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  4. Current Electricity Chapter-Wise Test 3

Current Electricity Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A wire has a resistance of \( 10 \, \Omega \) at \( 20^\circ \text{C} \) and \( 12 \, \Omega \) at \( 100^\circ \text{C} \). What is the temperature coefficient of resistivity?

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

Given: \( R_0 = 10 \, \Omega \), \( R_t = 12 \, \Omega \), \( T = 100^\circ \text{C} \), \( T_0 = 20^\circ \text{C} \).

Substitute: \( 12 = 10 [1 + \alpha (100 - 20)] \).

Solve: \( 12 = 10 + 80\alpha \Rightarrow 80\alpha = 2 \Rightarrow \alpha = \frac{2}{80} = 0.025 \times 10^{-2} = 2.5 \times 10^{-4} \, ^\circ\text{C}^{-1} \).

\( 2.0 \times 10^{-4} \, ^\circ\text{C}^{-1} \)
\( 2.5 \times 10^{-4} \, ^\circ\text{C}^{-1} \)
\( 3.0 \times 10^{-4} \, ^\circ\text{C}^{-1} \)
\( 3.5 \times 10^{-4} \, ^\circ\text{C}^{-1} \)
2

Why does a filament lamp deviate from Ohm’s law as current increases?

In a filament lamp, increased current raises the filament’s temperature, increasing its resistance due to a positive temperature coefficient. This makes the \( V \)-versus-\( I \) relationship non-linear, violating Ohm’s law.

Voltage drops to zero
Resistance increases with temperature
Current becomes constant
Filament length decreases
2

A wire of length \( 8 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) has a resistance of \( 16 \, \Omega \). What is the resistivity of the material?

Resistance: \( R = \frac{\rho l}{A} \).

Rearrange: \( \rho = \frac{R A}{l} \).

Substitute: \( \rho = \frac{16 \times 2 \times 10^{-6}}{8} = 4 \times 10^{-6} \, \Omega \text{m} \).

\( 4.0 \times 10^{-6} \, \Omega \text{m} \)
\( 4.5 \times 10^{-6} \, \Omega \text{m} \)
\( 5.0 \times 10^{-6} \, \Omega \text{m} \)
\( 5.5 \times 10^{-6} \, \Omega \text{m} \)
1

A Wheatstone bridge has \( R_1 = 25 \, \Omega \), \( R_2 = 50 \, \Omega \), \( R_3 = 20 \, \Omega \). What is \( R_4 \) for balance?

Balance condition: \( \frac{R_1}{R_2} = \frac{R_3}{R_4} \).

Substitute: \( \frac{25}{50} = \frac{20}{R_4} \).

Solve: \( 0.5 = \frac{20}{R_4} \Rightarrow R_4 = \frac{20}{0.5} = 40 \, \Omega \).

\( 35 \, \Omega \)
\( 40 \, \Omega \)
\( 45 \, \Omega \)
\( 50 \, \Omega \)
2

A \( 3 \, \text{m} \) long wire with a cross-sectional area of \( 1.5 \times 10^{-6} \, \text{m}^2 \) has a resistance of \( 6 \, \Omega \). What is the current density when a potential difference of \( 12 \, \text{V} \) is applied?

Current: \( I = \frac{V}{R} = \frac{12}{6} = 2 \, \text{A} \).

Current density: \( j = \frac{I}{A} = \frac{2}{1.5 \times 10^{-6}} = 1.33 \times 10^6 \, \text{A/m}^2 \).

\( 1.0 \times 10^6 \, \text{A/m}^2 \)
\( 1.33 \times 10^6 \, \text{A/m}^2 \)
\( 1.5 \times 10^6 \, \text{A/m}^2 \)
\( 2.0 \times 10^6 \, \text{A/m}^2 \)
2

A cell of emf \( 9 \, \text{V} \) and internal resistance \( 1 \, \Omega \) is connected to a \( 8 \, \Omega \) resistor. What is the terminal voltage?

Total resistance: \( R_{\text{total}} = 8 + 1 = 9 \, \Omega \).

Current: \( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{9}{9} = 1 \, \text{A} \).

Terminal voltage: \( V = \varepsilon - I r = 9 - 1 \times 1 = 8 \, \text{V} \).

\( 8.0 \, \text{V} \)
\( 8.5 \, \text{V} \)
\( 9.0 \, \text{V} \)
\( 9.5 \, \text{V} \)
1

What happens to the drift velocity of electrons in a conductor if the conductor’s temperature increases while the applied electric field remains constant?

Drift velocity \( v_d = e E \tau / m \). As temperature increases, \( \tau \) (relaxation time) decreases due to more collisions, reducing \( v_d \) if \( E \) is constant.

It increases
It remains constant
It decreases
It doubles
3

A copper wire of cross-sectional area \( 5 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 1 \, \text{A} \). 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: \( v_d = \frac{1}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-7}} \).

Calculate: \( v_d = \frac{1}{6.8 \times 10^3} \approx 1.47 \times 10^{-4} \, \text{m/s} \).

\( 1.2 \times 10^{-4} \, \text{m/s} \)
\( 1.47 \times 10^{-4} \, \text{m/s} \)
\( 1.6 \times 10^{-4} \, \text{m/s} \)
\( 1.8 \times 10^{-4} \, \text{m/s} \)
2

A wire has a resistance of \( 18 \, \Omega \) at \( 20^\circ \text{C} \) and \( 19.8 \, \Omega \) at \( 80^\circ \text{C} \). What is the temperature coefficient of resistivity?

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

Substitute: \( 19.8 = 18 [1 + \alpha (80 - 20)] \).

Solve: \( 19.8 = 18 + 1080\alpha \Rightarrow 1080\alpha = 1.8 \Rightarrow \alpha = \frac{1.8}{1080} \approx 1.67 \times 10^{-3} \, ^\circ\text{C}^{-1} \).

\( 1.67 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 1.8 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 2.0 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
\( 2.2 \times 10^{-3} \, ^\circ\text{C}^{-1} \)
1

A wire of length \( 9 \, \text{m} \) and cross-sectional area \( 1.5 \times 10^{-6} \, \text{m}^2 \) has a resistance of \( 18 \, \Omega \). What is the resistivity of the material?

Resistance: \( R = \frac{\rho l}{A} \).

Rearrange: \( \rho = \frac{R A}{l} \).

Substitute: \( \rho = \frac{18 \times 1.5 \times 10^{-6}}{9} = 3 \times 10^{-6} \, \Omega \text{m} \).

\( 3.0 \times 10^{-6} \, \Omega \text{m} \)
\( 3.5 \times 10^{-6} \, \Omega \text{m} \)
\( 4.0 \times 10^{-6} \, \Omega \text{m} \)
\( 4.5 \times 10^{-6} \, \Omega \text{m} \)
1

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