A capacitor is connected in the circuit shown with two resistors, A and B Resist

Question

A capacitor is connected in the circuit shown with two resistors, A and B Resistor A has a resistance of 2R, whle resistor B has a resistance of R. The emf of the battery, which has negligible internal resistance, is 12.0 v Immediately 8fter the switch is closed, the capacitor has a potential difference of 8.00 v, with the top plate having higher potential. immediately after the switch is closed, the current through resistor A 4202 xA 1.40 A. At that instant in time, caculate the magnitude of the current in the branch of the circuit that has the capactor ina:

Explanation / Answer

Applying KVL in left side loop,

e - I RA - Vc = 0

12 - (2 R x 1.40) - 8 = 0

R = 1.43 ohm

C and Rb are in paralllel so they will have same voltage drop across them

V_RB = Vc = 8 Volt

R ( IB) = 8

IB = 5.6 A

Applying KCL,

1.40 + Ic - IB = 0

Ic = 5.6 - 1.40

Ic = 4.2 A ........Ans

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