# Discharging a capacitor conceptual question

When a capacitor is discharging, why is the potential difference across it at any time $$t$$ equal to that of the resistor? Do the two potential difference's not have to add up to the EMF of the supply (the supply is the capacitor, right?) When we charge the capacitor, eventually the potential difference across the capacitor is the EMF and the potential difference across the resistor is $$0$$. Then when we discharge, initially the potential difference across both components is the EMF? How did the resistor gain so much voltage?

The potential difference (pd) across the resistor is the same as the pd across the capacitor since they are wired together, and we model the wires as having zero resistance. When the capacitor is charging, the resistor may or may not be in the circuit, but when it is fully charged there is no current flowing so there is no potential difference across the resistor, because the pd across the resistor is proportional to the current, $$V=IR$$. The pd across the capacitor is proportional to charge $$Q$$, $$V=Q/C$$, where $$C$$ is the capacitance, so if the circuit is open, that is current cannot flow, it retains its charge.

Imagine a series circuit consisting of a charged capacitor $$C$$, potential difference $$\mathcal E$$, a switch and a resistor $$R$$.

The switch is open so there is no current in the circuit and hence the potential difference across the resistor is zero.

How did the resistor gain so much voltage [instantaneously]?

You are asking, how is it that when the switch is closed the potential difference across the resistor instantaneously becomes $$\mathcal E$$ with a current in the circuit of $$\mathcal E/R$$.

In the real world the potential difference across the resister changes from zero to $$\mathcal E$$ as does the current change from zero to $$\mathcal E/R$$ in a finite amount of time which usually is very small compared with the time constant $$CR$$ of the circuit.
The reason is that the circuit has some (parasitic) inductance as the circuit is in the form of a loop, which prevents an instantaneous change in the current.
The inductance is small which means that the change of current from zero to $$\mathcal E/R$$ occurs over a very short interval of time and is often neglected.