Video: Finding the Equivalent Capacitance of a System of Capacitors

In the figure, what is the equivalent capacitance of the circuit between A and B? [A] 10 pF [B] 5.0 pF [C] 25 pF [D] 4.0 pF [E] 2.2 pF.


Video Transcript

In the figure, what is the equivalent capacitance of the circuit between A and B? a) 10 picofarads. b) 5.0 picofarads. c) 25 picofarads. d) 4.0 picofarads. e) 2.2 picofarads.

The question asks us to find an equivalent capacitance. So, to answer this question, we’ll need to address the idea of an equivalent circuit. In an equivalent circuit, we replace many components with a single representative component without changing the behavior of the rest of the circuit. In our problem, we’re looking to replace the three capacitors connected to the battery with a single equivalent capacitor.

Because this is an equivalent circuit, even though the capacitors are no different between A and B, the behavior of the rest of the circuit, such as values for current and voltage, remains unchanged. This makes equivalent circuits a very powerful tool since it is usually much easier to analyze a simple few-component circuit than a complex many-component circuit. So, being able to replace many components with a single component can reduce the problem from a complex one to a simple soluble one.

Every complex circuit is a combination of components connected in one of two ways, series and parallel. Components in series are connected serially, that is, one after another. In this circuit here, capacitors C one and C two are connected in series with a battery. Components in parallel, on the other hand, are both connected across the circuit the same way. In other words, they look like they’re parallel. As in this circuit here. More complex circuits are built up out of these basic building blocks. For example, one of the branches in a parallel combination could itself be a series combination.

Series and parallel each have their own rules for determining the equivalent circuit. For capacitors, specifically, the series equivalent capacitance is the reciprocal of the sum of the reciprocals of the capacitances that are connected in series. So, symbolically, C is equal to one divided by one over C one plus one over C two. For two capacitors, specifically, we can reduce this expression to C one times C two divided by C one plus C two. For more than two capacitors in series though, we have to stick to the sum of the reciprocals formula and in the denominator put one over C one plus one over C two plus one over C three, etcetera.

The equivalent capacitance for a parallel combination, on the other hand, is simply the sum of the capacitances of the capacitors in parallel. C is equal to C one plus C two. And again, for more than two capacitors, we simply extend the sum to C one plus C two plus C three, etcetera. We could’ve derived the series formula by recognizing that the charge on each capacitor in a series is the same. Similarly, we could’ve derived the parallel formula by recognizing that the voltage across each capacitor in parallel is the same.

Nevertheless, it’s a good idea to memorize these forms since they’re the same for all equivalent circuit components. The difference is that resistors and inductors use the direct sum for the series combination and the reciprocal sum for the parallel combination. But the forms are identical. Let’s now use these rules and formulas to find the equivalent capacitance for the problem.

The three capacitors are neither all in series nor are they all in parallel. Instead, the series combination of C two and C three is in parallel with the single capacitor C one. To address this, we’ll take the problem step-by-step and start by working with what we know. So, we’ll replace the series combination of C two and C three with an equivalent capacitor. Since there are only two capacitors here, we can use either of the two series capacitor formulas. We’ll use the simplification for two capacitors.

Plugging in 10 picofarads for both C two and C three, we get 10 picofarads times 10 picofarads divided by 10 picofarads plus 10 picofarads. We replace the denominator with two times 10 picofarads and cancel a factor of 10 picofarads from the numerator and denominator. 10 divided by two is five. So, C is five picofarads. With that, we reduce this circuit to an equivalent circuit with two capacitors connected in parallel. C one, the capacitor from before, and C, the new equivalent capacitor for C two and C three connected in series.

Using the formula for the equivalent capacitance of a parallel combination, we have that the total capacitance is C one plus C. C one is five picofarads and so is C. Five picofarads plus five picofarads is 10 picofarads, which is the total equivalent capacitance of the three capacitors that we started with. And 10 picofarads is choice a. We found this answer by successively applying the rules for series and parallel combinations of capacitors.

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