Answer to Question #126832 in Electric Circuits for Deema

Question

2. When a capacitor, charged to a p.d. of 400 V, is connected to a voltmeter having a resistance of 25 M-Ohm, the voltmeter reading is observed to have fallen to 50 V at the end of an interval of 2 minutes. Find the capacitance of the capacitor.  

Answer 1

The capacitor is discharging through the resistor therefore the value of charge is meant to decrease over time.

Therefore, the capacitor discharge formula is should be used in this case

$V=V_0e^{\frac{-t}{RC}}$

Making RC subject of the formula

$RC=\frac{t}{ln\frac{V_0}{V}}$

$RC=\frac{120}{ln\frac {400}{50}}$

RC= 57.08 seconds(This is the time constant)

now divide the time constant by the resistance to get the value of capacitance

$C=\frac{57.08}{25\times 10^{6}}$

Therefore capacitance is $2.2832\times 10^{-6}$ F