# NCERT Solutions for Class 12 Physics Chapter 3 - Current Electricity

## NCERT Solutions for Class 12 Physics Chapter 3 Free PDF Download

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**1. The storage battery of a car has an emf of 12V. If the internal resistance of the battery is 0.4W, what is the maximum current that can be drawn from the battery?**

**Sol.** Given, emf E = 12V, internal resistance r = 0.4 Ω

$$\therefore\space\text{Current drawn from the battery I =}\space\frac{\text{E}}{\text{R+r}}$$

In case of maximum current, R = 0

$$\therefore\space\text{I}_{max}=\frac{\text{E}}{\text{r}}=\frac{12}{0.4}=30\text{A}$$

**2. A battery of emf 10V and internal resistance 3W is connected to a resistor. If the current in the circuit is 0.5A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?**

**Sol.** Given, emf of battery, E = 10V

Internal resistance, r = 3W

Current in circuit, I = 0.5A

$$\text{The current in the circuit =}\frac{\text{emf of battery}}{\text{Total resistance}}\\\text{I}=\frac{\text{E}}{\text{R + r}}\\\therefore\space o.5=\frac{10}{\text{R+3}}$$

or R + 3 = 20

R = 17W

When the circuit is closed, the terminal voltage

V = E – Ir

= 10 – 0.5 × 3

= 10 – 1.5

= 8.5V

Thus, the resistance in the circuit is 17Ω and terminal voltage of the battery in case, is 8.5V.

**3. (a) Three resistors 1W, 2W and 3W are combined in series. What is the total resistance of the combination?**

**(b) If the combination is connected to a battery of emf 12V and negligible internal resistance, obtain the potential drop across each resistor.**

**Sol.** (a) R_{1} = 1Ω, R_{2} = 2Ω and R_{3} = 3Ω. Resultant resistance in series

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

R_{S} = 1 + 2 + 3 = 6Ω

(b) The potential drop across

each resistor is different when two or more resistances are connected in series combination.

Let V_{1}, V_{2} and V_{3} be the potential drops across resistances R_{1}, R_{2} and R_{3} respectively and the current flowing through the circuit.

$$\therefore\space\text{I}=\frac{\text{V}}{\text{R}_{s}}=\frac{12}{6}=2\text{A}$$

Current is same through each resistor as they are in series.

Potential drop across R_{1}, V_{1} = IR_{1} = 2 × 1 = 2V

Potential drop across R_{2}, V_{2} = IR_{2} = 2 × 2 = 4V

Potential drop across R_{1}, V_{3} = IR_{3} = 2 × 3 = 6V

Thus, the potential drop across resistance 1Ω is 2V, resistance 2Ω is 4V and resistance 3Ω is 6V.

**4. (a) Three resistors 2W, 4W and 5W are combined in parallel. What is the total resistance of the combination?**

**(b) If the combination is connected to a battery of emf 20V and negligible internal resistance, determine the current through each resistor and the total current drawn from the battery.**

**Sol.** Given, R_{1} = 2Ω, R_{2} = 4Ω and R_{3} = 5Ω

(a) Resultant resistance in parallel

$$\frac{1}{\text{R}_{p}}=\frac{1}{\text{R}_{1}} + \frac{1}{\text{R}_{2}} + \frac{1}{\text{R}_{3}}\\=\frac{1}{2}+ \frac{1}{4}+ \frac{1}{5}\\=\frac{10+5+4}{20}=\frac{19}{20}\\\text{R}_{p}=\frac{20}{19}Ω$$

(b) In parallel combination, the current flowing through each resistance is different, let I_{1}, I_{2} and I_{3} and the potential drop across each resistor which is same as the applied potential difference V = 20V

Current through resistance R_{1},

$$\text{I}_{1}=\frac{\text{V}}{\text{R}_{1}}\\=\frac{20}{2}=10\text{A}\\\text{Current through resistance R}_2,\\\text{I}_{2}=\frac{\text{V}}{\text{R}_{2}}=\frac{20}{4}=5\text{A}$$

Current through resistance R_{3},

$$\text{I}_{3}=\frac{\text{V}}{\text{R}_{3}}\\=\frac{20}{5}=4\text{A}$$

Total current drawn I = I_{1} +I_{2} + I_{3}

= 10 + 5 + 4 = 19A

Thus, total current drawn from the battery I =19A.

**5. At room temperature (27.0°C) the resistance of a heating element is 100Ω. What is the temperature of the element, if the resistance is found to be 117Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10 ^{–4}/°C.**

**Sol.** Given, the resistance of heating element at temperature 27°C,

R_{27} = 100Ω

Resistance of heating element at temperature t°C, R_{t} = 117Ω

Temperature coefficient α = 1.70 × 10^{–4}/°C.

Temperature coefficient of resistance Given, the resistance of heating element at temperature 27°C,

R_{27} = 100Ω

Resistance of heating element at temperature t°C,

R_{t} = 117Ω

Temperature coefficient a = 1.70 × 10^{–4}/°C.

