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

$2.8 = 12 e^{-3.90/\tau}$

$ln(\frac{2.8}{12}) = -\frac{3.90}{\tau}$

as we know that

$\tau = RC$

$1.45 = \frac{3.90}{RC}$

$1.45 = \frac{3.90}{3.60* 10^6*C}$

$C = 0.75 * 10^{-6} F$

so capacitance is 0.75 micro farad