Question 23.8: For the combination of capacitors shown in Fig. 23.13a, assu......
For the combination of capacitors shown in Fig. 23.13a, assume that C_{1} = 2 µF, C_{2} = 4µF , and C_{3} = 3µF, and ΔV = 12 V. (a) Find the equivalent capacitance of the combination. (b) What is the charge on C_{1}?
(a) Capacitors C_{1} and C_{2} in Fig. 23.13a are in parallel and their equivalent capacitance C_{12} is:
C_{12}=C_{1}+C_{2}=2 \ \mu{F}+4\ \mu{F}=6\mathrm{~}\mu{F}From Fig. 23.13b, we find that C_{12} and C_{3} form a series combination and their equivalent capacitance C_{123} is given by:
\frac{1}{C_{123}} = \frac{1}{C_{12}} + \frac{1}{C_{3}} = \frac{1}{6 µF} + \frac{1}{3 µF} = \frac{1}{2 µF} ⇒ C_{123} = 2 µF
(b) We first find the charge Q_{123} on C_{123} in Fig. 23.13c as follows:
Q_{123}=C_{123}\ \Delta V=(2 \ \mu{F})(12\ {V})=24 \ {\mu C}This same charge exists on each capacitor in the series combination of Fig. 23.13b.
Therefore, if Q_{12} represents the charge on C_{12},then Q_{12} = Q_{123} = 24 µC. Accordingly, the potential difference across C_{12} is:
This same potential difference exists across C_{1}, i.e. ΔV_{1} = ΔV_{12}. Thus:
Q_{1}=C_{1}\Delta V_{1} = (2 µF)(4 V) = 8 μC
