For the circuit diagram as Figure 4 of Assignment 7 epsilon = 15 kV; R = 4.5 k O

Question

For the circuit diagram as Figure 4 of Assignment 7 epsilon = 15 kV; R = 4.5 k Ohm, and C = 7800.0 mu F Determine the time constant for the circuit. 3.51 Times 10^(1) s. 2.58 Times^(-1) s 1.61 Times 10^(-3)s 7.51 Times 10^(-6)s 1.54 Times 10^(2) Determine the maximum charge that will appear on the capacitor 1.51 Times 10^(-2)C 2.58 Times 10^(-5)C 2.61 Times 10^(2) C 7.51 Times 10^(-9) C 1.17 Times 10^(2) C Assuming that the capacitor in initially uncharged, how will it take for the capacitor to charge to half the maximum charge? 1.51 Times 10^(-1)s 35.8 s 24.3 s 7.51 Times 10^(-2)s 15.4s

Explanation / Answer

A.

time constant is given by

tau = RC = 4.5*10^3*7800*10^-6 = 35.1 sec

Option A.

B.

maximum charge is given by

Q0 = C0*V0 = 7800*10^-6*15*10^3 = 117 C

Option E.

C.

relation between Q and t ia given by

Q = Q0*(1 - e^(-t/tau))

Q = Q0/2

Q0/2 = Q0*(1 - e^(-t/tau))

e^(-t/tau) = 1/2

-t/tau = ln (1/2)

t = tau*ln 2

t = 35.1*ln 2 = 24.33 sec

Option C.

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