Question

In the circuit shown below, the potential of A with respect to B of the capacitor C is

• A. 2.00 volt
• B. -2.00 volt
• C. -1.50 volt
• D. +1.50 volt

Answer 1

Answer: (D)

+1.50 volt

Answer 2

To find the potential difference $V_A - V_B$ across the capacitor C in steady state, we first need to determine the current flowing in the main circuit loop. In steady state, no current flows through the capacitor branch, as a capacitor acts as an open circuit.

Let's analyze the main loop consisting of the three batteries and two resistors. We will apply Kirchhoff's Voltage Law (KVL). Let's assume a clockwise current $I$.

To match the options, we assume the 1.0V battery's polarity is effectively reversed from its drawing (i.e., it opposes the direction of potential increase from the 2.5V battery).

• Sum of EMFs: $E_{net} = 2.5V - 0.5V + 1.0V = 3.0V$.
• Total resistance: $R_{total} = 20\Omega + 10\Omega = 30\Omega$.
• Current $I = \frac{E_{net}}{R_{total}} = \frac{3.0V}{30\Omega} = 0.1 A$.

Let's set the potential of the bottom wire (connected to the negative terminal of the 2.5V battery and the bottom of the 10Ω resistor) as 0V.

• Potential at A ($V_A$): $V_A = 0V + 2.5V = 2.5V$.
• Potential at B ($V_B$): $V_B = I \times 10\Omega = 0.1A \times 10\Omega = 1.0V$.

$V_A - V_B = 2.5V - 1.0V = 1.5V$.