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

Question

A capacitor is connected in the circuit shown with two resistors, A and B. Resistor A has a resistance of 2R, while resistor B has a resistance of R. The emf of the battery, which has negligible internal resistance, is 12.0 V. Immediately after the switch is closed, the capacitor has a potential difference of 10.0 V, with the top plate having a higher potential, and the current through resistor A is 1.00 A. At that instant in time, calculate the following.

(a) the resistance of resistor A ?

(b) the current through resistor B A

(c) the current through the capacitor A

(d) A long time after the switch is closed, the current through resistor A has reached a stable value. What is that value?

Explanation / Answer

a)

Potential difference across Ra = Va = Total battery Voltage - Voltage across capacitor = 12 - 10 = 2 volts

i = current through Ra = 1 A

Resistance Ra = Va / i = 2/1 = 2 ohm

B)

Vb = Voltage across Rb = Voltage across Capacitor C = 10 Volts

Rb = Ra/2 = 2 /2 = 1 ohm

using ohm's law

ib = current through Rb = Vb/Rb = 10 /1 = 10 A

C)

current in capacitor A = 11

D)

i = E/(Ra + Rb) = 12 / (1 + 2) = 4 A

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