, V_{B} = \(\frac { ξ }{ 3 }\)

Voltage across bulb C, V_{C} = \(\frac { ξ }{ 3 }\)

When switch is closed— Voltage across bulb A, V_{A} = I_{c}R = \(\frac { ξ }{ 2R }\) x \(\frac { ξ }{ 2 }\)

similarly:

Voltage across bulb B, V_{B} = I_{c}R \(\frac { ξ }{ 2 }\)

Voltage across bulb C, V_{C} = 0

(d) Power delivered to the circuit,

When switch is opened — Power P, = VI

When switch is closed — Power P, = VI

(e) Total power delivered to circuit increases.

Question 7.

The current through an element is shown in the figure. Determine the total charge that pass through the element at

(a) t = 0 s

(b) t = 2 s

(c) t = 5 s

Solution :

Rate of flow of charge is called current, I = \(\frac { dq }{ dt }\)

Total charge pass through element, dq = Idt

(a) At t= 0 s, I = 10 A

dq = Idt= 10 x 0 = 0 C.

(b) At t = 2 s, I = 5 A

dq = Idt = 5 x 2 = 10 C.

(c) At t = 5 s, I = 0

dq = Idt = 0 x 5 = 0 C.

Question 8.

An electronics hobbyist is building a radio which requires 150 Ω in her circuit, but she has only 220 Ω, 79 Ω and 92 Ω resistors available. How can she connect the available resistors to get desired value of resistance?

Solution:

Required effective resistance = 150 Ω

Given resistors of resistance, R_{1} = 220 Ω, R_{2} = 79 Ω, R_{3} = 92 Ω

Parallel combination R_{1} and R_{2}

\(\frac { 1 }{{ R }_{p}}\) = \(\frac { 1 }{{ R }_{1}}\) + \(\frac { 1 }{{ R }_{2}}\) = \(\frac { 1 }{ 220 }\) + \(\frac { 1 }{ 79 }\) = \(\frac { 79 + 220 }{ 220 × 79 }\)

R_{p} = 58 Ω

Parallel combination R_{p} and R_{3}

R_{s} = R_{p} + R_{3} = 58 + 92

R_{s} = 150 Ω

Parallel combination of 220 Ω and 79 Ω in series with 92 Ω.

Question 9.

A cell supplies a current of 0.9 A through a 2 Ω resistor and a current of 0.3 A through a 7 Ω resistor. Calculate the internal resistance of the cell.

Solution:

Current from the cell, I_{1} = 0.9 A

Resistor, R_{1} = 2 Ω

Current from the cell, I_{2} = 0.3 A

Resistor, R_{2} 7 Ω

Internal resistance of the cell, r = ?

Current in the circuit I_{1} = \(\frac { ξ }{{ r + R }_{1}}\)

ξ = I_{1} (r + R_{1}) …… (1)

Current in the circuit, I_{2} = \(\frac { ξ }{{ r + R }_{2}}\) …… (2)

Equating equation (1) and (2),

I_{1}r + I_{1}R_{1} = I_{2}R_{2} + I_{2}r

(I_{1} – I_{2})r = I_{2}R_{2} – I_{1}R_{1}

r = 0.5 Ω.

Question 10.

Calculate the currents in the following circuit.

Solution:

Applying Kirchoff’s 1^{st} Law at junction B

I_{1} – I_{1} – I_{3} = 0

I_{3} = I_{1} – I_{2} …… (1)

Applying Kirchoff’s II^{nd} Law at junction in ABEFA

100 I_{3} + 100 I_{1} = 15

100 (I_{3} + I_{1}) = 15

100 I_{1} – 100 I_{2 }+ 100 I_{1} =15

200 I_{1} – 100 I_{2} = 15 …… (2)

Applying Kirchoff’s IInd Law at junction in BCDED

-100I_{2} + 100 I_{3} = 9

-100 I_{2}+ 100(I_{1} – I_{2}) = 9

100I_{1} – 200 I_{2} = 9

Solving equating (2) and (3)

Substitute I_{1} values in equ (2)

200(0.07) – 100 I_{2} = 15

14 – 100 I_{2} = 15

– 100 I_{2} = 15 – 14

I_{2} = \(\frac { -1 }{ 100 }\)

I_{2} = -0.01A

Substitute I_{1} and I_{2} value in equ (1), we get

I_{3} = I_{1} – I_{2} = 0.07 – (-0.01)

I_{3} = 0.08 A.

Question 11

A potentiometer wire has a length of 4 m and resistance of 20 Ω. It is connected in series with resistance of 2980 Ω and a cell of emf 4 V. Calculate the potential along the wire.

