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
Show Hint
The voltage across R and L together is iR plus L di/dt, with the sign of di/dt flipping.
Step 1: Understanding the Concept:
The figure shows a resistor \(R\) and an inductor \(L\) in series between A and B, with current \(i\) flowing from A to B. The potential difference is \(V_{AB} = iR + L\frac{di}{dt}\).
Step 2: Key Formula or Approach:
Case 1: \(i = 3\) A, increasing at \(1\) A/s: \(12 = 3R + L(1)\). Case 2: \(i = 3\) A, decreasing at \(1\) A/s: \(6 = 3R - L(1)\).
Step 3: Detailed Explanation:
Add the two equations: \(18 = 6R\).
\[ R = 3\ \Omega \]
Then \(L = 12 - 9 = 3\) H, and the second equation gives \(9 - 3 = 6\), which matches.
Final Answer:
The resistance is \(3\ \Omega\), option (A).
\[ \boxed{3\ \Omega} \]
