00:02
So here we're looking for the voltage across each capacitor.
00:06
And first, let's take the 2 micro -faird capacitor.
00:10
So we know that around this loop, the total voltage drop has to be 0.
00:16
And so if 26 volts are supplied by the battery or whatever is connected to this circuit, 26 volts has to drop at the capacitor, since that's the only other thing in that loop.
00:28
So for the 2 micro -faird capacitor, the voltage drop is going to be 26 volts.
00:37
Now, the other two are going to take a little bit more work, but we can start the same way by looking at this loop.
00:44
26 volts are supplied, and then we need to subtract the voltage from the 3 micro -faird, which i'll call a, and from the 4 -micro -faird capacitor, which i'll call b.
01:01
And the next thing we want to do is write out another expression for the voltage drop across each of these capacitors.
01:09
We know that capacitance is equal to q times v, and so we can write the voltage drop across a is equal to q over the capacitance of a, and the voltage drop over b is equal to the charge over capacitance from b.
01:32
And we can substitute these into this equation up here, and we find the charge over the capacitance of a plus the charge over the capacitance of b is equal to 26 volts.
01:50
Now, we know the capacitance of a and b...