Solution:

Length of the potential wire, l = 4 m

Resistance of the wire, r = 20 Ω

Resistance connected series with potentiometer wire, R = 2980 Ω

Emf of the cell, ξ = 4 V

Effective resistance, R_{s} = r + R = 20 + 2980 = 3000 Ω

Current flowing through the wire, I = \(\frac { ξ }{{ r + R }_{s}}\) = \(\frac { 4 }{ 3000 }\)

I = 1.33 x 10^{-3} A

Potential drop across the wire, V = Ir

= 1.33 x 10^{-3}3 x 20

V = 26.6 x 10^{-3} V

= 6.65 x 10^{-3}

Potential garadient = 0.66 x 10^{-2} Vm^{-1}

Question 12.

Determine the current flowing through the galvanometer (G) as shown in the figure.

Solution:

Current flowing through the circuit, I = 2A

Applying Kirchoff’s I^{st} law at junction P, I = I_{1} + I_{2}

I_{2} = I – I_{1} …(1)

Applying Kirchoff’s I^{nd} law at junction PQSP

5 I_{1} + 10 I_{g} – 15 I_{2} = 0

5 I_{1} + 10 I_{g} -15(I – I_{1}) = 0

20 I_{1} + 10 I_{g} = 15 I

20 I_{1} + 10 I_{g} = 15 x 2

÷ by 10 21 I_{1} + I_{g} = 3 … (2)

Applying Kirchoff’s II^{nd} law at junction QRSQ

10(I_{1} – I_{g}) – 20(I – I_{1} – I_{g}) – 10 I_{g} = 0

10 I_{1} – 10_{g} – 20(I – I_{1} – I_{g}) – 10 I_{g} = 0

10 I_{1} – 10_{g} – 20 I + 20 I_{1} -20I_{g} – 10 I_{g} = 0

30 I _{1} – 40 I_{g} = 20 I

÷ by 10 ⇒ 3 I_{1} – 4 I_{g} = 20 I

3 I_{1} – 4 I_{g} = 2 I

3 I_{1} – 4 I_{g} = 2 x 2 = 4

Question 13.

Two cells each of 5 V are connected in series across a 8 Ω resistor and three parallel resistors of 4 Ω, 6 SI and 12 Ω. Draw a circuit diagram for the above arrangement. Calculate (i) the current drawn from the cell (ii) current through each resistor.

Solution:

V_{1} = 5 V; V_{2} = 5 V

R_{1} = 8 Ω; R_{2} = 4 Ω; R_{3} = 6 Ω; R_{4} = 12 Ω

Three resistors R_{2}, R_{3} and R_{4} are connected parallel combination

Resistors R_{1} and R_{p} are connected in series combination

R_{s} = R_{1} +R_{p} = 8 + 2 = 10

R_{s} = 10Ω

Total voltage connected series to the circuit

V = V_{1} + V_{2}

= 5 + 5 = 10

V = 10 V.

(i) Current through the circuit, I = \(\frac { V }{{ R }_{s}}\) = \(\frac { 10 }{ 10 }\)

I = 1 A

Potential drop across the parallel combination,

V’ = IR_{p} = 1 x 2

V’ 2 V

(ii) Current in 4 Ω resistor, I = \(\frac { V’ }{{ R }_{2}}\) = \(\frac { 2 }{ 4 }\) = 0.5 A

Current in 6 Ω resistor, I = \(\frac { V’ }{{ R }_{3}}\) = \(\frac { 2 }{ 6 }\) = 0.33 A

Current in 12 Ω resistor, I = \(\frac { V’ }{{ R }_{4}}\) = \(\frac { 2 }{ 12 }\) = 0.17 A

Question 14.

Four light bulbs P, Q, R, S are connected in a circuit of unknown arrangement. When each bulb is removed one at a time and replaced, the following behavior is observed.

Solution:

Question 15.

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm, what is the emf of the second cell?

Solution:

Emf of the cell_{1}, ξ_{1} = 1.25 V

Balancing length of the cell, l_{1} = 35 cm = 35 x 10^{-2} m

Balancing length after interchanged, l_{2} = 63 cm = 63 x 10^{-2} m

Emf of the cell_{1}, ξ_{2} = ?

The ration of emf’s, \(\frac {{ ξ }_{1}}{{ ξ }_{2}}\) = \(\frac {{ l }_{1}}{{ l }_{2}}\)

The ration of emf’s, ξ_{2} = ξ_{1} = \(\left( \frac { { l }_{ 2 } }{ { l }_{ 1 } } \right) \)

= 1. 25 x \(\left( \frac { { 63×10 }^{ -2 } }{ { 35×10 }^{ -2 } } \right) \) = 12.5 x 1.8

ξ_{2} = 2.25 V.

