Question

In the circuit shown in the following figure both batteries have insignificant internal resistance and the idealized ammeter reads 1.00 A in the direction shown. (Assume R_1 = 47.0 ohm.) A 4.55-muF capacitor that is initially uncharged is connected in series with a 7.90-kohm resistor and an emf source with e = 120 V and negligible internal resistance. Just after the circuit is completed, what is the voltage drop across the capacitor? Just after the circuit is completed, what is the voltage drop across the resistor? Just after the circuit is completed, what is the charge on the capacitor? Just after the circuit is completed, what is the current through the resistor? (e) A long time after the circuit is completed (after many time constants) what are the values of the quantities in parts (a) - (d)?

Answer 1

Say, I₁ flows through R₁ and I₂ through the other branch

then I₂ = 1-I₁ lets write eqn in the loop passing through R₁

I₂ = -0.34 A (0.34A from bottom to top)

Now, apply KVL on loop with R₁ and 15ohm

a) Just after the circuit is completed,Voltage drop will be 0V across capacitor as it cannot be charged in 0 time with finite current.

b) Since Voltage drop across cap = 0, => Voltage drop across resistor = 120V

c) With finite current, charge cannot accumulate immediately, i.e. q=0 immediately after circuit is closed.

d) I_R = V_R/R = 120/7.9 = 15.19 mA (mA since R is in kohm)

e) After a long time, steady state is reached in which, current through capacitor = 0 => current through resistor = 0

=> Voltage drop is only across capacitor, i.e.

V_C = 120V

V_R = 0V

q = CV = 4.55*120 = 546 uC = 0.546 mC

I = 0A