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 3.20 M? . After a time of 3.90 s the voltmeter reads 2.8 V

Problem 19.60 Part A A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.20 MS2 . After a time of 3.90 s the voltmeter reads 2.8 V . What is the capacitance of the circuit? Express your answer in farads to three significant figures. c= Submit My Answers Give Up Part B What is the time constant of the circuit? Express your answer in seconds to three significant figures. Submit My Answers Give Up

Explanation / Answer

V = Vo*e^-t/RC

Vo = 12 V

V = 2.8 V after 3.9 s

2.8 = 12 * e^-3.9/RC

(2.8/12) = e^-3.9/RC

take natural log both side

-1.455 = -3.9/RC

given R = 3.2 M ohm = 3.2 x 10^6 ohm

C = 3.9/(1.455 * 3.2 x 10^6)

C = 8.37 x 10^-7 F

part b )

time constant = t = RC

t = 2.68 s

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