Question

A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40 MΩ. After a time of 4.00 s, the voltmeter reads 3.0 V.
A) What are the capacitance?
B) What is the time constant of the circuit?

Answer 1

Answer:

a. 0.849 micro farad

b. 2.89 s

Explanation:

a) V=V0 e^-t/RC

3=12*e^-4/3.4*10^6*C

3/12=e^-4/3.4*10^6*C

-1.3863 =-4/3.4*10^6*C

C=8.49*10^-7 F

=0.849 micro farad

B) time constant= R*C

=3.4*10^6*8.49*10^-7

=2.89 S

Answer 2

a. The capacitance is 0.849 micro farad

b. The  time constant of the circuit is 2.89 s

Calculation of capacitance & time constant:

a)

We know that

V=V0 e^-t/RC

3=12*e^-4/3.4*10^6*C

3/12=e^-4/3.4*10^6*C

-1.3863 =-4/3.4*10^6*C

C=8.49*10^-7 F

=0.849 micro farad

B)

Now

time constant= R*C

=3.4*10^6*8.49*10^-7

=2.89 S

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