We are going to see what happens when we charge a capacitor and why it blocks DC.

I am not going to go into all the theory behind capacitor, but lets dig in to the experiment setup.

In the beginning, when the switch S1 is open, lets assume the capacitor is discharged. (i.e. the plates of the capacitor have zero potential across them.)

We are probing the setup at

1. the input of the capacitor C1 and (channel 2)

2. across the resistor R1. (channel 1)

3. We are going to subtract the channel 1 from channel 2 to get the voltage across the capacitor in our Math channel of the scope.

So we know what our setup is. Lets wait no longer and press the switch, S1.

A surge of current equal to about V1/R1 starts to flow and we get v1 (here which is 5v) across the resistor R1; as if the capacitor is a short circuit! At this very point we can see in the Math channel that the voltage across the capacitor is indeed zero. (a signature of a short circuit.)

This current that is now flowing will decrease gradually as the capacitor gets charged, (i.e. the between the the two plates becomes equal to the source voltage (here that is 5v).

Now as the current decreases, charging the capacitor more and more, we see a the voltage across the resistor decrease at the same rate. The current this time becomes,

Soon the capacitor will be fully charged, with the left plate at around 5v (we see it to be around 4.8v) and the right plate at zero, at this point very little current flows and therefore we get very little voltage across the resistor.

To find the time for this to happen, we use the capacitor charging equation.

Our experimental data checks out with our math. No further current will flow, until the capacitor gets discharged.

Hope you like this little, ‘back to the basics’ post. I am waiting to hear back from you.