Temperature coefficient of resistance

$$\alpha=\frac{\text{R}_{t}-\text{R}_{27}}{\text{R}_{27}(t-27)}\\1.70×10^{\normalsize-4}=\frac{117-100}{100(t-27)}\\\text{or}\space t-27=\frac{17}{100×1.70×10^{\normalsize-4}}$$

or t = 1000 + 27 = 1027°C

Thus, the temperature of element is 1027 °C.

**6. A negligible small current is passed through a wire of length 15m and uniform cross-section 6.0 ×10–7m2 and its resistance is measured to be 5.0W. What is the resistivity of the material at the temperature of the experiment?**

**Sol.** Given, area of cross-section of wire (A) = 6.0 × 10^{-7} m^{2}, Length of the wire l = 15 m,

Resistance of wire R = 5Ω

Let the resistivity of the material be ρ.

$$\text{Resistance of wire R =}\rho\frac{\text{l}}{\text{A}}\\\text{or\qquad}\space ρ=\frac{\text{RA}}{l}\\=\frac{5×6.0×10^{\normalsize-7}}{15}$$

= 2 × 10^{–7} Ω-m

Thus, the resistivity of the material is 2 × 10^{–7}Ω-m.

**7. A silver wire has a resistance of 2.1W at 27.5 °C and a resistance of 2.7W at 100°C. Determine the temperature coefficient resistivity of silver.**

**Sol.** Given, resistance of silver wire at 27.5°C = R_{27.5} = 2.1W

Resistance of silver wire at 100°C = R_{100} = 2.7Ω

The temperature coefficient of silver.

$$\alpha=\frac{\text{R}_{t_{2}}-\text{R}_{t_{2}}}{\text{R}_1(t_{2}-t_{1})}\\\alpha=\frac{\text{R}_{100}-\text{R}_{27.5}}{\text{R}_{27.5}(100-27.5)}\\\alpha=\frac{2.7-2.1}{2.1×72.5}$$

α= 0.0039/°C

Thus, the temperature coefficient of resistivity of silver is 0.0039/°C.

**8. A heating element using nichrome connected to a 230V supply draws an initial current of 3.2. A which settles after a few seconds to a steady value of 2.8A. What is the steady temperature of the heating element, if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10 ^{–4} °C^{–1}.**

**Sol.** Given, potential difference = 230V

Initially current at 27°C = I_{27°C} = 3.2A

Finally current at t°C = I_{t°C} = 2.8A

Temperature coefficient α = 1.70 × 10^{–4}/°C

$$\text{Resistance at 27t°C, R}_{t°C} =\frac{\text{V}}{t_{27}\degree\text{C}}=\frac{230}{3.2}=\frac{2300}{32}Ω\\\text{Resistance at t°C, R}_{t°C} =\frac{\text{V}}{\text{t}_{t\degree C}}=\frac{230}{2.8}= \frac{2300}{28}Ω$$

Temperature of coefficient

$$\alpha=\frac{\text{R}_{t}-\text{R}_{27}}{\text{R}_{27}(t-27)}\\\Rarr\space 1.7×10^{\normalsize-4}=\frac{\frac{2300}{28}-\frac{2300}{32}}{\frac{2300}{32}(t-32)}\\\text{or}\space t – 27 =\frac{82.143-71.875}{71.875×1.7×10^{\normalsize-4}}=840.347$$

or t = 840.3 + 27 = 867.3°C

Thus, the steady temperature of heating element is 867.3°C.

**9. Determine the current in each branch of the network shown in given figure.**

**Sol.** From Kirchhoff’s’s Ist law, i.e., loop law,

ΣV = ΣIR

In loop ABDA, Distributing the current

10I_{1} + 5I_{2} – 5(I – I_{1}) = 0

or

3I_{1} + I_{2} = I …(i)

In loop BCDB,

5(I_{1} – I_{2}) – 10 (I – I_{1} + I_{2}) – 5I_{2} = 0

I_{1} – I_{2} – 2I + 2I_{1} – 2I_{2} – I_{2} = 0

or

3I_{1} – 4I_{2} = 2I …(ii)

By solving the equations (i) and (ii), we get

$$\text{I}_{1}=\frac{2I}{5}\text{and}\space\text{I}_{2}=-\frac{1}{5}\qquad...(\text{iii})$$

In loop ABCEFA

10 = 10I + 10I_{1} + 5(I_{1} – I_{2})

2 = 2I + 3I_{1} – I_{2} …(iv)

