Question

7) A circuit consists of a capacitor (capacitance of 1μF) and a resistor (resistance 1000 Ω) in series, with a battery supplying a potential difference of 12.0 V. At time t=0 a switch is closed to allow current to flow in the circuit for the first time. Remember that the potential difference supplied by the battery ε is not the same as the potential across the capacitor ΔV unless a lot of time has elapsed. Also an RC circuit is not the same as an RL circuit.
A) What is the voltage across the capacitor immediately after the switch is closed?
B) What is the current in the circuit immediately after the switch is closed?
C) What value does the voltage across the capacitor approach after a lot of time?
D) What value does the current in the circuit approach after a lot of time?
E) Express the quantity ??0????????? across the capacitor in terms of the current in the circuit I.

Answer 1

After the switch is closed the capacitor gets charging with time aa the current flows through the circuit.

A) immediately after the switch is closed the charge on capacitor is zero. Thus the potential difference across it. V = Q/C = 0V

B) Thus the capacitor acts as a conducting wire of no resistance immediately after the switch is closed. By ohm's law current, I = E/R = 12/1000. = 0.0012 A = 12 mA

C) After a very long time, the capacitor gets fully charged and no more current passes through the circuit. Thus potential difference across capacitor becomes equal to emf of battery E =12V

D) Current = 0

E) this part is mistyped. Please write down the correct version in the comments so that I can help you with the solution of this part.