\( Q = \int_{0}^{2} I(t) dt = \int_{0}^{2} (4t^3 + 2) dt = [t^4 + 2t]_{0}^{2} \).
\( Q = (2^4 + 2(2)) – 0 = 16 + 4 = 20 \text{ C} \).
Correct Answer: C

\( I = nqv_d A \). Since \( I \) and \( nq \) are constant, \( v_d \propto \displaystyle\frac{1}{A} \).
If diameter \( d \rightarrow d/2 \), then Area \( A \propto d^2 \rightarrow A/4 \).
Since area is 1/4th, drift velocity must be 4 times larger to maintain the same current.
Correct Answer: C

In the absence of a field, electrons undergo thermal motion at very high speeds, but because the motion is random, the average velocity (drift velocity) is zero.
Correct Answer: D

\( I = \int J dA = \int_{0}^{R} \left(\displaystyle\frac{J_0 r}{R}\right) (2\pi r dr) = \displaystyle\frac{2\pi J_0}{R} \int_{0}^{R} r^2 dr \).
\( I = \displaystyle\frac{2\pi J_0}{R} \left[\displaystyle\frac{r^3}{3}\right]_{0}^{R} = \displaystyle\frac{2}{3} J_0 \pi R^2 \).
Correct Answer: B

\( Q = I \Delta t = (20,000 \text{ A})(100 \times 10^{-6} \text{ s}) = 2 \text{ C} \).
Number of electrons \( n = Q/e = 2 / (1.6 \times 10^{-19}) = 1.25 \times 10^{19} \).
Correct Answer: A

\( V = \varepsilon – Ir = I R \).
Since \( V = \varepsilon/2 \), the voltage drop across the internal resistance must also be \( \varepsilon/2 \).
Therefore, \( Ir = IR \), which implies \( R = r \).
Correct Answer: B

The Loop Rule states that the sum of potential differences around a closed loop is zero. Since potential is energy per unit charge, this is a direct application of the law of conservation of energy.
Correct Answer: C

Parallel part: \( R_p = 12 / 3 = 4 \text{ \(\Omega\)} \).
Total: \( R_{eq} = R_p + 12 = 4 + 12 = 16 \text{ \(\Omega\)} \).
Correct Answer: B

Adding a resistor in parallel provides an additional path for current, which decreases the total equivalent resistance of the circuit. By Ohm’s Law (\( I = V/R \)), as \( R \) decreases, \( I \) increases.
Correct Answer: A

An ammeter has zero resistance and will short out the component it is parallel to, drawing massive current. A voltmeter has infinite resistance and will block nearly all current in the series branch, measuring the full source voltage.
Correct Answer: B

Volume \( V = AL \) is constant. If \( L \rightarrow 3L \), then \( A \rightarrow A/3 \).
Resistance \( R = \rho \displaystyle\frac{L}{A} \). New resistance \( R’ = \rho \displaystyle\frac{3L}{A/3} = 9 \left(\rho \displaystyle\frac{L}{A}\right) = 9R \).
Correct Answer: C

\( R \propto \displaystyle\frac{L}{r^2} \).
\( \displaystyle\frac{R_X}{R_Y} = \left(\displaystyle\frac{L_X}{L_Y}\right) \left(\displaystyle\frac{r_Y}{r_X}\right)^2 = (2) \times (2)^2 = 8 \).
Correct Answer: C

Resistance \( R = V/I \). On an \( I \) vs. \( V \) graph, the slope is \( 1/R \). If resistance increases, the slope must decrease. Thus, the curve is concave downward.
Correct Answer: C

Higher temperatures cause ions in the lattice to vibrate more violently, which increases the probability that a drifting electron will collide with one, thereby increasing resistivity.
Correct Answer: B

\( R = V/I = 12/2 = 6 \text{ \(\Omega\)} \).
\( \rho = \displaystyle\frac{RA}{L} = \displaystyle\frac{6 \times \pi (0.001)^2}{2} = 3\pi \times 10^{-6} \approx 9.42 \times 10^{-6} \text{ \(\Omega\)\(\cdot\)m} \).
Correct Answer: A

\( R = V^2 / P \). The \( 60 \text{ W} \) bulb has higher resistance. In series, current \( I \) is the same. Since \( P = I^2 R \), the higher resistance (\( 60 \text{ W} \)) bulb dissipates more power and shines brighter.
Correct Answer: B

Power in internal resistance is \( P_r = I^2 r \). This is maximized when the current \( I \) is maximized. Since \( I = \varepsilon / (R + r) \), current is maximum when external resistance \( R = 0 \).
Correct Answer: A

\( P \propto V^2 \).
\( P_{new} = 1200 \times \left(\displaystyle\frac{200}{240}\right)^2 = 1200 \times \left(\displaystyle\frac{5}{6}\right)^2 = 1200 \times \displaystyle\frac{25}{36} = 833.3 \text{ W} \).
Correct Answer: A

\( P_{total} = \sum \displaystyle\frac{V^2}{R_i} = \displaystyle\frac{144}{10} + \displaystyle\frac{144}{20} + \displaystyle\frac{144}{30} \).
\( P_{total} = 14.4 + 7.2 + 4.8 = 26.4 \text{ W} \).
Correct Answer: C

Maximum power transfer occurs when \( R = r \). Total power is \( I^2(R+r) = I^2(2R) \). Load power is \( I^2R \). Efficiency \( = (I^2R) / (2I^2R) = 0.5 \) or \( 50\% \).
Correct Answer: B

