Question 16.6: Three capacitors are combined as in Figure 16.20a. (a) What'...
Three capacitors are combined as in Figure 16.20a. (a) What’s the equivalent capacitance of this combination? (b) What s the charge on each capacitor when the combination is connected across a 14-V battery?
ORGANIZE AND PLAN This problem involves a network of capacitors, so it’s necessary to apply both parallel and series rules. The parallel rule will give the equivalent capacitance of the combination of and C_{1} \text { and } C_{2}. Call this C_{4}=C_{1}+C_{2} . \text { Next, redraw the circuit, with } C_{4} replacing C_{1} \text { and } C_{2} \text { (Figure } 16.20 b \text { ). }
\text { Now } C_{4} \text { and } C_{3} are in series, so the equivalent capacitance follows from
\frac{1}{C_{ s }}=\frac{1}{C_{4}}+\frac{1}{C_{3}}.
To find the charge on each capacitor, we’ll then use Q = CV.
(a) For the parallel combination,
C_{4}=C_{1}+C_{2}=1.0 \mu F +4.0 \mu F =5.0 \mu F.
\text { Now for the series combination of } C_{4}=5.0 \mu F \text { and } C_{3}=2.0 \mu F \text { : }\frac{1}{C_{ s }}=\frac{1}{C_{4}}+\frac{1}{C_{3}}=\frac{1}{5.0 \mu F }+\frac{1}{2.0 \mu F }=\frac{7}{10 \mu F }.
Thus the equivalent capacitance for the three-capacitor network is C=10 / 7 \mu F \approx 1.43 \mu F.
(b) Consider first the charge on the equivalent capacitance. With the 14-V battery across the combination, Q=C V=(10 / 7 \mu F ) (14 V )=20 \mu F \cdot V =20 \mu C \text {. This } C \text { actually consists of the } 2-\mu F \text { capacitor } C_{3} \text { in series with the parallel combination } C_{4} . Since the
charge on series capacitors is the same Q_{3}=20 \mu C \text { and } Q_{4}=20 \mu C .
So we’ve got our answer for the charge on C_{3} \text {, but for } C_{1} \text { and } C_{2} \text { we're } first going to need the potential difference. For the parallel combination C_{4}.
V_{4}=\frac{Q_{4}}{C_{4}}=\frac{20 \mu C }{5.0 \mu F }=4.0 V.
The potential difference across each of the individual parallel capacitors is the same as across the combination, so the charges are
Q_{1}=C_{1} V_{1}=(1.0 \mu F )(4.0 V )=4 \mu C.
Q_{2}=C_{2} V_{2}=(4.0 \mu F )(4.0 V )=16 \mu C.
REFLECT Do these answers seem about right? It’s not as easy to tell as in the case of a single series or parallel combination. The equivalent capacitance is less than 2.0-μF the series capacitor, so that much is as it should be. Note again how we used C = Q/V, alternately solving for Q or V, to burrow down into the circuit and end up with the charges on the individual capacitors.