### Samacheer Kalvi 12th Physics Current Electricity Additional Questions Solved

I. Choose the Correct Answer

Question 1.

From which part of an atom, the electric current originates?

a) nucleus

b) entire atom

c) positively charged protons

d) negatively charged electrons

Answer:

d) negatively charged electrons

Question 2.

A copper wire of length 2 m and area of cross-section 1.7 x 10^{-6} m^{2} has a resistance of 2 x 10^{-2} Ω. The resistivity of copper is

(a) 1.7 x 10^{-8} Ωm

(b) 1.9 x ^{-8} Ωm

(c) 2.1 x 10^{-7} Ωm

(d) 2.3 x 10^{-7} Ωm

Answer:

(a) 1.7 x 10^{-8} Ωm

Hint:

Resistivity, ρ = \(\frac { RA }{ l }\) = \(\frac{2 \times 10^{-2} \times 1.7 \times 10^{-6}}{2}\) = 1.7 x 10^{-8} Ωm.

Question 3.

If the length of a wire is doubled and its cross-section is also doubled, then its resistance will

(a) become 4 times

(b) become 1 / 4

(c) becomes 2 times

(d) remain unchanged

Answer:

(d) remain unchanged

Question 4.

Resistance is a measure of a material’s opposition to _______.

a) voltage

b) current

c) electric force

d) movement of protons

Answer:

b) current

Question 5.

The resistivity of a wire

(a) varies with its length

(b) varies with its mass

(c) varies with its cross-section

(d) does not depend on its length, cross-section and mass.

Answer:

(d) does not depend on its length, cross-section and mass.

Question 6.

The electric intensity E, current density and conductivity a are related as

(a) j = σE

(b) j = \(\frac { E }{ σ }\)

(c) JE = s

(d) j = σ^{2}E

Answer:

(a) j = σE

Question 7.

For which of the following dependences of drift velocity V_{d} on electric field E, is Ohm’s law obeyed?

(a) v_{d} ∝ E

(b) v_{d} ∝ E^{2}

(c) v_{d} ∝ √E

(d) v_{d} = constant

Answer:

(a) v_{d} ∝ E

Hint:

v_{d} = \(\frac { 1 }{ nAe }\) = \(\frac { j }{ ne }\) = \(\left( \frac { σ }{ ne } \right) \)E ⇒ v_{d} ∝ E.

Question 8.

When one of the 3 series resistors is removed from a circuit and the circuit is reconnected, the current ________

a) increases by half

b) increases

c) decreases by half

d) None of the above

Answer:

b) increases

Question 9.

The resistance, each of 1 Ω, is joined in parallel. Three such combinations are put in series. The resultant resistance is

(a) 9 Ω

(b) 3 Ω

(c) 1 Ω

(d) \(\frac { 1 }{ 3 }\) Ω

Answer:

(c) 1 Ω

Hint:

R_{p} = 1 + 1 + 1 = 3 Ω;

\(\frac { 1 }{{ R }_{s}}\) = \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ 3 }\) = \(\frac { 3 }{ 3 }\) = 1

⇒ R_{s} = 1 Ω.

Question 10.

Constantan is used for making standard resistance because it has

(a) high resistivity

(b) low resistivity

(c) negligible temperature coefficient of resistance

(d) high melting point

Answer:

(c) negligible temperature coefficient of resistance

Question 11.

The unit of emf is ________

a) Newton

b) Volt

c) Joule

d) Ampere

Answer:

b) Volt

Question 12.

The resistance R_{0} and R_{t} of a metallic wire at temperatures 0° C and t° C are related as (a is the temperature coefficient of resistance).

(a) R_{t} = R_{0}(1 + αt)

(b) R_{t} = R_{0}(1 – αt)

(c) R_{t} = R_{0}(1 + αt)^{2}

(d) R_{t} = R_{0}(1 – αt) ^{2}

Answer:

(a) R_{t} = R_{0}(1 + αt)

Question 13.

A cell of emf 2 V and internal resistance 0.1 Ω is connected with a resistance of 3.9 Ω. The voltage across the cell terminals will be

(a) 0.5 V

(b) 1.9 V

(c) 1.95 V

(d) 2 V

Answer:

(c) 1.95 V

Hint:

V = \(\frac { ER }{ R + r }\) = \(\frac { 2 x 3.9 }{ 3.9 + 0.1 }\) = 1.95 V.

Question 14.

Resistance of resistor having colour code Yellow – Green – Red – Gold

a) (4700 ± 5%)Ω

b) (4500 ± 5%)Ω

c) (4700 ± 10%)Ω

d) (4500 ± 10%)Ω

Answer:

b) (4500 ± 5%)Ω

Question 15.

