Question

A capacitor charging circuit consists of a battery, an uncharged 20 μF capacitor, and a 5.6 kΩ resistor. At t = 0 s the switch is closed; 0.15 s later, the current is 0.62 mA .

What is the battery's emf?

Answer 1

Given,

Capacitor, C = 20 $\mu$ F = 20 * 10^-6 F

Resistor, R = 5.6 k$\Omega$ = 5.6 * 10³$\Omega$

The current, I = 0.62 mA = 0.62 * 10^-3 A

At t = 0 sec, the switch is closed.

Time, t = 0.15 sec

I = (V₀ / R) * e^{- t /RC}

0.62 * 10^-3 = (V₀ / 5.6 * 10³) * e^{- 0.15 / (5.6 * 10^3 * 20 * 10^-6)}

0.62 * 10^-3 = (V₀ / 5.6 * 10³) * 0.262

(V₀ / 5.6 * 10³) = 2.37 * 10^-3

V₀ = 13.27 V