Consider three capacitors C1, C2, and C3 and a battery. If...

Consider three capacitors $C_{1}, C_{2},$ and $C_{3}$ and a battery. If only $C_{1}$ is connected to the battery, the charge on $C_{1}$ is 30.8$\mu \mathrm{C}$ . Now $C_{1}$ is disconnected, discharged, and connected in series with $C_{2} .$ When the series combination of $C_{2}$ and $C_{1}$ is connected across the battery, the charge on $C_{1}$ is 23.1$\mu \mathrm{C}$ . The circuit is disconnected, and both capacitors are discharged. Next, $C_{3}, C_{1},$ and the battery are connected in series, resulting in a charge on $C_{1}$ of 25.2$\mu \mathrm{C}$ . If, after being disconnected and discharged, $C_{1}, C_{2},$ and $C_{3}$ , are connected in series with one another and with the battery, what is the charge on $C_{1} ?$

Key Concepts

Capacitance and Charge Relationship

This concept explains that the charge stored on a capacitor is directly proportional to the voltage across it, expressed by the equation Q = C × V. It is fundamental to understanding how capacitors operate in circuits, as it relates the physical property of capacitance to the electrical quantities of charge and voltage.

Capacitors in Series and Equivalent Capacitance

When capacitors are connected in series, they all share the same charge regardless of their individual capacitances. The overall or equivalent capacitance of the series connection is given by the reciprocal of the sum of the reciprocals of the individual capacitances. This concept is important because it determines how the total voltage supplied by a battery is divided among the capacitors.

Voltage Distribution in Series Capacitive Circuits

In a series arrangement of capacitors, the total voltage supplied by the battery is distributed across the capacitors in inverse proportion to their capacitances. This means that a smaller capacitor (with less capacitance) will have a higher voltage drop across it compared to a larger one, even though the same charge flows through each. Understanding this concept is key to analyzing and designing circuits involving series-connected capacitors.

Transcript

00:01 Well, this is a really interesting problem, and it's a bit tricky to see how to get to the solution at first.

00:08 But let's try.

00:13 So what we're looking for at the end is the charge on capacitor c1, and luckily they're all combined in series for all the parts.

00:25 So that means the charge on any capacitor is going to be the same, and we can figure out the charge on any individual capacitor by taking the entire combined capacitance.

00:35 The equivalent capacitance, multiplying it by the whole voltage, and that gives us the charge on all of the capacitors that are equal.

00:46 So for the final combination, we're combining three capacitors in series, and the way to do that is like this, c1, c2, c3, and then you have to multiply in all permutations.

01:01 So c1, c2, c2, c3, c2, or c1, c1, or c1, i guess that's just combinations.

01:13 And then multiply by the voltage, and this is going to give us the final cue that we're looking for.

01:20 So now let's walk through what the problem gives us.

01:26 We know that when c1 is connected to the battery alone, we have a charge of 30 .8 microculems.

01:36 And when c1 and c2 are combined in series, that's how we combine them in series, multiply by the voltage, we have a charge of 23 .1.

01:48 But we already know one of these terms.

01:50 If we stick the voltage right there, we have c1 times v, which is 30 .8.

01:58 So we can substitute that in.

02:03 That equals 23 .1.

02:06 Okay, now i'll multiply the denominator.

02:11 So 23 .1c1 plus 23 .1c2.

02:17 Now let me combine the terms with c2.