Question

Consider a series circuit containing a resistor of resistance R and a capacitor of capacitance C connected to a source of EMF E with negligible internal resistance. The wires are also assumed to have zero resistance. Initially, the switch is open and the capacitor discharged. (Figure 1)

A Immediately after the switch is closed, what is the voltage across the capacitor?

B Complete previous part(s) 

C Immediately after the switch is closed, what is the direction of the current in the circuit? 

E Eventually, the process approaches a steady state. In that steady state, the charge of the capacitor is not changing. What is the current in the circuit in the steady state? 

F In the steady state, what is the charge of the capacitor? 

G How much work W is done by the voltage source by the time the steady state is reached? 

Answer 1

Initially capacitor is shorted. Hence the voltage across capacitor will be zero. The current emerges from positive terminal of the battery, hence the current will be clockwise. In steady state, the capacitor is fully charged and no current flows through the circuit. In steady-state the voltage across capacitor is equal to voltage across battery.