A capacitor is charged to a potential of 12.0 V and is then connected to a voltm

Question

A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40M? . After a time of 4.10s the voltmeter reads 2.9V . Part A What is the capacitance? C = F SubmitMy AnswersGive Up Part B What is the time constant of the circuit? ? = s SubmitMy AnswersGive Up A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40M? . After a time of 4.10s the voltmeter reads 2.9V . A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40M? . After a time of 4.10s the voltmeter reads 2.9V . A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40M? . After a time of 4.10s the voltmeter reads 2.9V . A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40M? . After a time of 4.10s the voltmeter reads 2.9V . Part A What is the capacitance? C = F SubmitMy AnswersGive Up Part B What is the time constant of the circuit? ? = s SubmitMy AnswersGive Up Part A What is the capacitance? C = F SubmitMy AnswersGive Up Part B What is the time constant of the circuit? ? = s SubmitMy AnswersGive Up Part A What is the capacitance? C = F SubmitMy AnswersGive Up Part B What is the time constant of the circuit? ? = s SubmitMy AnswersGive Up Part A What is the capacitance? C = F SubmitMy AnswersGive Up Part B What is the time constant of the circuit? ? = s SubmitMy AnswersGive Up Part A What is the capacitance? C = F SubmitMy AnswersGive Up Part A What is the capacitance? C = F SubmitMy AnswersGive Up C = F C = F SubmitMy AnswersGive Up Part B What is the time constant of the circuit? ? = s SubmitMy AnswersGive Up Part B What is the time constant of the circuit? ? = s SubmitMy AnswersGive Up ? = s ? = s SubmitMy AnswersGive Up A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40M? . After a time of 4.10s the voltmeter reads 2.9V . Part A What is the capacitance? C = F SubmitMy AnswersGive Up Part B What is the time constant of the circuit? ? = s SubmitMy AnswersGive Up

Explanation / Answer

use charring eqn V   = vo (1-e^-t/Rc)

1-e^-t/RC = 2.9/12 = 0.241

e^-t/RC = 1- 0.241 = 0.758

-t/RC = ln ( 0.758) = -0.27707

RC = 4.1/0.27707 = 14.797

C = 14.797/3.4 *10^6

C = 4.35 uF

b. time constant T = RC

T = 3.4*10^6 * 4.35*10^-6

T = 14.79 secs

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