Question

The potential difference \((V_A - V_B)\) between the points A and B in the given figure is

• A. 6 V
• B. -3 V
• C. 9 V
• D. 3 V

Hint:

Moving with current? $-IR$. Moving against current? $+IR$. Crossing a battery from $-$ to $+$? $+V$.

Answer 1

The Correct Option is C

Step 1: Concept
Use Kirchhoff’s Voltage Law (KVL) along the path from point A to point B: $V_A + \sum \Delta V = V_B$.

Step 2: Meaning
As you move in the direction of current, potential drops across resistors ($ -IR $) and increases when moving from negative to positive terminals of a battery ($ +E $).

Step 3: Analysis
Following the path from A to B with current $ I = 2\text{ A} $:
$V_A - (2 \times 2) + 3 - (2 \times 1) - 6 = V_B$
$V_A - 4 + 3 - 2 - 6 = V_B$
$V_A - 9 = V_B \implies V_A - V_B = 9\text{ V}$.

Step 4: Conclusion
The potential difference $V_A - V_B$ is $9\text{ V}$. Final Answer: (C)