The diagram below depicts an RC -circuit where C1 = 7.50 muF, C2 = 1.95 mu F, R1

Question

The diagram below depicts an RC -circuit where C1 = 7.50 muF, C2 = 1.95 mu F, R1 = 28.0 Ohm, R2 = 19.5 Ohm, R3 = 10.0 Ohm, and 1 = 1.50 Volts . The current in the diagram reflect the situation some finite time after the switch has been closed. Note that when a capacitor is charging or discharging , it carries with it a voltage Vc(t) equaling the amount of charge on it at that particular moment in time divided by its capacitance . In other words, Vc(t) = Q(t)/c. What is the current the battery provides to the circuit immediately after the switch S has been closed?

Explanation / Answer

assumption:

before switch S is closed, the capacitors are completely discharged.

solution:

when the switch is just closed, the capacitors will behave like short circuit.

then R2 and R3 will be in series.

equivalent resistance=R2+R3=19.5+10=29.5 ohms

it is in parallel with R1.

equivalent resistance=29.5*R1/(29.5+R1)

=29.5*28/(29.5+28)
=14.365 ohms

then current supplied by the emf source=E1/total resistance

=1.5/14.365

=0.10422 A

=104.22 mA

hence the battery provides 104.22 mA immediately after the switch is closed.

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