Current Electricity Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A nichrome wire has a resistance of \( 50 \, \Omega \) at \( 20^\circ \text{C} \) and \( \alpha = 1.7 \times 10^{-4} \, ^\circ\text{C}^{-1} \). What is its resistance at \( 380^\circ \text{C} \)?

Use: \( R_t = R_0 [1 + \alpha (T - T_0)] \).

Substitute: \( R_t = 50 [1 + 1.7 \times 10^{-4} (380 - 20)] \).

Calculate: \( R_t = 50 [1 + 1.7 \times 10^{-4} \times 360] = 50 [1 + 0.0612] = 50 \times 1.0612 = 53.06 \, \Omega \).

\( 52 \, \Omega \)
\( 53.06 \, \Omega \)
\( 54 \, \Omega \)
\( 55 \, \Omega \)
2

A nichrome wire has a resistance of \( 80 \, \Omega \) at \( 25^\circ \text{C} \) and \( \alpha = 1.7 \times 10^{-4} \, ^\circ\text{C}^{-1} \). What is its resistance at \( 225^\circ \text{C} \)?

Use: \( R_t = R_0 [1 + \alpha (T - T_0)] \).

Substitute: \( R_t = 80 [1 + 1.7 \times 10^{-4} (225 - 25)] \).

Calculate: \( R_t = 80 [1 + 1.7 \times 10^{-4} \times 200] = 80 [1 + 0.034] = 80 \times 1.034 = 82.72 \, \Omega \).

\( 81 \, \Omega \)
\( 82.72 \, \Omega \)
\( 83 \, \Omega \)
\( 84 \, \Omega \)
2

Two cells in parallel have emf \( 6 \, \text{V} \) and \( 4 \, \text{V} \) with internal resistances \( 2 \, \Omega \) and \( 1 \, \Omega \). What is the equivalent emf?

For parallel: \( \varepsilon_{\text{eq}} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} \).

Substitute: \( \varepsilon_{\text{eq}} = \frac{6 \times 1 + 4 \times 2}{2 + 1} = \frac{6 + 8}{3} = \frac{14}{3} \approx 4.67 \, \text{V} \).

\( 4.0 \, \text{V} \)
\( 4.67 \, \text{V} \)
\( 5.0 \, \text{V} \)
\( 5.33 \, \text{V} \)
2

A conductor has a resistivity of \( 1.0 \times 10^{-7} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\text{C}^{-1} \) at \( 25^\circ \text{C} \). What is its resistivity at \( 85^\circ \text{C} \)?

Use: \( \rho_t = \rho_0 [1 + \alpha (T - T_0)] \).

Substitute: \( \rho_t = 1.0 \times 10^{-7} [1 + 4 \times 10^{-3} (85 - 25)] \).

Calculate: \( \rho_t = 1.0 \times 10^{-7} [1 + 0.24] = 1.0 \times 10^{-7} \times 1.24 = 1.24 \times 10^{-7} \, \Omega \text{m} \).

\( 1.2 \times 10^{-7} \, \Omega \text{m} \)
\( 1.22 \times 10^{-7} \, \Omega \text{m} \)
\( 1.24 \times 10^{-7} \, \Omega \text{m} \)
\( 1.3 \times 10^{-7} \, \Omega \text{m} \)
3

In a Wheatstone bridge with \( R_1 = 30 \, \Omega \), \( R_2 = 60 \, \Omega \), \( R_3 = 15 \, \Omega \), and \( R_4 = 32 \, \Omega \), a \( 10 \, \text{V} \) battery is connected across AC. What is the current through the galvanometer (\( R_g = 10 \, \Omega \))?

Apply Kirchhoff’s rules. Let currents be \( I_1 \) (AB), \( I_2 \) (AD), \( I_g \) (BD).

Junction B: \( I_1 = I_g + I_4 \), Junction D: \( I_2 = I_g + I_3 \).

Loop BADB: \( 30 I_1 + 10 I_g - 60 I_2 = 0 \Rightarrow 3 I_1 + I_g - 6 I_2 = 0 \).

Loop BCDB: \( 60 (I_1 - I_g) - 10 I_g - 15 (I_2 + I_g) = 0 \Rightarrow 4 I_1 - 2 I_2 - 2 I_g = 0 \).

Loop ADCEA: \( 60 I_2 + 15 (I_2 + I_g) = 10 \Rightarrow 75 I_2 + 15 I_g = 10 \Rightarrow 5 I_2 + I_g = \frac{2}{3} \).

Solve: From (3) \( I_g = \frac{2}{3} - 5 I_2 \), substitute in (1): \( 3 I_1 + \frac{2}{3} - 5 I_2 - 6 I_2 = 0 \Rightarrow 3 I_1 - 11 I_2 = -\frac{2}{3} \).

