Question

In the circuit shown, when the current '\(i\)' is \(3\text{A}\) and increasing at the rate of \(1\text{A/s}\) the measurement of the potential difference between A and B is \(12 \text{V}\). But when the same current \(3\text{A}\) is decreasing at the rate of \(1\text{A/s}\), the measured potential difference \(V_{AB}\) between A and B is \(6\text{V}\). The value of 'R' in the circuit is

• A. \(3 \Omega\)
• B. \(4 \Omega\)
• C. \(6 \Omega\)
• D. \(8 \Omega\)

Hint:

The voltage across R and L together is iR plus L di/dt, with the sign of di/dt flipping.

Answer 1

The Correct Option is A

Step 1: Eliminate $L$:
Adding the equations cancels the $L\frac{di}{dt}$ terms, which have opposite signs: $12 + 6 = 2 \times 3R$.

Step 2: Result:
$R = \frac{18}{6} = 3\ \Omega$.

Final Answer:
$R = 3\ \Omega$, option (A). \[ \boxed{3\ \Omega} \]