A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 53.0 Ω, R3 = 91.0 Ω and R4 = 90.0 Ω. The capacitance is C = 55.0 μF and the battery voltage is V = 12.0 V. The positive terminal of the battery is indicated with a + sign. The switch has been open for a long time when at time t = 0, the switch is closed. What is I4(0), the magnitude of the current through the resistor R4 just after the switch is closed?
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Question

A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 53.0 Ω, R3 = 91.0 Ω and R4 = 90.0 Ω. The capacitance is C = 55.0 μF and the battery voltage is V = 12.0 V. The positive terminal of the battery is indicated with a + sign. The switch has been open for a long time when at time t = 0, the switch is closed. What is I4(0), the magnitude of the current through the resistor R4 just after the switch is closed?
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anonymous · Answer

Slightly different than last time. This time I'm pretty sure i need to use kirchhoff's laws but i can't get the right unknowns.

anonymous · Answer

Since the max current if flowing the instant the switch is closed acting like the capacitor isn't there. short it out and combine the resistance into one and use Ohms law since the current leaving the battery is the same as through R4.

anonymous · Answer

How about this to go along with it? What is Q(∞), the charge on the capacitor after the switch has been closed for a very long time?

anonymous · Answer

After a long time no current flows through R2 because C acts as an open circuit so all the voltage drop across R3 appears across C.  
Q=C*V*(R3/(R1+R3+R4))