The sensitivity of a potentiometer can be increased by

(a) increasing the emf of the cell

(b) increasing the length of the wire

(c) decreasing the length of the wire

(d) none of the above

Answer:

(b) increasing the length of the wire

Question 16.

The potential gradient is defined as

(a) fall of potential per unit length of the wire.

(b) fall of potential per unit area of the wire.

(c) fall of potential between two ends of the wire.

(d) none of the above.

Answer:

(a) fall of potential per unit length of the wire.

Question 17.

n equal resistors are first connected in series and then in parallel. The ratio of the equivalent resistance in two cases is

(a) n

(b) \(\frac { 1 }{{ n }^{2}}\)

(c) n^{2}

(d) \(\frac { 1 }{ n }\)

Answer:

(c) n^{2}

Hint:

Required ratio = \(\frac{n \mathrm{R}}{\left(\frac{\mathrm{R}}{n}\right)}\) = n(c) n^{2}

Question 18.

The conductors who do not obey Ohm’s law is called as _________

a) un-ohmic conductors

b) non-ohmic conductors

c) low-ohmic conductors

d) zero-ohmic conductors

Answer:

b) non-ohmic conductors

Question 19.

The reciprocal of resistance is

(a) conductance

(b) resistivity

(c) conductivity

(d) none of the above

Answer:

(a) conductance

Question 20.

A student has 10 resistors, each of resistance r. The minimum resistance that can be obtained by him using these resistors is

(a) 10r

(b) \(\frac { r }{ 10 }\)

(c) \(\frac { r }{ 100 }\)

(d) \(\frac { r }{ 5 }\)

Answer:

(b) \(\frac { r }{ 10 }\)

Question 21.

The drift velocity of electrons in a wire of radius r is proportional to

(a) r

(b) r^{2}

(c) r^{3}

(d) none of the above

Answer:

(d) none of the above

Question 22.

Kirchhoff’s first law, i.e. ∑I = 0 at a junction deals with conservation of

(a) charge

(b) energy

(c) momentum

Answer:

(a) charge

Question 23.

The resistance of a material increases with temperature. It is a

(a) metal

(b) insulator

(c) semiconductor

(d) semi-metal

Answer:

(a) metal

Question 24.

The melting point of tungsten is _______.

a)3280°C

b) 3580°C

c) 3380°C

d) 3680°C

Answer:

c) 3380°C

Question 25.

A carbon resistance has colour bands in order yellow, brown, red. Its resistance is

(a) 41 Ω

(b) 41 x 10^{2} Ω

(c) 41 x 10^{3} Ω

(d) 4.2 Ω

Answer:

(b) 41 x 10^{2} Ω

Question 26.

The conductivity of a superconductor is

(a) infinite

(b) very large

(c) very small

(d) zero

Answer:

(a) infinite

Question 27.

The resistance of an ideal voltmeter is

(a) zero

(b) very high

(c) very low

(d) infinite

Answer:

(d) infinite

Question 28.

Carriers of electric current in superconductors are

(a) electrons

(b) photons

(c) holes

Answer:

(c) holes

Question 29.

Thermophile works on the principle of _______.

a) Joule effect

b) Seebeck effect

c) Peltier effect

d) Thomson effect

Answer:

b) Seebeck effect

Question 30.

Electromotive force is most closely related to

(a) electric field

(b) magnetic field

(c) potential difference

(d) mechanical force

Answer:

(c) potential difference

Question 31.

The capacitance of a pure capacitor is 1 farad. In DC circuit, the effective resistance will be

(a) zero

(b) infinite

(c) 1 Ω

(d) 0.5 Ω

Answer:

(b) infinite

Question 32.

The resistance of an ideal ammeter is

(a) zero

(b) small

(c) high

(d) infinite

Answer:

(a) zero

Question 33.

A milliammeter of range 10 mA has a coil of resistance 1 Ω. To use it as a voltmeter of range 10 V, the resistance that must be connected in series with it is

(a) 999 Ω

(b) 1000 Ω

(c) 9 Ω

(d) 99 Ω

Answer:

(a) 999 Ω

Hint:

R = \(\frac { V }{{ I }_{g}}\)-R_{g} = \(\frac { 10 }{{ 10 × 10 }^{-3}}\)-1 = 999 Ω.

Question 34.

A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. The current in the circuit is. 0.5 A. The terminal voltage of the battery when the circuit is closed is

(a) 10 V

(b) zero

(c) 8.5 V

(d) 1.5 V

Answer:

(c) 8.5 V

Hint:

V = ξ – Ir = 10 – (0.5 x 3) = 8.5 V.

