Question

Find the following. (In the figure use C1 = 14.40 MuF and C2 = 8.40 MuF.) (a) the equivalent capacitance of the capacitors in the figure above MuF (b) the charge on each capacitor on the right 14.40 MuF capacitor____________ MuC on the left 14.40 MuF capacitor _____________ MuC on the 8.40 Muf capacitor _____________ MuC on the 6.00 MuF capacitor _____________ MuC (c) the potential difference across each capacitor on the right 14.40 MuF capacitor V on the left 14.40 MuF capacitor V on the 8.40 MuF capacitor V on the 6.00 MuF capacitor V

Answer 1

6*10^-6F and 8.4*10^-6 F are in parallel then

equivalent capacitance is (6+8.4) $\mu F$ = 14.4 $\mu F$

now the capacitor on right , left and equivalent capacitor are in series

total capacitance = $\frac{1}{C_{t}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$

$C_{t}=\frac{14.4}{3}$

$C_{t}=4.8 \mu F$

To find the charge we have to know the potential diffrence across the each capacitor.

as 6 $\mu F$ and 8.4 $\mu F$ are in parallel so voltage across them will be equal with equivalent capacitance.

therefore

$V_{1}=\frac{C_{t}}{C_1}*V_{t}$

$V_{1}=\frac{4.8 \mu }{14.4 \mu }*90$

as the capacitance is equal for other capacitor so the

voltage across them also equal with 30V

V₂= 30 V

V₃=30 V

therefore voltage across right , left 14.4 $\mu F$ ,8.4 $\mu F$ and 6 $\mu F$ is 30 V

now charge across each capacitor as

on right and left 14.4 $\mu F$ capacitor are $Q=CV$

Q_{14.4 $\mu F$} =14.4 $\mu F$ *30 V

$Q_{14.4\mu F}$ =432 $\mu C$

$Q_{8.4\mu F}$ =8.4 $\mu F$ *30 V

$Q_{8.4\mu F}$ =252 $\mu C$

$Q_{6\mu F}$ = 6 $\mu F$ *30 V

$Q_{6\mu F}$ =180 $\mu C$