# Why does the capacitor charge through a resistor?

I was working on a question where there was a circuit and the switch was open, there was one capacitor and one resistor. It said: immediately after the switch is closed, what is the current in the circuit, the pd across the resistor and the pd across the capacitor. The answer was: the current was 6A, the pd across the resistor was 12V and the pd across the capacitor is 0V.

I don't understand why the resistor would have have all the voltage and why wouldn't the capacitor have the voltage. Is this because the capacitor is being charged 'through the resistor'? And what does this mean?

I understand that the capacitor has no resistance, and V/I = R so it must not have voltage. But this is not the correct logic because I know the capacitor stores voltage. Can someone tell me why this answer was correct?

And also, why does a capacitor charge through a resistor? Is it because charging a capacitor happens so quickly and so the resistor increases the time is takes to charge so it can be measured via an experiment?

• Phoooebe, here is a KEY concept for this type of circuit: the instant the switch is closed, the capacitor acts like a short circuit. After a "long" time, the capacitor acts like an open circuit. Your question is testing this key concept. Sep 23, 2020 at 20:55

I don't understand why the resistor would have have all the voltage and why wouldn't the capacitor have the voltage. Is this because the capacitor is being charged 'through the resistor'? And what does this mean?

The short answer is the voltage across an ideal capacitor cannot change instantaneously, i.e., in zero time. It therefore looks like a short-circuit (zero voltage) when the switch is first closed. That in turn means all the voltage must appear across the resistor immediately after closing the switch, per Kirchhoff's voltage law.

To fully understand, you need to consider the basic relationship between current and voltage for an ideal capacitor, which is

$$i(t)=C\frac{dv(t)}{dt}$$

or

$$v(t)=\frac{1}{C}\int i(t)dt$$

The second equation tells you that the voltage across an ideal capacitor cannot change in zero time, i.e., $$v_{C}(t)$$ = 0 for $$dt=0$$. It takes time to deliver charge to the capacitor plates so that a voltage appears across the plates. So the instant after the switch closes, the ideal capacitor looks like a short circuit. That means all the voltage drop is across the resistor the instant after the switch closes.

Then, as net charge builds up on the capacitor plates the current goes down, eventually becoming zero. From the first equation if the current is zero, the voltage across the capacitor is no longer changing in time, i.e., it is fully charged and equals the voltage of the source.

The applicable DC transient equations after closing a switch on a series RC circuit where there is no net charge initially on the capacitor (no initial Voltage) are

$$v_{C}(t)=V(1-e\large^{-\frac{t}{RC}})$$

$$i(t)=\frac{V}{R} {e\large^{-\frac{t}{RC}}}$$

$$v_{r}(t)=i(t)R=V{e\large^{-\frac{t}{RC}}}$$

The first equation shows that at time $$t=0$$, the instant after closing the switch, the voltage across the capacitor is zero. The second equation shows that at $$t=0$$ the current is $$V/R$$. Finally, the third equation shows that the voltage across the resistor is $$V$$ at $$t=0$$.

After a long time ($$t$$ = infinity), the voltage across the capacitor is $$V$$ (first equation), current in the circuit is zero (second equation), and the voltage across the resistor is zero (third equation).

Hope this helps.

• Yes, I've never looked at it this way. Thanks! Sep 24, 2020 at 6:04

The potential drop across a capacitor is directly proportional to the excess charge stored on each plate of the capacitor: $$V_C=q/C$$. If there capacitor starts off discharged then it has no potential drop across it.

By Kirchoff's loop rule, this means that initially the potential drop across the resistor has to be equal in magnitude to the potential across the power supply (battery).