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When the current in the portion of the circuit shown in the figure is 2A flowing from a to b and is increasing at the rate of 1 A/s, the measured potential difference V_a–V_b = 8V. However when the current is 2A and decreasing at the rate of 1 A/s, the measured potential difference V_a–V_b = 4V. The values of R and L are

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10 ohm and 6 henry respectively

2 ohm and 3 henry respectively

3 ohm and 2 henry respectively

6 ohm and 1 henry respectively

answer is A.

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Detailed Solution

$\begin{array}{l} V_{A} - L (1) - 2 (R) = V_{B}; 8 = L + 2 R - (1) \\ V_{A} + L - 2 (R) = V_{B}; 4 = - L + 2 R - (2) \\ 12 = 4 R \Rightarrow R = 3 Ω, L = 2 H \end{array}$

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