Question

A capacitor of capacitance 10 μF is connected to a battery of emf 2 V. It is found that it takes 50 ms for the charge of the capacitor to become 12.6 μC. Find the resistance of the circuit.

Answer 1

Given:
Capacitance of the capacitor, C = 10 μF = 10⁻⁵ F
Emf of the battery, E= 2 V
Time taken to charge the capacitor completely, t = 50 ms = 5 × 10⁻² s
The charge growth across a capacitor,
q = $Q{e}^{-\frac{t}{RC}}$
Q = CE = 10⁻⁵ × 2 C
and q = 12.6 × 10⁻⁶ C
$$\begin{gathered} \Rightarrow 12.6\times {10}^{-6}=2\times {10}^{-5}{\mathrm{e}}^{-\frac{5\times {10}^{-2}}{R\times {10}^{-5}}} \\ \Rightarrow \frac{12.6\times {10}^{-6}}{2\times {10}^{-5}}={\mathrm{e}}^{-\frac{5\times {10}^3}{R}} \\ \Rightarrow 0.63={\mathrm{e}}^{-\frac{5\times {10}^3}{R}} \\ \Rightarrow \log 0.63=-\frac{5\times {10}^3}{R} \\ \Rightarrow R=\frac{-5\times {10}^3}{-0.2}=2.5\mathrm{k\Omega } \end{gathered}$$