Question

A capacitor charging circuit consists of a battery, an uncharged 20 μF capacitor, and a 6.0 kΩ resistor. At t = 0 s the switch is closed; 0.15 s later, the current is 0.46 mA . Part A What is the battery's emf?

Answer 1

Answer:

Battery voltage will be equal to 9.65 volt

Explanation:

We have given capacitance $C=20\mu F=20\times 10^{-6}F$

Resistance $R=6kohm=6000ohm$

Time constant of RC circuit is

$\tau =RC=20\times 10^{-6}\times 6000=0.12sec$

Time is given t = 0.15 sec

Current i = 0.46 mA

Current in RC circuit is given by

$i=\frac{V}{R}e^{\frac{-t}{\tau }}$

$0.00046=\frac{V}{6000}e^{\frac{-0.15}{0.12 }}$

$0.00046=\frac{V}{6000}\times 0.286$

V = 9.65 volt

So battery emf will be equal to 9.65 volt