Regarding the given circuit, the correct statement is

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$(V_{a} - V_{b})$ is increasing with time

$(V_{a} - V_{b}) = z e r o$

$(V_{a} - V_{b})$ is decreasing with time

$(V_{a} - V_{b}) = 10 V$

answer is D.

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Detailed Solution

Charge on capacitor $q = 5 (1 - e^{- t / 2})$
Current through inductor $I = 5 (1 - e^{- t / 2})$
Now, $V_{a} + \frac{q}{0.6} - I \times 2 = V_{b}$
$∴ (V_{a} - V_{b}) = 0$

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