## What is the equivalent capacitance for the circuit of the figure?

**What is capacitance**

The capacitance of a material is the amount of charge that can be stored on its surface. This can be expressed in Farads (F) or coulombs (C). The SI unit of capacitance is the coulomb (C).

The SI unit for capacitance is the coulomb (C). It is also expressed in Farads (F).

Materials with a high capacitance store a large amount of charge. This is why capacitors are used to store energy in electronics. Materials with a low capacitance do not have a lot of room to store charge, and are used in applications like batteries.

**Process**

Today we're finding the equivalent capacitance of two capacitors that are wired in series and We can tell these are wired in series because there is no branching in this circuit. There is no option if we are traveling through the circuit. But to pass through c1 and c2 as it go from the high side of the battery to the low side. So part a asks for the equivalent capacitance of this combination and then we get into some deeper questions about the charge on each capacitor.

The voltage drop over each individual capacitor. We'll go ahead and label everything and when we are simplifying circuits. We like to use this notation so just a big arrow and we would like to just keep track of what WE did at each step.

So what we're going to do is simplify a series combination of capacitors to do that we use the formula

Compute that equivalent capacitance and

C1 was 2.5 microfarads keep in mind that's the units here

And C2 is 7.0 microfarads

Then We hit enter and then we say 1 over the previous answer that gives me the reciprocal and we get the equivalent capacitance. In this case 1.84 microfarads,

So let's redraw the circuit with the equivalent capacitance in there so when we're simplifying circuits whether it's a capacitor circuit or a resistor circuit. The end goal is to get all the way down to a single equivalent capacitor or a single equivalent resistor. In this case it only took one step. But we'll start to look at more complex networks of capacitors. As we move on and once you get down to the simplest possible capacitor circuit. You're able to solve for something new and in this case it's the charge on the capacitor. So we can get the charge on the equivalent capacitor.

The strategy here is to use the definition of capacitance that's charge per volt and then solve for Q and that's given by V. So across my single equivalent capacitor my charge is going to be C eq times the potential. Difference across that capacitor We'll just call it v to leave things symbolic for a second and my equivalent capacitance is 1.84 microfarads. My potential difference across this is the same as the potential difference across the battery the high side of the capacitor is connected directly to the positive terminal of the battery. In the low side to the negative terminal so we get a 12volt potential difference.

When We run the numbers on this we get 22.1 micro coulombs of charge on the equivalent capacitor now things get a little bit conceptually. Tricky here because we asked for the charge on each individual capacitor so We have to be very clear about what we have done if we draw these two nodes. In the original circuit what we are doing here is cutting.

Those two capacitors out of the circuit and replacing it with an equivalent capacitor at those two nodes now the total charge on the high side of that equivalent capacitor is the same as the total charge on the high side Of C1. That's the total charge that we see right after that node the total charge on the low side of c2 is going to be the same as the total charge. On the low side of the equivalent capacitor so what happens in a series combination of capacitors. That plus Q on the high side of C1 is going to attract to - Q from the top plate of C2 leaving behind a plus Q on the top plate of c2 so our two original capacitors actually have the same charge magnitude as the equivalent capacitor.

So if we look at the question in part b compute the charge in each capacitor, we've already answered the question it's 22.1 micro coulombs. Same as it is for the equivalent capacitor finally i want to get the voltage drop over each individual capacitor and then verify that it adds up to what it should so what should it add up to.

We are going to call this V1 voltage drop over C1 and then V2 the voltage drop over C2 Well we know the total voltage drop over this capacitor combination is 12 volts. Because at that upper node we are sitting at 12 volts and at the lower node. We are sitting at zero volts so whatever these two drops are V1 and V2 they better add up to 12. Again we appeal to the definition of capacitance that's Q over V. We solve for v that's Q over C and then We can get V1 and V2 so V1 is going to be Q over C1

That's 22.1 micro coulombs over 2.5 microfarads micro stands for 10 to the negative six. Those are going to cancel out. When we run the numbers on this we get 8.84 volts to get V2. We do Q over C2.

That's 22.1 micro coulombs over 7.0 microfarads.We find a voltage drop of 3.16 volts across that capacitor. Finally we are going to check V1 plus V2 that gives me 8.84 plus 3.16 and that is indeed.

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