If I take a 1V battery and a 1 farad capacitor and a bulb and connect them in a circuit. Then will after the capacitor gets charged (not fully charged as that is impossible) will the bulb gradually stop glowing? Then what would happen if you change the battery and make it a 2V battery?

The question I want to ask is, that if there is a capacitor in any circuit, does current flow in the circuit stop once the capacitor is charged? Why or why not?

  • 2
    \$\begingroup\$ Mathematically, if there's any resistance R (such as the bulb resistance) the current never quite gets to zero. In reality it gets close enough for most purposes after RC*5 or RC*10 seconds. As Andy says, a real capacitor has leakage (and leakage has noise) so at some point it's effectively fully charged. \$\endgroup\$ Oct 24, 2014 at 10:40
  • \$\begingroup\$ To add --- if the 1V battery could be suddenly changed to 2V then the capacitor would charge first to 1V the current would go close to zero. Then when the voltage stepped to 2V current would flow again as the capacitor charged toward 2V. At the 2V charge level the current would again drop to close to zero. \$\endgroup\$ Oct 24, 2014 at 11:03

3 Answers 3


For a capacitor charge Q = capacitance C multiplied by voltage V.

Rate of change of charge is current and this leads to: -

\$\dfrac{dq}{dt} = C\dfrac{dv}{dt}\$ = current

This quite simply means that a rate of change of voltage gives rise to a current. If the voltage is rising linearly with time, the capacitor will take a constant current and once the voltage stops changing the current is zero.

In a real capacitor there will always be a small leakage current and this can be modelled by a resistor in parallel with the "perfect" capacitor.


If you have a perfectly flat DC voltage source, and an ideal capacitor, then yes, when the capacitor is fully charged then no current will flow.

However, DC voltage sources are seldom perfectly flat, and capacitors are far from ideal.

  • Ripple on the DC supply will pass through the capacitor

The ripple is basically a small AC component superimposed over the DC voltage. This AC component can pass through the capacitor quite happily. That's basically how smoothing capacitors work - they fill up with the DC, which can't then get through to ground, but the ripple can pass through the capacitor to ground, effectively getting rid of it.

  • Capacitors have a certain amount of self-discharge

Charge a capacitor up and it won't stay charged up forever. It will gradually discharge. So a very small amount of current will flow to keep the capacitor filled up.


yes, because it is the steady state(t= infinity) condition and after this capacitor act as a voltage source.

also, as the (impedance)Z=-jXc(reactance), where Xc= 1/WC and W=2(pi)f and f=1/t. so at t=infinity, Z=infinity. hence (current)I=0.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.