Question

For a circuit having equivalent resistance, $R=5\text{M}\Omega$ and capacitance $C=2\mu \text{F}$, what time will be elapsed before the power delivered by the battery drops to $25\%$ of its maximum value?

• A. $6.9\text{s}$
• B. $69\text{s}$
• C. $138\text{s}$
• D. $13.8\text{s}$

Answer 1

The correct option is D $13.8\text{s}$

Power delivered by a battery is,
$$P=EI$$Where, $I$ is the current supplied by the battery and $E$ is the emf of the battery.$$\Rightarrow P\propto I$$
So, if power becomes $25\%$, current also becomes $25\%$.

In the case of charging, at any instant time $t$, $$I=I_{0}e{}^{-t/RC}$$
$\Rightarrow \frac{25I_0}{100}=I_0{e}^{-t/RC}$

$\Rightarrow 4=e^{\frac{t}{RC}}$

Taking $\ln$ on both sides,

$\Rightarrow \ln 4=\frac{t}{RC}$

$\Rightarrow 2\ln 2=\frac{t}{RC}$

$\therefore t=2RC\ln 2=2\times 5\times {10}^6\times 2\times {10}^{-6}\times \ln 2\approx 13.8\text{s}$

Hence, option $\left(D\right)$ is correct.