1). Assume the resistance values are R1 = 2,300 %u03A9 , R2 = 1,200 %u03A9 , R3

Question

1). Assume the resistance values are R1 = 2,300%u03A9, R2 = 1,200%u03A9, R3 = 4,600%u03A9, and R4 = 6,200%u03A9, and the battery emf is emf2 = 3.0 V.   

In the figure below, what is the current through the battery and resistor R1? Hint: First redraw the circuit so that the parallel and series resistor combinations are more obvious.

2). In the circuit in the figure below, assume the resistance values are R1 = 1,800%u03A9 and R2 = 2,300%u03A9, and with capacitance C1 = 55 %u03BCF. The emf is = 5.5 V. The switches are labeled S. The capacitor is initially uncharged, and both switches are open. What is the current through the battery the instant after switch S1 is closed?

Explanation / Answer

1) Req = R1 + (R2||(R3||R4))

=> Req = 2300+ R2|| (4600*6200/(4600+6200)) = 2300+ 1200|| 2640.74074074 = 2300+825.072324 = 3125.072324 ohms

thus current in R1 = current through battery = V/Req = 3/3125.072324 = 9.5997778*10^-4 A

2)

current (at just closed instant the capacitor is shorted) = 5.5/(1800+2300) = 0.0013414634 A

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