Question 35.

Good resistance coils are made of

(a) copper

(b) manganin

(c) iron

(d) aluminium

Answer:

(b) manganin

Question 36.

A wire of resistance R is stretched to three times its original length. The new resistance is

(a) 3R

(b) 9R

(c) R/3

(d) R/9

Answer:

(b) 9R

Question 37.

n resistances, each of r Ω, when connected in parallel give an equivalent resistance of R Ω. If these resistances were connected in series, the combination would have resistance in horns equal to

(a) n^{2}R

(b) \(\frac { R }{{n}^{ 2 }}\)

(c) \(\frac { R }{ n }\)

(d) nR

Answer:

(a) n^{2}R

Hint:

Resistance in parallel combination, R = \(\frac { r }{ n }\) ⇒ r = Rn

Resistance in series combination, R’ = nr = n^{2}R

Question 38.

When a wire of uniform cross-section, having resistance R, is bent into a complete circle, the resistance between any two of diametrically opposite points will be

(a) \(\frac { R }{ 8 }\)

(b) \(\frac { R }{ 2 }\)

(c) 4R

(d) \(\frac { R }{ 4 }\)

Answer:

(d) \(\frac { R }{ 4 }\)

Hint:

It becomes two resistors each of (d) \(\frac { R }{ 2 }\), connected in parallel.

Question 39.

A steady current is set up in a metallic wire of a non-uniform cross-section. How is the rate of flow K of electrons related to the area of cross-section A?

(a) K is independent of A

(b) K ∝ A

(c) K ∝ A^{-1}

(d) K ∝ A^{2}

Answer:

(c) K ∝ A^{-1}

Question 40.

Ohm’s Law is not obeyed by

(a) electrolytes

(b) discharge tubes

(c) vacuum tubes

(d) all of these

Answer:

(d) all of these

Question 41.

Which of the following has a negative temperature coefficient of resistance?

(a) Copper

(b) Aluminium

(c) Germanium

(d) Iron

Answer:

(c) Germanium

II. Fill in the blanks

Question 1.

The material through which electric charge can flow easily is ……………….

Answer:

Copper.

Question 2.

A toaster operating at 240 V has a resistance of 120 Ω. The power is ……………….

Answer:

480 W.

Question 3.

In the case of insulators, as the temperature decreases, resistivity ……………….

Answer:

Increases.

Question 4.

When n resistors of equal resistance (R) are connected in series, the effective resistance is ……………….

Answer:

nR.

Question 5.

The net flow of charge at any point in the conductor is ……………….

Answer:

Zero.

Question 6.

The flow of free electrons in a conductor constitutes ……………….

Answer:

Electric current.

Question 7.

The rate of flow of charge through any wire is called ……………….

Answer:

Current.

Question 8.

The drift velocity acquired per unit electric field is the ……………….

Answer:

Mobility.

Question 9.

The reciprocal of resistance is ……………….

Answer:

Conductance.

Question 10.

The unit of specific resistance is ……………….

Answer:

Ohm meter.

Question 11.

The reciprocal of electrical resistivity is called ……………….

Answer:

Electrical conductivity.

Question 12.

With increase in temperature the resistivity of metals ……………….

Answer:

Increases.

Question 13.

The resistivity of insulators is of the order of ……………….

Answer:

10^{8} 10^{14} Ωm.

Question 14.

The resistivity of semiconductors is of the order of ……………….

Answer:

10^{-2} -10^{2} Ωm.

Question 15.

The materials which conduct electricity at zero resistance are called ……………….

Answer:

Superconductors.

Question 16.

Conductors turn into superconductors at ……………….

Answer:

Low temperatures.

Question 17.

The resistance of superconductors is ……………….

Answer:

Zero.

Question 18.

The phenomenon of superconductivity was discovered by ……………….

Answer:

Kammerlingh onnes.

Question 19.

Mercury becomes a superconductor at ……………….

Answer:

4.2 K.

Question 20.

With the increase of temperature, the resistance of conductors ……………….

Answer:

increases

Question 21.

In insulators and semiconductors, as temperature increases, resistance ……………….

Answer:

Decreases.

Question 22.

A material with a negative temperature coefficient is called a ……………….

Answer:

Thermistor.

Question 23.

The temperature coefficient for alloys is ……………….

Answer:

Low.

Question 24.

The electric current in an external circuit flows from the ……………….

Answer:

Positive to the negative terminal.

Question 25.

In the electrolyte of the cell, current flows from ……………….

Answer:

Negative to the positive terminal.

Question 26.

A freshly prepared cell has ………………. internal resistance.

