Question 16.8: FOUR CAPACITORS CONNECTED IN SERIES GOAL Find an equivalent ...
FOUR CAPACITORS CONNECTED IN SERIES
GOAL Find an equivalent capacitance of capacitors in series, and the charge and voltage on each capacitor.
PROBLEM Four capacitors are connected in series with a battery, as in Figure 16.21. (a) Calculate the capacitance of the equivalent capacitor. (b) Compute the charge on the 12-\mu \mathrm{F} capacitor. (c) Find the voltage drop across the 12-\mu \mathrm{F} capacitor.
STRATEGY Combine all the capacitors into a single, equivalent capacitor using Equation 16.15.
\frac{1}{C_{eq}} =\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+ \cdots [16.15]
Find the charge on this equivalent capacitor using C=Q / \Delta V. This charge is the same as on the individual capacitors. Use this same equation again to find the voltage drop across the 12-\mu \mathrm{F} capacitor.
(a) Calculate the equivalent capacitance of the series.
Apply Equation 16.15:
\begin{aligned}\frac{1}{C_{eq}} =&\frac{1}{3 \mu \mathrm{F}}+\frac{1}{6.0 \mu \mathrm{F}}+\frac{1}{12 \mu \mathrm{F}}+\frac{1}{24 \mu \mathrm{F}} \\C_{\mathrm{eq}} =& 1.6 \mu \mathrm{F} \end{aligned}
(b) Compute the charge on the 12-\mu \mathrm{F} capacitor.
The desired charge equals the charge on the equivalent capacitor:
Q=C_{\text {eq }} \Delta V=\left(1.6 \times 10^{-6} \mathrm{~F}\right)(18 \mathrm{~V})=29 \mu \mathrm{C}
(c) Find the voltage drop across the 12-\mu \mathrm{F} capacitor.
Apply the basic capacitance equation:
C=\frac{Q}{\Delta V} \rightarrow \Delta V=\frac{Q}{C}=\frac{29 \mu \mathrm{C}}{12 \mu \mathrm{F}}=2.4 \mathrm{~V}
REMARKS Notice that the equivalent capacitance is less than that of any of the individual capacitors. The relationship C=Q / \Delta V can be used to find the voltage drops on the other capacitors, just as in part (c).