Capacitance: 5microfarad R1: 10 Ohms , R2:20 Ohms E: 30 Volts Please explain eve

Question

Capacitance: 5microfarad

R1: 10 Ohms , R2:20 Ohms

E: 30 Volts

Please explain every step of the answer.

Consider the RC circuit shown on the right. You do not how the capacitor was charged at the initial time. But, at a precise moment t, you measure the current I that passes over the emf device to be I = 4 A. (a) Find the currents that pass over each resistor at this moment. (b) Find the charge on the plates of the capacitor at this moment. Which plate is positively charged? (c) Find the rate-of-change of the charge on the upper plate, dq_upper/dt, at this moment.

Explanation / Answer

a) The current through the battery = 4A

The current through resistance R1 = E/R1 = 30/10 = 3 A

hence the current through RC circuit will be = 4A - 3A = 1A

b) The voltage drop across R2 = i*R2 = 1*20 = 20 V.

Hence the voltage across capacitor = 30V - 20V = 10 V

The lower plate is positive here and the charge on the plate Q=CV = 5*10 = 50 mu C.

c)

The rate of change of charge is basically the current through the capacitor which is 1 C/s.

dQ/dt = 1 C/s at this moment.

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