A 12.0? ? F capacitor is connected through a 0.890?M? resistor to a constant pot

Question

A 12.0??F capacitor is connected through a 0.890?M? resistor to a constant potential difference of 60.0 V

Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0 s.

Express your answers using two significant figures. Enter your answers numerically separated by commas.

q0, q5.0, q10, q20, q100 = ____C

Compute the charging currents at the same instants.

Express your answers using two significant figures. Enter your answers numerically separated by commas.

i0, i5.0, i10, i20, i100 = _____A

Explanation / Answer

a)

Time Constant

T=RC=0.89*12=10.68 s

Maximum Charge

Qo=CV=60*12=7.2*10-4 C

In a RC Cirucit for charging Capacitor as a function of time is given by

Q=Qo[1-e-t/T]

at t=0

Q=(7.2*10-4)[1-e-0/10.68]=0 C

at t=5 s

Q=(7.2*10-4)[1-e-5/10.68]=2.69*10-4 C

at t=10 s

Q=(7.2*10-4)[1-e-10/10.68]=4.38*10-4 C

at t=20 s

Q=(7.2*10-4)[1-e-20/10.68]=6.09*10-4 C

at t=100 s

Q=(7.2*10-4)[1-e-100/10.68]=7.2*10-4 C

b)

Maximum Current

Io=V/R=60/0.89*106=6.74*10-5 A

In a RC circuit current as a function of time is given by

I=Ioe-t/RC

at t=0

I=(6.74*10-5)e-0/10.68=6.74*10-5A

at t=5 s

I=(6.74*10-5)e-5/10.68=4.22*10-5A

at t=10 s

I=(6.74*10-5)e-10/10.68=2.64*10-5A

at t=20 s

I=(6.74*10-5)e-20/10.68=1.04*10-5A

at t=100 s

I=(6.74*10-5)e-100/10.68=5.78*10-9A

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