Deciding voltage of a capacitor in a circuit?

If I have a circuit with a battery #ɛ=3.0 V#, a resistor #R=0.40 Ω# and a capacitor #C= 10 μF#. Will the voltage over the capacitor be #ɛ=3.0 V# if I were to for example use Kirchoff's voltage laws to decide #ΣV=0# over the whole circuit?

1 Answer
Jul 14, 2018

Yes, the voltage across the capacitor would be 3 V shortly after (a few time constants after) making the circuit connections.


I will assume that the R and C are connected in series across the battery terminals.

When this circuit is first connected as described, the voltage across the capacitor will be 0 V. It will quickly, but not instantaneously, increase to a final value of 3.0 V. A way to predict charging time of a series RC circuit uses a calculated factor called "time constant", #tau#.

#tau = R*C = 0.40 Omega*10*10^-6 F = 4.0*10^-6 s#

That means that in one time constant, or #4.0 mus#, the voltage across the capacitor will be 63% or the way to being fully charged to 3 V.

During the short time before the capacitor is fully charged, there is a current flowing thru the resistor, so your application of Kirchoff's voltage laws would need to register a voltage across the resistor. But as I indicated in the previous paragraph, the capacitor would be fully charged to 3 V significantly before 1 s had expired. And therefore, there would be no current flowing and no voltage across the resistor.

Full charge, as measured to the number of significant digits implied by the data, would be achieved before 3 time constants #(12 mu"seconds"#) have expired. Once full charge on the capacitor has been reached, this would be your application of Kirchoff's voltage laws:

#epsilon + i*R + V_C = 0#

#3 V - 0*0.4 Omega - 3 V = 0#

Note: This resistor is small. If it had been 1000 times larger (#400 Omega#), it would take 4.0 ms to reach 63% of full charge. In either case, the capacitor would be fully charged well before 1 second had passed.

I hope this helps,