Question

A capacitor of capacitance 10 $\mu F$ is connected to a battery of emf 2 V. It is found that it takes 50 ms for the charge on the capacitor to become 12.6 $\mu C$. Find the resistance of the circuit.

Answer 1

$C=10\mu f={10}^{-5}f,E_{emf}=2V,$

$t=50ms=5\times {10}^{-2}S$

$\text{Since,}q=Q{e}^{\frac{-t}{RC}}$

$Q=CV={10}^{-3}\times 2$

$q=12.6\times {10}^{-6}C$

$\text{Hence}12.6\times {10}^{-6}=2\times {10}^{-5}{e}^{\frac{5\times {10}^{-2}}{R\times {10}^{-5}}}$

$\Rightarrow \frac{12.6\times {10}^{-6}}{2\times {10}^{-5}}=e^{-\frac{5\times {10}^3}{R}}$

$\Rightarrow 0.63=e^{-\frac{5\times {10}^3}{R}}$

$\Rightarrow \text{log}0.63=\frac{5\times {10}^3}{R}$

$\Rightarrow R=\frac{5\times {10}^3}{0.79-2}=4k\Omega$