Question 16.9: EQUIVALENT CAPACITANCE GOAL Solve a complex combination of s...
EQUIVALENT CAPACITANCE
GOAL Solve a complex combination of series and parallel capacitors.
PROBLEM (a) Calculate the equivalent capacitance between a and b for the combination of capacitors shown in Figure 16.22a. All capacitances are in microfarads. (b) If a 12-\mathrm{V} battery is connected across the system between points a and b, find the charge on the 4.0-\mu \mathrm{F} capacitor in the first diagram and the voltage drop across it.
STRATEGY For part (a), use Equations 16.12
C_{eq} = C_1 + C_2 + C_3 + \cdots [16.12]
and 16.15
PE = k_e \frac{q_1 q_2}{r} [16.15]
to reduce the combination step by step, as indicated in the figure. For part (b), to find the charge on the 4.0-\mu \mathrm{F} capacitor, start with Figure 16.22 \mathrm{c}, finding the charge on the 2.0-\mu \mathrm{F} capacitor. This same charge is on each of the 4.0-\mu \mathrm{F} capacitors in the second diagram, by fact 5 \mathrm{E} of the Problem-Solving Strategy. One of these 4.0-\mu \mathrm{F} capacitors in the second diagram is simply the original 4.0-\mu \mathrm{F} capacitor in the first diagram.
(a) Calculate the equivalent capacitance.
Find the equivalent capacitance of the parallel 1.0-\mu \mathrm{F} and 3.0- \mu \mathrm{F} capacitors in Figure 16.22a:
C_{\mathrm{eq}} =C_{1}+C_{2}=1.0 \mu \mathrm{F}+3.0 \mu \mathrm{F}=4.0 \mu \mathrm{F}
Find the equivalent capacitance of the parallel 2.0-\mu \mathrm{F} and 6.0-\mu \mathrm{F} capacitors in Figure 16.22a:
C_{\mathrm{eq}} =C_{1}+C_{2}=2.0 \mu \mathrm{F}+6.0 \mu \mathrm{F}=8.0 \mu \mathrm{F}
Combine the two series 4.0-\mu \mathrm{F} capacitors in Figure 16.22 \mathrm{~b} :
\begin{aligned}\frac{1}{C_{\mathrm{eq}}} &=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{4.0 \mu \mathrm{F}}+\frac{1}{4.0 \mu \mathrm{F}} \\ &=\frac{1}{2.0 \mu \mathrm{F}} \rightarrow C_{\mathrm{eq}}=2.0 \mu \mathrm{F}\end{aligned}
Combine the two series 8.0-\mu \mathrm{F} capacitors in Figure 16.22 \mathrm{~b} :
\begin{aligned}\frac{1}{C_{\mathrm{eq}}} &=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{8.0 \mu \mathrm{F}}+\frac{1}{8.0 \mu \mathrm{F}} \\&=\frac{1}{4.0 \mu \mathrm{F}} \quad \rightarrow \quad C_{\mathrm{eq}}=4.0 \mu \mathrm{F}\end{aligned}
Finally, combine the two parallel capacitors in Figure 16.22c
C_{\mathrm{eq}} =C_{1}+C_{2}=2.0 \mu \mathrm{F}+4.0 \mu \mathrm{F}=6.0 \mu \mathrm{F}
(b) Find the charge on the 4.0-\mu \mathrm{F} capacitor and the voltage drop across it.
Compute the charge on the 2.0-\mu \mathrm{F} capacitor in Figure 16.22c, which is the same as the charge on the 4.0-\mu \mathrm{F} capacitor in Figure 16.22a:
\begin{gathered}C=\frac{Q}{\Delta V} \quad \rightarrow \quad Q=C \Delta V=(2.0 \mu \mathrm{F})(12 \mathrm{~V})=24 \mu \mathrm{C} \end{gathered}
Use the basic capacitance equation to find the voltage drop across the 4.0-\mu \mathrm{F} capacitor in Figure 16.22 \mathrm{a} :
\begin{gathered}C=\frac{Q}{\Delta V} \quad \rightarrow \quad \Delta V=\frac{Q}{C}=\frac{24 \mu \mathrm{C}}{4.0 \mu \mathrm{F}}=6.0 \mathrm{~V}\end{gathered}
REMARKS To find the rest of the charges and voltage drops, it’s just a matter of using C=Q / \Delta V repeatedly, together with facts 5 \mathrm{C} and 5 \mathrm{E} in the Problem-Solving Strategy. The voltage drop across the 4.0-\mu \mathrm{F} capacitor could also have been found by noticing, in Figure 16.22b, that both capacitors had the same value and so by symmetry would split the total drop of 12 volts between them.