In fig. 1 is disconnected from 2 and connected to 3 at time $t=0 .$ The time taken by the power dissipated through $R_{2}$ to decrease to one half the initial power is

Question

Aicharya · Accepted Answer

The current due to discharging of the capacitor decreases with time $t$ as
$I=I_{0} e^{-t / \tau_{2}}$
where $I_{0}=\frac{V}{R_{2}}$ is the maximum current and $\tau_{2}=R_{2} C$. The power dissipated through $R_{2}$ is
$P=I^{2} R_{2}=I_{0}^{2} R_{2} e^{-2 t / \tau_{2}}=P_{0} e^{-2 t / \tau_{2}}$
where $P_{0}=I_{0}^{2} R_{2}$ is the maximum power dissipated. For $P$ to become $P_{0} / 2$, the time $t$ required is given by
$\Rightarrow \frac{P_{0}}{2} =P_{0} e^{-2 t / \tau_{2}}$
$\Rightarrow e^{-2 t / \tau_{2}} =\frac{1}{2}$
$\Rightarrow e^{2 t / \tau_{2}}=2$
$\Rightarrow \frac{2 t}{\tau_{2}}=\ln (2)$
$\Rightarrow t=\frac{\tau_{2} \ln (2)}{2}=\frac{R_{2} C \ln (2)}{2}$
So the correct choice is (a). .

In fig. 1 is disconnected from 2 and connected to 3 at time \(t=0 .\) The time taken by the power dissipated…

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