## SAT Physics Subject Test

**Chapter 9 ****Electric Potential and Capacitance**

**COMBINATIONS OF CAPACITORS**

When a capacitor charges up, work must be done by an external force (for example a battery). This increases the potential energy stored by the capacitor. The potential energy stored is given by the formula

Capacitors are often arranged in combination in electric circuits. Let’s review two types of arrangements: the parallel combination and the series combination.

A collection of capacitors are said to be in **parallel** if they all share the same potential difference. The following diagram shows two capacitors wired in parallel.

The top plates are connected by a wire and form a single equipotential; the same is true for the bottom plates. Therefore, the potential difference across one capacitor is the same as the potential difference across the other capacitor.

If we want to find the capacitance of a *single* capacitor that would perform the same function as this combination, and if the capacitances are *C*_{1} and *C*_{2}, then the charge on the first capacitor is *Q*_{1} = *C*_{1}Δ*V* and the charge on the second capacitor is *Q*_{2} = *C*_{2}Δ*V*. The total charge on the combination is *Q*_{1} + *Q*_{2 }, so the equivalent capacitance, *C** _{P}*, must be

so

*C*_{p} = *C*_{1} + *C*_{2}

which can be extended to more than 2 capacitors. So the **equivalent** capacitance of a collection of capacitors in parallel is found by adding the individual capacitances.

A collection of capacitors are said to be in **series** if they all share the same charge magnitude. The following diagram shows two capacitors wired in series.

When a potential difference is applied, as shown, negative charge will be deposited on the bottom plate of the bottom capacitor; this will push an equal amount of negative charge away from the top plate of the bottom capacitor toward the bottom plate of the top capacitor. When the system has reached equilibrium, the charges on all the plates will have the same magnitude.

If the top and bottom capacitors have capacitances of *C*_{1} and *C*_{2}, respectively, then the potential difference across the top capacitor is Δ*V*_{1} = *Q*/*C*_{1}, and the potential difference across the bottom capacitor is Δ*V*_{2} = *Q*/*C*_{2}. The total potential difference across the combination is Δ *V*_{1} + Δ *V*_{2}, which equals Δ *V*. Therefore, the equivalent capacitance, *C** _{S}*, must be

This can be written as

which can also be extended to more than 2 capacitators. As you can see, the *reciprocal* of the capacitance of a collection of capacitors in series is found by adding the reciprocals of the individual capacitances.

8. Given that *C*_{1} = 2 µF, *C*_{2} = 4 µF, and *C*_{3} = 6 µF, calculate the equivalent capacitance for the following combination:

Here’s How to Crack It

Notice that *C*_{2} and *C*_{3} are in series, and they are in parallel with *C*_{1}. That is, the capacitor equivalent to the series combination of *C*_{2} and *C*_{3} (which we’ll call *C*_{2–3}) is in parallel with *C*_{1}. We can represent this as follows:

So the first step is to find *C*_{2–3}.

Now this is in parallel with *C*_{1}, so the overall equivalent capacitance (*C*_{1–2–3}) is

Substituting in the given numerical values, we get