Question

A capacitor 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?

Answer 1

(A). The capacitance of the circuit is
`2.31*10^(-6)\ F`.

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

Step-by-step 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

`V_(C)=V_(0)(e^{(-t)/(RC)})`

Put the value into the formula

`3.2=12.0e^{(-4.10)/(3.10*10^(6)C)}`

`e^{(-4.10)/(3.10*10^(6)C)}=0.267`

`(-4.10)/(3.10*10^(6)C)=log 0.267`

`C=(-4.10)/(3.10*10^(6)* log0.267)`

`C=0.000002306\ F`

`C=2.31*10^(-6)\ F`

The capacitance of the circuit is
`2.31*10^(-6)\ F`

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

Using formula of time constant

`\tau=R* C`

Put the value into the formula

`\tau=3.10*10^(6)*2.31*10^(-6)`

`\tau=7.16\ sec`

The time constant of the circuit is 7.16 sec.

Hence, (A). The capacitance of the circuit is
`2.31*10^(-6)\ F`.

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