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Current Electricity

CBSE · Class 12 · Physics

NCERT Solutions for Current Electricity — CBSE Class 12 Physics.

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Graphs showing the variation of resistivity with temperature for different types of materials: metals, semiconductors, and alloys.
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EXERCISES

3.1The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?Show solution
The maximum current is obtained when the external resistance is zero, so

Imax⁡=εr I_{\max}=\frac{\varepsilon}{r}

Given ε=12 V\varepsilon=12\,\text{V} and r=0.4 Ωr=0.4\,\Omega,

Imax⁡=120.4=30 A I_{\max}=\frac{12}{0.4}=30\,\text{A}

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3.2A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?Show solution
Using V=ε−IrV=\varepsilon-Ir, the terminal voltage is

V=10−(0.5)(3)=8.5 V V=10-(0.5)(3)=8.5\,\text{V}

Also, for the external resistor,

V=IR V=IR

so

R=VI=8.50.5=17 Ω R=\frac{V}{I}=\frac{8.5}{0.5}=17\,\Omega

But the battery drop across internal resistance is Ir=1.5 VIr=1.5\,\text{V}, so the total emf equation is

ε=I(R+r) \varepsilon=I(R+r)

10=0.5(R+3) 10=0.5(R+3)

R+3=20 R+3=20

R=17 Ω R=17\,\Omega

So the resistance of the resistor is 17 Ω17\,\Omega, and the terminal voltage is 8.5 V8.5\,\text{V}.

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3.3At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70×10−41.70 \times 10^{-4} °C −1^{-1} .Show solution
Use

RT=R0[1+α(T−T0)] R_T=R_0[1+\alpha(T-T_0)]

Given R0=100 ΩR_0=100\,\Omega at T0=27∘CT_0=27^\circ\text{C}, RT=117 ΩR_T=117\,\Omega, and α=1.70×10−4 ∘C−1\alpha=1.70\times10^{-4}\,^{\circ}\text{C}^{-1}.

117=100[1+1.70×10−4(T−27)] 117=100[1+1.70\times10^{-4}(T-27)]

1.17=1+1.70×10−4(T−27) 1.17=1+1.70\times10^{-4}(T-27)

0.17=1.70×10−4(T−27) 0.17=1.70\times10^{-4}(T-27)

T−27=0.171.70×10−4=1000 T-27=\frac{0.17}{1.70\times10^{-4}}=1000

T=1027∘C T=1027^\circ\text{C}

So, the temperature is approximately 1028 °C.

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3.4A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10⁻⁷ m², and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?Show solution
Resistivity is

ρ=RAl \rho=\frac{RA}{l}

Given R=5.0 ΩR=5.0\,\Omega, A=6.0×10−7 m2A=6.0\times10^{-7}\,\text{m}^2, and l=15 ml=15\,\text{m}:

ρ=5.0×6.0×10−715 \rho=\frac{5.0\times 6.0\times10^{-7}}{15}

ρ=30×10−715=2.0×10−7 Ω m \rho=\frac{30\times10^{-7}}{15}=2.0\times10^{-7}\,\Omega\text{ m}

So the resistivity is **2.0×10−7 Ω m2.0\times10^{-7}\,\Omega\text{ m}**.

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3.5A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.Show solution
For a small temperature range,

RT=R0[1+α(T−T0)] R_T=R_0[1+\alpha(T-T_0)]

Given R0=2.1 ΩR_0=2.1\,\Omega at T0=27.5∘CT_0=27.5^\circ\text{C}, and RT=2.7 ΩR_T=2.7\,\Omega at 100∘C100^\circ\text{C}.

2.7=2.1[1+α(100−27.5)] 2.7=2.1[1+\alpha(100-27.5)]

2.72.1=1+72.5α \frac{2.7}{2.1}=1+72.5\alpha

1.2857=1+72.5α 1.2857=1+72.5\alpha

72.5α=0.2857 72.5\alpha=0.2857

α=0.285772.5≈0.00394 ∘C−1 \alpha=\frac{0.2857}{72.5}\approx 0.00394\,^{\circ}\text{C}^{-1}

But the book’s standard method for temperature coefficient of resistivity from two resistance readings is to use

α=R2−R1R1(T2−T1) \alpha=\frac{R_2-R_1}{R_1(T_2-T_1)}

So,

α=2.7−2.12.1(100−27.5)=0.62.1×72.5≈3.94×10−3 ∘C−1 \alpha=\frac{2.7-2.1}{2.1(100-27.5)} =\frac{0.6}{2.1\times72.5} \approx 3.94\times10^{-3}\,^{\circ}\text{C}^{-1}

Thus the temperature coefficient is **3.94×10−3 ∘C−13.94\times10^{-3}\,^{\circ}\text{C}^{-1}**.

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3.6A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. 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.
3.7Determine the current in each branch of the network shown in Fig. 3.20:
3.8A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
3.9The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10²⁸ 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 m² and it is carrying a current of 3.0 A.

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Frequently Asked Questions

What are the important topics in Current Electricity for CBSE Class 12 Physics?
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