Question

In the circuit shown, the potential difference between A and B is:

• A. 92V
• B. 46V
• C. 30V
• D. 0V

Answer 1

Answer: (A)

92V

Answer 2

The parallel combination of resistors and voltage sources between nodes D and C can be simplified using Millman's theorem. The equivalent resistance is $R_{eq} = \frac{1}{\frac{1}{1\Omega} + \frac{1}{1\Omega} + \frac{1}{1\Omega}} = \frac{1}{3}\Omega$. The equivalent voltage source is $E_{eq} = \frac{\frac{1V}{1\Omega} + \frac{2V}{1\Omega} + \frac{3V}{1\Omega}}{\frac{1}{1\Omega} + \frac{1}{1\Omega} + \frac{1}{1\Omega}} = \frac{6V}{3} = 2V$, directed from D to C.

The circuit then becomes a series combination of a $5\Omega$ resistor, the equivalent $2V$ source and $\frac{1}{3}\Omega$ resistor (both from D to C), and a $10\Omega$ resistor. The total resistance is $R_{total} = 5\Omega + \frac{1}{3}\Omega + 10\Omega = \frac{46}{3}\Omega$.

Let $I$ be the current flowing from A to B. The potential difference $V_A - V_B$ can be expressed as the sum of voltage drops and rises in the loop: $V_A - V_B = (V_A - V_D) + (V_D - V_C) + (V_C - V_B)$ $V_A - V_B = I \times 5\Omega + E_{eq} + I \times 10\Omega$ $V_A - V_B = 15I + 2V$.

Alternatively, $V_A - V_B = I \times R_{total} = I \times \frac{46}{3}\Omega$.

Equating the two expressions for $V_A - V_B$: $15I + 2 = \frac{46}{3}I$ $2 = \left(\frac{46}{3} - 15\right)I = \left(\frac{46 - 45}{3}\right)I = \frac{1}{3}I$ $I = 6A$.

Substituting $I$ back into the equation $V_A - V_B = \frac{46}{3}I$: $V_A - V_B = \frac{46}{3} \times 6 = 46 \times 2 = 92V$.