Step 1: Parallel \( R_2, R_3 \Rightarrow R_{23} = \displaystyle\frac{6 \times 3}{6+3} = 2 \text{ \(\Omega\)} \).
Step 2: Series with \( R_1 \Rightarrow R_{123} = 4 + 2 = 6 \text{ \(\Omega\)} \).
Step 3: Parallel with \( R_4 \Rightarrow R_{eq} = \displaystyle\frac{6 \times 6}{6+6} = 3 \text{ \(\Omega\)} \).
Correct Answer: B

\( I_{total} = \displaystyle\frac{12 \text{ V}}{3 \text{ \(\Omega\)}} = 4 \text{ A} \). The current splits equally between the two main branches (\( R_{123}=6 \text{ \(\Omega\)} \) and \( R_4=6 \text{ \(\Omega\)} \)), so \( 2 \text{ A} \) flows through \( R_1 \).
Voltage drop across \( R_1 = 2 \text{ A} \times 4 \text{ \(\Omega\)} = 8 \text{ V} \). Remaining voltage for \( R_2 \) and \( R_3 = 12 – 8 = 4 \text{ V} \).
Power in \( R_2 = \displaystyle\frac{V^2}{R_2} = \displaystyle\frac{4^2}{6} = \displaystyle\frac{16}{6} = 2.67 \text{ W} \).
Correct Answer: C

A balanced bridge (like a Wheatstone bridge) means the ratio of resistances in the two branches is equal, resulting in identical potentials at the junction points. Thus, the potential difference \( \Delta V \) between them is zero.
Correct Answer: C

By shorting a resistor, you provide a path with zero resistance. This effectively removes that resistor from the network, decreasing the overall \( R_{eq} \). Since \( I = \displaystyle\frac{V}{R_{eq}} \), the total current increases.
Correct Answer: D

Original \( R_{eq} = 2R \). After adding parallel resistor: \( R_{eq}’ = R + \left(\displaystyle\frac{R \times R}{R+R}\right) = 1.5R \). Since resistance decreased, total current increases, and since \( P = \displaystyle\frac{V^2}{R_{eq}} \), total power increases.
Correct Answer: A

Current flows from higher potential to lower potential. Traversing a resistor in the direction of the current results in a potential drop, denoted by \( -IR \).
Correct Answer: B

The positive terminal is at a higher potential than the negative terminal by an amount equal to the EMF (\( \varepsilon \)). Moving from negative to positive is a potential gain.
Correct Answer: B

Loop equation: \( +10 – I(2) – I(3) – 5 = 0 \Rightarrow 5 = 5I \Rightarrow I = 1 \text{ A} \).
Correct Answer: B

\( V_B = V_A – IR – \varepsilon = 12 – (3 \times 2) – 6 = 12 – 6 – 6 = 0 \text{ V} \).
Correct Answer: A

The electric force is a conservative force. This means the work done in moving a charge around a closed path is zero, which implies the total change in potential energy (and thus potential) is zero.
Correct Answer: B

\( \sum I_{in} = \sum I_{out} \). If we assume Wire 3 is “out”: \( 5 = 2 + I_3 \Rightarrow I_3 = 3 \text{ A} \). Positive result confirms the “away” direction.
Correct Answer: B

The Equation of Continuity states that the mass flow rate into a volume equals the flow rate out, assuming no accumulation. This is exactly analogous to the flow of charge (current) at a junction.
Correct Answer: C

In Kirchhoff’s analysis, we assume directions for currents. A negative numerical result simply means the physical flow of charge is opposite to the direction initially chosen for the calculation.
Correct Answer: B

\( \text{In: } 2 + 4 = 6 \text{ A} \). \( \text{Out: } 1 + 3 = 4 \text{ A} \). To balance, \( I_5 \) must be \( 2 \text{ A} \) leaving.
Correct Answer: B

Current is defined as the rate of flow of charge. If charge were accumulating at the junction, the net current would not be zero. For steady-state DC circuits, we assume \( \displaystyle\frac{dq}{dt} = 0 \) at the node itself.
Correct Answer: B

At \( t=0 \), the uncharged capacitor has zero voltage (\( V_c = q/C = 0 \)). It acts like a short circuit (a simple wire). Thus, the entire battery voltage drops across the resistor: \( I = \displaystyle\frac{\varepsilon}{R} \).
Correct Answer: B

As \( t \rightarrow \infty \), the capacitor is fully charged and current stops (\( I = 0 \)). The potential difference across the capacitor then equals the battery EMF. Thus, \( q = C V_c = C \varepsilon \).
Correct Answer: C

Discharge formula: \( q(t) = Q_0 e^{-t/\tau} \). At \( t = \tau \), \( q = Q_0 e^{-1} \approx 0.368 Q_0 \), which is roughly \( 37\% \).
Correct Answer: B

\( P_R = I^2 R \). Since current \( I(t) = \left(\displaystyle\frac{\varepsilon}{R}\right)e^{-t/\tau} \) decays exponentially, the power \( P_R \propto e^{-2t/\tau} \) also decays exponentially to zero.
Correct Answer: C

\( \tau = RC \). If \( R \rightarrow 2R \) and \( C \rightarrow 2C \), then \( \tau’ = (2R)(2C) = 4RC = 4\tau \).
Correct Answer: C