Question

Consider the circuit shown in the figure. The potential difference between points A and B is: \includegraphics[width=0.4\linewidth]{872image.png}

• A. $6 \, \text{V}$
• B. $8 \, \text{V}$
• C. $9 \, \text{V}$
• D. $12 \, \text{V}$

Hint:

By using Kirchhoff’s Voltage Law and Ohm’s law, you can calculate the potential difference in a simple series circuit with resistors and voltage sources.

Answer 1

The Correct Option is B

We will apply Kirchhoff’s Voltage Law (KVL) to find the potential difference between points A and B. According to KVL, the sum of the potential differences around any closed loop must be zero. The resistors and voltage sources are arranged as follows: - There is a 12 V source and a 1 \(\Omega\) resistor in series. - A 0.5 \(\Omega\) resistor is in series with a 6 V source. First, calculate the current flowing through the circuit. The total resistance in the circuit is: \[ R_{\text{total}} = 1 \, \Omega + 0.5 \, \Omega = 1.5 \, \Omega \] Now, calculate the total voltage in the circuit: \[ V_{\text{total}} = 12 \, \text{V} - 6 \, \text{V} = 6 \, \text{V} \] Using Ohm's Law, the current \(I\) flowing through the circuit is: \[ I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{6}{1.5} = 4 \, \text{A} \] Now, calculate the potential drop across the 0.5 \(\Omega\) resistor: \[ V_{\text{drop}} = I \times 0.5 = 4 \times 0.5 = 2 \, \text{V} \] Thus, the potential difference between points A and B is: \[ \text{Potential difference} = 6 \, \text{V} + 2 \, \text{V} = 8 \, \text{V} \] Thus, the correct answer is: \[ \text{(B) } 8 \, \text{V}. \]