Question

A capacitor is charged to a potential of 12.0 v and is then connected to a voltmeter having an internal resistance of 3.60 mω . after a time of 3.90 s the voltmeter reads 2.8 v . part a what is the capacitance?

Answer 1

As we know that when a charged capacitor is connected across a resistor then its voltage will decreased

This is known as discharging of capacitor

as we know that the equation of discharging is given as

`V = V_0 e^(-t/\tau)`

here we know that

`V = 2.8 Volts`

`V_0 = 12 volts`

t = 3.90 s

now from above equation we have

`2.8 = 12 e^(-3.90/\tau)`

`ln((2.8)/(12)) = -(3.90)/(\tau)`

as we know that

`\tau = RC`

`1.45 = (3.90)/(RC)`

`1.45 = (3.90)/(3.60* 10^6*C)`

`C = 0.75 * 10^(-6) F`

so capacitance is 0.75 micro farad