Acapacitor is charged to a potential of 12.0 vand is then connected to a voltmeter having an internal resistance of 3.10 mω. after a time of 4.10 sthe voltmeter reads 3.2 v.
part a: what is the capacitance of the circuit?
part b: what is the time constant of the circuit?

(A). The capacitance of the circuit is .

(B). The time constant of the circuit is 7.16 sec.

Explanation:

Given that,

Potential = 12.0 V

Internal resistance = 3.10 MΩ

Time = 4.10 s

Voltage = 3.2 V

(A). We need to calculate the capacitance of the circuit

Using formula of capacitance

Put the value into the formula

The capacitance of the circuit is

(B). We need to calculate the time constant of the circuit

Using formula of time constant

Put the value into the formula

The time constant of the circuit is 7.16 sec.

Hence, (A). The capacitance of the circuit is .

(B). The time constant of the circuit is 7.16 sec.