Answer:

Low.

Question 27.

Kirchhoff’s first law is ……………….

Answer:

Current law.

Question 28.

The current law states that the algebraic sum of the currents meeting at any junction in a circuit is ……………….

Answer:

Zero.

Question 29.

Current law is a consequence of conservation of ……………….

Answer:

Charges.

Question 30.

Kirchhoff’s second law is ……………….

Answer:

Voltage law.

Question 31.

Kirchhoff’s second law is a consequence of conservation of ……………….

Answer:

Energy.

Question 32.

Wheatstone bridge is an application of ……………….

Answer:

Kirchhoff’s Law.

Question 33

………………. is a form of Wheatstone’s bridge.

Answer:

Metre bridge.

Question 34.

The temperature coefficient of manganin wire is ……………….

Answer:

Low.

Question 35

………………. is an instrument to measure the potential difference.

Answer:

Potentiometer.

Question 36.

Unit of electrical energy is ……………….

Answer:

Joule.

Question 37.

An instrument to measure electrical power consumed is ……………….

Answer:

Wattmeter.

Question 38.

………………. first introduced the electrochemical battery

Answer:

Volta.

Question 39.

Charging is a process of reproducing ……………….

Answer:

Active materials.

III. Match the following.

Question 1.

Answer:

(i) → (b)

(ii) → (a)

(iii) → (d)

(iv) →(c)

Question 2.

Answer:

(i) → (b)

(ii) → (c)

(iii) → (d)

(iv) → (a)

Question 3.

Answer:

(i) → (d)

(ii) → (c)

(iii) → (b)

(iv) → (a)

Question 4.

Answer:

(i) → (b)

(ii)→ (d)

(iii) → (a)

(iv) → (c)

IV.Assertion and reason type

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.

(b) If both assertion and reason are true but the reason is not the correct explanation of the assertion.

(c) If the assertion is true but the reason is false.

(d) If the assertion and reason both are false.

(e) If the assertion is false but the reason is true.

Question 1.

Assertion: Fuse wire must have high resistance and low melting point.

Reason: Fuse is used for small current flow only

Answer:

(c) If the assertion is true but the reason is false.

Question 2.

Assertion: In practical application, the power rating of resistance is not important.

Reason: Property of resistance remains same even at high temperature

Answer:

(d) If the assertion and reason both are false.

Question 3.

Assertion: Electric appliances with metallic body e.g. heaters, presses, etc, have three-pin connections, whereas an electric bulb has two pins.

Reason: Three-pin connection reduces the heating of connecting cables.

Answer:

(c) If the assertion is true but the reason is false.

**Samacheer Kalvi 12th Physics Current Electricity Short Answer Questions**

Question 1.

What are conductors?

Answer:

- Atoms in metals have one or more electrons which are loosely bound to the nucleus.
- These electrons are called free electrons and can be easily detached from the atoms.
- The substances which have an abundance of these free electrons are called conductors.

Question 2.

Define instantaneous current?

Answer:

The instantaneous current I is defined as the limit of the average current, as ∆t → 0.

Question 3.

What is resistance? Give its unit?

Answer:

The resistance is the ratio of potential difference across the given conductor to the current passing through the conductor V.

R = \(\frac { V }{ I }\).

Question 4.

Can positive ions give rise to current? How?

Answer:

- The atoms after losing their outermost electrons will have more positive charges and hence are called positive ions.
- These ions will not move freely within the material like the free electrons.
- Hence the positive ions will not give rise to current.

Question 5.

What is Joule’s heating effect?

Answer:

When current flows through a resistor, some of the electrical energy delivered to the resistor is converted into heat energy and it is dissipated. This heating effect of current is known as Joule’s heating effect.

Question 6.

What is meant by thermoelectric effect?

Answer:

Conversion of temperature differences into electrical voltage and vice versa is known as the thermoelectric effect.

Question 7.

What is a thermopile? On what principle does it work?

Answer:

Thermopile is a device used to detect thermal radiation. It works on the principle of the Seebeck effect.

Question 8.

How do you classify the materials based on resistivity?

Answer:

- Based on the resistivity, materials are classified as conductors, insulators, and semiconductors.
- The conductors have the lowest resistivity.
- Insulators have the highest resistivity.
- Semiconductors have a resistivity greater than conductors but less than insulators.

Question 9.

State principle of a potentiometer?

Answer:

The principle of a potentiometer states that the emf of the cell is directly proportional to its balancing length.

ξ ∝ l

ξ = Irl.

**Samacheer Kalvi 12th Physics Current Electricity Long Answer Questions**

Question 1.

How electric bulbs glow as soon as we switch on the battery?

