Current Electricity Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A wire of length \( 2 \, \text{m} \) and resistance \( 4 \, \Omega \) is stretched to \( 4 \, \text{m} \). What is the new resistance?

Volume constant: \( l A = l' A' \Rightarrow A' = \frac{A}{2} \).

New resistance: \( R' = \frac{\rho l'}{A'} = \frac{\rho (2l)}{\frac{A}{2}} = 4 \frac{\rho l}{A} = 4R = 4 \times 4 = 16 \, \Omega \).

\( 12 \, \Omega \)
\( 14 \, \Omega \)
\( 16 \, \Omega \)
\( 18 \, \Omega \)
3

What ensures that the total voltage drop across a series circuit equals the source voltage?

Kirchhoff’s loop rule (conservation of energy) ensures that the sum of potential drops across all elements in a closed loop equals the supplied voltage, as energy is conserved in the circuit.

Current cancellation
Resistance equality
Charge accumulation
Energy conservation
4

Why does the resistance of a metallic conductor increase with temperature?

Resistance (\( R = \rho l / A \)) depends on resistivity (\( \rho \)), which is \( \rho = m / (n e^2 \tau) \). As temperature increases, the average time between collisions (\( \tau \)) decreases due to increased lattice vibrations, leading to higher \( \rho \) and thus higher \( R \).

Number of free electrons decreases
Collision frequency of electrons increases
Electron charge increases
Conductor length increases
2

A nichrome wire has a resistance of \( 60 \, \Omega \) at \( 30^\circ \text{C} \) and \( \alpha = 1.7 \times 10^{-4} \, ^\circ\text{C}^{-1} \). What is its resistance at \( 330^\circ \text{C} \)?

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

Substitute: \( R_t = 60 [1 + 1.7 \times 10^{-4} (330 - 30)] \).

Calculate: \( R_t = 60 [1 + 1.7 \times 10^{-4} \times 300] = 60 [1 + 0.051] = 60 \times 1.051 = 63.06 \, \Omega \).

\( 62 \, \Omega \)
\( 63.06 \, \Omega \)
\( 64 \, \Omega \)
\( 65 \, \Omega \)
2

A Wheatstone bridge has \( R_1 = 14 \, \Omega \), \( R_2 = 28 \, \Omega \), \( R_3 = 10 \, \Omega \). What is \( R_4 \) for balance?

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

Substitute: \( \frac{14}{28} = \frac{10}{R_4} \).

Solve: \( 0.5 = \frac{10}{R_4} \Rightarrow R_4 = \frac{10}{0.5} = 20 \, \Omega \).

\( 18 \, \Omega \)
\( 20 \, \Omega \)
\( 22 \, \Omega \)
\( 24 \, \Omega \)
2

In a circuit with a battery and a resistor, if the internal resistance of the battery equals the external resistance, what fraction of the total power is dissipated in the external resistor?

Let emf = \( \varepsilon \), internal resistance = \( r \), external resistance = \( R = r \). Total resistance = \( r + R = 2r \). Current = \( I = \varepsilon / (2r) \). Power in external resistor = \( I^2 R = (\varepsilon / 2r)^2 r = \varepsilon^2 r / (4 r^2) = \varepsilon^2 / (4 r) \). Total power = \( \varepsilon I = \varepsilon \cdot \varepsilon / (2r) = \varepsilon^2 / (2r) \). Fraction = \( (\varepsilon^2 / 4r) / (\varepsilon^2 / 2r) = 1/2 \).

1/4
1/3
1/2
2/3
3

A \( 16 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance delivers a current of \( 2 \, \text{A} \) to a resistor. What is the resistance of the resistor?

Terminal voltage: \( V = \varepsilon - I r = 16 - 2 \times 2 = 12 \, \text{V} \).

Resistance: \( R = \frac{V}{I} = \frac{12}{2} = 6 \, \Omega \).

\( 5.0 \, \Omega \)
\( 5.5 \, \Omega \)
\( 6.0 \, \Omega \)
\( 6.5 \, \Omega \)
3

What causes the resistivity of a conductor to deviate from Ohm’s law at very high electric fields?

At high fields, resistivity (\( \rho = m / (n e^2 \tau) \)) may change as \( \tau \) or \( n \) varies (e.g., due to electron saturation or heating), making the \( I \)-versus-\( V \) relationship non-linear and violating Ohm’s law.

Current stops flowing
Voltage becomes zero
Resistivity becomes field-dependent
Charge carriers vanish
3

What causes the potential difference across a resistor to drop when current flows through it?

Current through a resistor (\( I = V / R \)) causes a potential drop (\( V = I R \)) as electrical energy is converted to thermal energy due to collisions with lattice ions.

Current stops
Voltage increases
Energy conversion to heat
Resistance disappears
2

A conductor has a resistivity of \( 4 \times 10^{-8} \, \Omega \text{m} \) at \( 20^\circ \text{C} \) and a temperature coefficient of \( 4 \times 10^{-3} \, ^\circ\text{C}^{-1} \). What is its resistivity at \( 60^\circ \text{C} \)?

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

Given: \( \rho_0 = 4 \times 10^{-8} \, \Omega \text{m} \), \( \alpha = 4 \times 10^{-3} \, ^\circ\text{C}^{-1} \), \( T = 60^\circ \text{C} \), \( T_0 = 20^\circ \text{C} \).

Substitute: \( \rho_t = 4 \times 10^{-8} [1 + 4 \times 10^{-3} (60 - 20)] \).

Calculate: \( \rho_t = 4 \times 10^{-8} [1 + 0.16] = 4 \times 10^{-8} \times 1.16 = 4.64 \times 10^{-8} \, \Omega \text{m} \).

\( 4.48 \times 10^{-8} \, \Omega \text{m} \)
\( 4.64 \times 10^{-8} \, \Omega \text{m} \)
\( 4.80 \times 10^{-8} \, \Omega \text{m} \)
\( 5.00 \times 10^{-8} \, \Omega \text{m} \)
2

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