From (2): \( 4 I_1 - 2 I_2 - 2 (\frac{2}{3} - 5 I_2) = 0 \Rightarrow 4 I_1 - 2 I_2 - \frac{4}{3} + 10 I_2 = 0 \Rightarrow 4 I_1 + 8 I_2 = \frac{4}{3} \).

Solve: \( 12 I_1 - 33 I_2 = -2 \), \( 12 I_1 + 24 I_2 = 4 \). Subtract: \( -57 I_2 = -6 \Rightarrow I_2 = \frac{6}{57} = \frac{2}{19} \, \text{A} \).

\( I_g = \frac{2}{3} - 5 \times \frac{2}{19} = \frac{38}{57} - \frac{30}{57} = \frac{8}{57} \approx 0.14 \, \text{A} \).

\( 0.10 \, \text{A} \)
\( 0.14 \, \text{A} \)
\( 0.18 \, \text{A} \)
\( 0.22 \, \text{A} \)
2

A battery of emf \( 12 \, \text{V} \) and internal resistance \( 1 \, \Omega \) is connected to a \( 5 \, \Omega \) resistor. What is the power dissipated in the external resistor?

Current: \( I = \frac{\varepsilon}{R + r} = \frac{12}{5 + 1} = 2 \, \text{A} \).

Power: \( P = I^2 R = (2)^2 \times 5 = 20 \, \text{W} \).

\( 16 \, \text{W} \)
\( 20 \, \text{W} \)
\( 24 \, \text{W} \)
\( 28 \, \text{W} \)
2

A \( 15 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance delivers a current of \( 2 \, \text{A} \) to a resistor. What is the resistance of the resistor?

Terminal voltage: \( V = \varepsilon - I r = 15 - 2 \times 3 = 9 \, \text{V} \).

Resistance: \( R = \frac{V}{I} = \frac{9}{2} = 4.5 \, \Omega \).

\( 3.5 \, \Omega \)
\( 4.0 \, \Omega \)
\( 4.5 \, \Omega \)
\( 5.0 \, \Omega \)
3

A platinum wire has a resistance of \( 6 \, \Omega \) at \( 0^\circ \text{C} \) and \( 6.48 \, \Omega \) at \( 100^\circ \text{C} \). What is the temperature coefficient of resistivity?

Use: \( R_t = R_0 [1 + \alpha (T - T_0)] \).

Substitute: \( 6.48 = 6 [1 + \alpha (100 - 0)] \).

Solve: \( 6.48 = 6 + 600\alpha \Rightarrow 600\alpha = 0.48 \Rightarrow \alpha = \frac{0.48}{600} = 8 \times 10^{-4} \, ^\circ\text{C}^{-1} \).

\( 8.0 \times 10^{-4} \, ^\circ\text{C}^{-1} \)
\( 9.0 \times 10^{-4} \, ^\circ\text{C}^{-1} \)
\( 1.0 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 1.2 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
1

Two cells of emf \( 3 \, \text{V} \) and \( 5 \, \text{V} \) with internal resistances \( 1 \, \Omega \) and \( 3 \, \Omega \) are connected in series with a \( 4 \, \Omega \) resistor. What is the current through the circuit?

Equivalent emf: \( \varepsilon_{\text{eq}} = 3 + 5 = 8 \, \text{V} \).

Total resistance: \( R_{\text{total}} = r_1 + r_2 + R = 1 + 3 + 4 = 8 \, \Omega \).

Current: \( I = \frac{\varepsilon_{\text{eq}}}{R_{\text{total}}} = \frac{8}{8} = 1 \, \text{A} \).

\( 1.0 \, \text{A} \)
\( 1.5 \, \text{A} \)
\( 2.0 \, \text{A} \)
\( 2.5 \, \text{A} \)
1

A copper wire of cross-sectional area \( 3 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 0.9 \, \text{A} \). If \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \) and \( e = 1.6 \times 10^{-19} \, \text{C} \), what is the drift speed?

Drift speed: \( v_d = \frac{I}{n e A} \).

Substitute: \( v_d = \frac{0.9}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 3 \times 10^{-7}} \).

Calculate: \( v_d = \frac{0.9}{4.08 \times 10^3} \approx 2.21 \times 10^{-4} \, \text{m/s} \).

\( 2.0 \times 10^{-4} \, \text{m/s} \)
\( 2.21 \times 10^{-4} \, \text{m/s} \)
\( 2.4 \times 10^{-4} \, \text{m/s} \)
\( 2.6 \times 10^{-4} \, \text{m/s} \)
2

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