Answer:

- When the battery is switched on, the electrons begin to move away from the negative terminal of the battery and this electron exerts a force on the nearby electrons.
- This process creates a propagating influence (electric field) that travels through the wire at the speed of light.
- The energy is transported from the battery to the light bulb at the speed of light through propagating influence.

Question 2.

Explain in detail the temperature dependence of resistivity.

Answer:

Temperature dependence of resistivity:

The resistivity of a material is dependent on temperature. It is experimentally found that for a wide range of temperatures, the resistivity of a conductor increases with an increase in temperature according to the expression,

ρ_{T} = ρ_{0} [1 + α(T -T_{0})] ……. (1)

where ρ_{T} is the resistivity of a conductor at T_{0}C, ρ_{0} is the resistivity of the conductor at some reference temperature To (usually at 20°C) and a is the temperature coefficient of resistivity. It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity atT_{0}.

From the equation (1), we can write

ρ_{T} – ρ_{0} = αρ_{0} (T -T_{0})

∴ α = \(\frac{\rho_{\mathrm{T}}-\rho_{0}}{\rho_{0}\left(\mathrm{T}-\mathrm{T}_{0}\right)}\) = \(\frac{\Delta p}{\rho_{0} \Delta T}\)

where ∆_{ρ} = ρ_{T} – ρ_{0} is change in resistivity for a change in temperature ∆T = T – T_{0}. Its unit is per °C.

1. α of conductors:

For conductors a is positive. If the temperature of a conductor increases, the average kinetic energy of electrons in the conductor increases. This results in more frequent collisions and hence the resistivity increases. Even though the resistivity of conductors like metals varies linearly for a wide range of temperatures, there also exists a nonlinear region at very low temperatures. The resistivity approaches some finite value as the temperature approaches absolute zero. As the resistance is directly proportional to the resistivity of the material, we can also write the resistance of a conductor at temperature T °C as

R_{T} -R_{0} = [1 + α(T -T_{0})] ……. (2)

The temperature coefficient can be also be obtained from the equation (2), +

where ∆R = R_{T} -R_{0} changes in resistance during the change in temperature ∆T = T – T_{0}

2. α of semiconductors:

For semiconductors, the resistivity decreases with an increase in temperature. As the temperature increases, more electrons will be liberated from their atoms (Refer to unit 9 for conduction in semi-conductors). Hence the current increases and therefore the resistivity decreases. A semiconductor with a negative temperature coefficient of resistance is called a thermistor.

We can understand the temperature dependence of resistivity in the following way. The electrical conductivity, σ = \(\frac{n e^{2} \tau}{m}\) \(\frac{m}{n e^{2} \tau}\). As the resistivity is inverse of σ, it can be written as,

σ = \(\frac{n e^{2} \tau}{m}\) \(\frac{m}{n e^{2} \tau}\) …… (4)

The resistivity of materials is

- inversely proportional to the number density (n) of the electrons
- inversely proportional to the average time between the collisions (τ).

In metals, if the temperature increases, the average time between the collision (τ) decreases and n is independent of temperature. In semiconductors when temperature increases, n increases, and τ decreases, but an increase in n is dominant than decreasing x so that overall resistivity decreases.

Question 3.

Explain the effective internal resistance of cells connected in series combination. Compare the results to the external resistance.

Answer:

Cells in series Several cells can be connected to form a battery. In a series connection, the negative terminal of one cell is connected to the positive terminal of the second cell, the negative terminal of the second cell is connected to the positive terminal of the third cell, and so on. The free positive terminal of the first cell and the free negative terminal of the last cell become the terminals of the battery.

Suppose n cells, each of emf ξ volts and internal resistance r ohms are connected in series with an external resistance R.

The total emf of the battery = nξ

The total resistance in the circuit = nr + R

By Ohm’s law, the current in the circuit is

Case (a) If r << R, then,

I = \(\frac { nξ }{ R }\)nI_{1} ≈ nI_{1} ……. (2)

where, I, is the current due to a single cell \(\left(\mathrm{I}_{1}=\frac{\xi}{\mathrm{R}}\right)\)

Thus, if r is negligible when compared to R the current supplied by the battery is n times that supplied by a single cell.

Case (b) If r >> R, I = \(\frac { nξ }{ nr }\) ≈ \(\frac { ξ }{ R }\) …….. (3)

It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells. Thus the series connection of cells is advantageous only when the effective internal resistance of the cells is negligibly small compared with R.

Question 4.

Explain the effective internal resistance of cells connected in parallel combination. Compare the results to the external resistance.

Answer:

Cells in parallel: In a parallel connection, all the positive terminals of the cells are connected to one point and all the negative terminals to a second point. These two points form the positive and negative terminals of the battery.

