Question

A capacitor is charged to a potential of 12.0 V and then connected to a voltmeter having an

internal resistance of 3.40 MΩ. After a time of 4.00 s the voltmeter reads 3.0 V. What are

(a) the capacitance and (b) the time constant of the circuit?

Answer 1

We know that

$Q=q_0e^\frac{-t}{RC}$

Q=CV

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

$V=12\times e^\frac{-4}{3.40\times10^6\times C}$

(2)

Time constant

$\tau =RC$

$\tau=3.40\times10^6\times C$