A 12.8? ? F capacitor is connected through a 0.890?M? resistor to a constant pot

Question

A 12.8??F capacitor is connected through a 0.890?M? resistor to a constant potential difference of 60.0 V.

A)

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.

B)

Compute the charging currents at the same instants.

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

Explanation / Answer

The charge = Q = CV

C = 12.8 uF

tau = RC = (0.890)(12.8) = 11.392 sec

To find the charge Q(t), we need to find V(t) using the basic equation:

V(t) = (60 volts)(1 - 1/e to the t/tau)


For example, at t = 5.0 sec:

- t/tau = - 5.0/11.392

e to the -0.4545 =1.54, and plugging these into the basic equation above:

V = (60.0)(1 - 1/1.54) = 60(0.42) = 21.03 volts

Q = CV = (12.8 uF)(21.03 V) = 269.18 uCoulombs

Q=I*t therefore,,current for 5 sec I=269.18 uCoulombs /5s=53.83amp

The charge and current on the capacitor at t= 10, 20, 100 etc can be calculated in the same way.

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