Substituting the values of I_{1} and I_{2} from Eq. (iii) in Eq. (iv), we get

$$\text{2=2I} + 3\bigg(\frac{2\text{I}}{5}\bigg)-\bigg(-\frac{\text{I}}{5}\bigg)\\2=\frac{17}{5}\text{t}\\\therefore\space\text{I}=\frac{10}{17}\text{A}\\\text{Current in branch AB,}\\\text{I}_{1}=\frac{2}{5}×\frac{10}{7}=\frac{4}{17}\text{A}\\\text{and}\space\text{I}_{2}=-\frac{1}{5}=-\frac{2}{17}\text{A}\\\text{Current in branch AB = I}_1=\frac{4}{17}\text{A}\\\text{Current in branch BC = I}_{1}-\text{I}_{2}=\frac{4}{17}-\bigg(-\frac{2}{17}\bigg)$$

$$=\frac{6}{17}\text{A}\\\text{Current in branch AD = I – I}_1 =\frac{10}{17}-\frac{4}{17}=\frac{6}{17}\text{A}\\\text{Current in branch DC = (I – I}_1) + \text{I}_2 =\frac{6}{17}+\bigg(-\frac{2}{17}\bigg)\\=\frac{4}{17}\text{A}$$

**10. (a) In a meter bridge, the balance point is found to be at 39.5 cm from the end A, when the resistor S is of 12.5 W. Determine the resistance of R. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?**

**(b) Determine the balance point of the bridge above, if R and S are interchanged.**

**(c) What happens, if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?**

**Sol.** (a) Balance point from end A,

l = 39.5 cm

Resistance Y = 12.5 Ω

Resistance X = ?

Applying the condition of balanced Wheatstone bridge.

$$\frac{\text{X}}{\text{Y}}=\frac{l}{100-l}\\\text{X}=\frac{l}{100-l}.\text{Y}\\\text{X}=\frac{39.5×12.5}{100-39.5}=8.16Ω$$

The resistance of resistor X is 8.16 Ω

In meter bridge, the resistance at the connections is not taken in the consideration that’s why the connections between resistors in a Wheatstone bridge or meter bridge made of thick copper strips because higher is the thickness, lesser be the resistance as

$$\bigg(\text{as R}∝\frac{l}{\text{A}}\bigg)$$

, so due to thick copper stripes, the resistance at the connections becomes minimum.

(b) On interchanging X and Y the balance length will also interchanged. Thus, the balance length becomes.

100 – 39.5 = 60.5 cm

(c) If the galvanometer and cell are interchanged at the balance point of the bridge, the balance point is not obtained. The galvanometer shows no deflection.

**11. A storage battery of emf 8.0V and internal resistance 0.5Ω is being charged by a 120V DC supply using a series resistor of 15.5Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?**

**Sol.** Emf e = 8V, voltage of DC supply V = 120V

The effective emf in the circuit

E = V – e = 120 – 8 = 112V

Current flowing in circuit,

$$\text{I}=\frac{\text{Effectiveemf}}{\text{Total resistance}}=\frac{\text{E}}{\text{r+R}}\\\frac{112}{0.5 + 15.5}=\frac{112}{6}$$

Since, the battery of 8V is being charged by 120V, so the terminal potential across battery of 8V will be greater than its emf

Terminal voltage V = E + Ir = 8 + 7(0.5) = 11.5V

The purpose of the series resistance is to control the current drawn from the external supply, other the current in the circuit will be very high.

**12. In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, when is the emf of the second cell?**

**Sol.** Given, E_{1} = 1.25 V, l_{1} = 35 cm and l_{2} = 63 cm

By the principle of potentiometer, the potential gradient remains constant, i.e.,

E ∝ l

$$\therefore\space\frac{\text{E}_{1}}{\text{E}_{2}}=\frac{\text{l}_{1}}{\text{l}_{2}}\\\frac{1.25}{\text{E}}=\frac{35}{\text{63}}\\\text{or}\space\text{E}=\frac{1.25×63}{35}=2.25\text{V}$$

Thus, the emf of the second cell is 2.25V.

**13. The number density of free electrons in a copper conductor estimated at 8.5 × 10 ^{28} m^{–3}. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10^{–6} m^{2} and it is carrying a current of 3.0 A.**

**Sol.** Given, number density of electrons n = 8.5 × 10^{28} m^{3}, length of wire l = 3 m

Area of cross-second of wire

A = 2 × 10^{–6} m^{2}

Current I = 3A and e = 1.6 × 10^{–19}C.

Time taken by electron to drift from one end to another of the wire.

$$t=\frac{\text{Length of thewire}}{\text{Drift velocity}}\\=\frac{l}{v_{d}}\space...(\text{i})\\\text{Using the relation, I = neAv}_{d}\\\text{or}\space v_{d}=\frac{l}{\text{neA}}\qquad\text{...(ii)}\\\text{Putting the value in Eq. (ii) from Eq. (i),}\\\text{I}=\frac{\text{IneA}}{\text{I}}\\\frac{3×8.5×10^{28}×1.6×10^{\normalsize-19}×10^{\normalsize-6}}{1}$$

or t = 2.72 × 10^{4} s = 7 hr 33 min

Thus, the time taken by an electron to drift from one end to another end is 7h 33 min.