Video: Capacitance of a Parallel Combination

Three capacitors, with capacitances of ๐ถโ‚ = 1.5 ๐œ‡F, ๐ถโ‚‚ = 5.0 ๐œ‡F, and ๐ถโ‚ƒ = 8.0 ๐œ‡F, are connected in parallel. A 720 V potential difference is applied across the combination. What is the voltage across the ๐ถโ‚ capacitor? What is the charge across the ๐ถโ‚‚ capacitor?

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Video Transcript

Three capacitors, with capacitances of ๐ถ one equals 1.5 microfarads, ๐ถ two equals 5.0 microfarads, and ๐ถ three equals 8.0 microfarads, are connected in parallel. A 720-volt potential difference is applied across the combination. What is the voltage across the ๐ถ one capacitor? What is the charge across the ๐ถ two capacitor?

Weโ€™re told here that these three capacitors are connected in parallel and that they have a potential difference power supply. We may as well go ahead and draw this as a circuit. So here, we have it: capacitors ๐ถ one, ๐ถ two, and ๐ถ three are arranged in parallel in the circuit with a 720-volt potential difference across it.

In our first question, weโ€™re asked to solve for the voltage across the ๐ถ one capacitor, where the values for ๐ถ one, ๐ถ two, and ๐ถ three are all given to us in the problem statement. Interestingly though, for this particular question, we wonโ€™t need to know the value of the ๐ถ one capacitor. To see why, letโ€™s consider this circuit that weโ€™ve drawn.

We know that 720 volts of potential difference is set up across the positive and negative terminals of our cell. Furthermore, in this idealized circuit, the only elements that use up that potential difference are the capacitors ๐ถ one, ๐ถ two, and ๐ถ three. And since they are arranged in parallel, that means the voltage drop across any one of our three capacitors must be equal to the total voltage supplied to the circuit.

There is nowhere else for voltage to drop in the circuit. And so that tells us the answer to this first question. The voltage drop across the ๐ถ one capacitor โ€” weโ€™ll call it ๐‘‰ sub ๐ถ one โ€” is equal to 720 volts, the same for ๐ถ two and ๐ถ three by the way.

Letโ€™s now move on to part two, which asks not for the voltage, but the charge across the ๐ถ two capacitor. To figure out what this charge is, letโ€™s recall a relationship between charge, capacitance, and potential difference.

The capacitance ๐ถ of a capacitor is equal to the charge on the plates divided by the potential difference across them. Applying this relationship to our scenario, we can say ๐‘„ two โ€” the charge across the ๐ถ two capacitor โ€” is equal to that capacitance times the potential across the capacitor.

Now remember we saw in part one that 720 volts is yes, the potential difference across ๐ถ one, but itโ€™s also the potential difference across the other capacitors. That takes care of ๐‘‰ sub ๐ถ two. And then, for ๐ถ two, the capacitance value itself, thatโ€™s told to us in the problem statement; itโ€™s 5.0 microfarads.

When we plug in and then multiply these two terms together, we find a result of 3.6 millicoulombs. Thatโ€™s the amount of charge across the ๐ถ two capacitor.

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