Let n cells be connected in parallel between points A and B and a resistance R is connected between points A and B. Let ξ, be the emf and r the internal resistance of each cell.

The equivalent internal resistance of the battery is \(\frac { 1 }{{ r }_{eq}}\) = \(\frac { 1 }{ r }\) + \(\frac { 1 }{ r }\) + ….. \(\frac { 1 }{ r }\) (n terms) = \(\frac { n }{ r }\).

So r_{eg} = \(\frac { r }{ n }\) and the total resistance in the circuit = R + \(\frac { r }{ n }\). The total emf is the potential difference between the points A and B, which is equal to ξ. The current in the circuit is given by

Case (a) If r << R, then,

I = \(\frac { nξ }{ R }\) = nI_{1} …….. (2)

where II is the current due to a single cell and is equal to \(\frac { ξ }{ R }\) when R is negligible. Thus, the

current through the external resistance due to the whole battery is n times the current due to a single cell.

Case (b) If r << R. I = \(\frac { ξ }{ R }\) …… (3)

The above equation implies that the current due to the whole battery is the same as that due to a single cell. Hence it is advantageous to connect cells in parallel when the external resistance is very small compared to the internal resistance of the cells.

**Samacheer Kalvi 12th Physics Current Electricity Numerical Problems**

Question 1.

Show that one ampere is equivalent to a flow of 6.25 x 10^{18} elementary charges per second.

Solution:

Here I = 1 A, t= 1s, e = 1.6 x 10^{-19} C

As I = \(\frac { q }{ t }\) = \(\frac { ne }{ t }\)

Number of electrons, n = \(\frac { It }{ e }\) = \(\frac { 1 × 1 }{{ 1.6 × 10 }^{19}}\) = 6.25 × 10^{18}

Question 2.

Calculate the resistivity of a material of a wire 10 m long. 0.4 mm in diameter and having a resistance of 2.0 Ω.

Solution:

Here l = 10 cm, r = 0.2 mm = 0.2 x 10^{-3} m, R = 2 Ω

\(\left[ r\quad =\quad \frac { d }{ 2 } \right] \)

= 2.513 x 10^{-8} Ω.

Question 3.

A wire of 10-ohm resistance is stretched to thrice its original length. What will be its (i) new resistivity and (ii) new resistance?

Solution:

(i) Resistivity p remains unchanged because it is the property of the material of the wire.

(ii) In both cases, the volume of wire is the same, so

Question 4.

A copper wire has a resistance of 10 Ω. and an area of cross-section 1 mm^{2}. A potential difference of 10 V exists across the wire. Calculate the drift speed of electrons if the number of electrons per cubic meter in copper is 8 x 10^{28} electrons.

Solution:

Here, R = 10 Ω, A = 1 mm^{2} = 10^{-6} m^{2}, V = 10 V, n = 8 x 10^{28} electrons/m^{3}

Now,

I = enAv_{d}

∴ \(\frac { V }{ R }\) = enAv_{d} (or) v_{d} = \(\frac { V }{ enAR }\)

= \(\frac{10}{1.6 \times 10^{-19} \times 8 \times 10^{28} \times 10^{-6} \times 10}\) = 0.078 x 10^{-3} ms^{-1}

= 0.078 x ms^{-1}

Question 5.

(i) At what temperature would the resistance of a copper conductor be double its resistance at 0°C.

(ii) Does this temperature hold for all copper conductors regardless of shape and size? Given a for Cu = 3.9 x 10^{-3} °C^{-1}.

Solution:

Thus the resistance of copper conductor becomes double at 256 °C.

(ii) Since a does not depend on the size and shape of the conductor. So the above result holds for all copper conductors.

Question 6.

Find the value of current I in the circuit shown in the figure.

Solution:

In the circuit, the resistance of arm ACB (30 + 30 = 60 Ω) is parallel with the resistance of arm AB (= 30 Ω).

Hence, the effective resistance of the circuit is

R = \(\frac { 30 × 60 }{ 30 + 60 }\) = 20 Ω

Current, I = \(\frac { V }{ R }\) = \(\frac { 2 }{ 20 }\) = 0.1 A.

Common Errors and Its Rectifications:

Common Errors:

- Sometimes students think that charge and current are the same.
- In doing the calculation part students can’t give the importance to mention the units.
- They may confuse the parallel and series network of the resistance.

Rectifications:

- Charge q = ne Current I = q/t
- Unit is very important to every physical quantity.
- If the resistors are series, their resultant is the sum of the all the reciprocal of individual resistance. If the resistors are parallel their resultant is the sum of the individual resistances.

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