A 12.2? ? F capacitor is connected through a 0.895?M?resistor to a constant pote

Question

A 12.2??F capacitor is connected through a 0.895?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,q10,q20,q100 (________________)

Compute the charging currents at the same instants.

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

i0,i5,i10,i20,i100 (_____________)

Explanation / Answer

time constant is

T = RC = 0.895 * 10 ^6 ohm (12.2 * 10^-6 F) = 10.919 s

final charge is Qf = CV = (12.2 * 10^-6 F) (60 V) = 7.32 * 10 ^-4 C

the charge on the capacitor after time t is

Q = Qf ( 1- e^ -t/T)

when t = 0

Q = Qf ( 1- e^ -0/T)

Q = 0

t = 5.0 s

Q = 7.32 * 10 ^-4 C( 1- e^ -5 s/10.919 s)

=2.68 * 10 ^-4 C

when t = 10 s

Q = 7.32 * 10 ^-4 C( 1- e^ -10 s/10.919 s)

= 4.39 * 10 ^-4 C

when t = 20 s

Q = 7.32 * 10 ^-4 C( 1- e^ -20 s/10.919 s)

= 6.14 * 10 ^-4 C

when t = 100 s

Q = 7.32 * 10 ^-4 C( 1- e^ -100 s/10.919 s)

= 7.319 * 10 ^-4 C

-------------------------------------------------------------------------------------------

the growth of current is

I= I0 ( e^-t/T)

I0 = V/R = 60 V/0.895 * 10 ^6 ohm = 67 * 10 ^-6 A

when t= 0

I= 67 * 10 ^-6 A ( e^-0/T) = 67 * 10 ^-6 A

when t= 5

I= 67 * 10 ^-6 A ( e^-5/10.919 s) = 42.38 * 10 ^-6 A

when t= 10

I= 67 * 10 ^-6 A ( e^-10 s/10.919 s) = 26.81 * 10 ^-6 A

when t= 20

I= 67 * 10 ^-6 A ( e^-20 s/10.919 s) = 10.72 * 10 ^-6 A

when t= 100 s

I= 67 * 10 ^-6 A ( e^-100 s/10.919 s) = 7.05 * 10 ^-9A

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