Capacitor Current

Amy connects a pair of metal plates to an electric circuit made up of a battery, a switch, and a resistor. The metal plates are separated by a small distance by air.

What happens when she closes the switch?

Constant current flows in the circuit Current flows for a short while, and then drops to zero The circuit is open, so no current flows

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Steven Chase
Nov 6, 2017

Capacitor differential equation:

i = C d V d t i = C \frac{dV}{dt}

Here, the voltage is essentially a step, so the current in the beginning is either infinite or undefined. However, in reality, there will be some resistance in the circuit, as well as some parasitic inductance. These additional elements prevent the infinities from arising. In DC steady-state, the derivative of the voltage across the capacitor is zero by definition, and so is the current.

Peyton Bell
Nov 6, 2017

After Amy closes the switch, current will flow through the circuit until the capacitor (the metal plates) becomes fully charged. As the charge on the plates increases, resistance of the capacitor also increases, and therefore the resistance of the circuit increases, causing overall current to decrease. After some time, the resistance in the capacitor will be so great that current stops flowing altogether.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...