Question

1. Three capacitors with individual capacitances of 2 μF, 5 μF, and 10 μF respectively are connected in series with a 12V battery. What are the total capacitance and total charge in the network? 2. Same capacitors in number 1 connected in a parallel. If the combined charge in the network is 50 μC, what are the total voltage and total capacitance in the network?

Answer 1

Series combination

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\frac{1}{C}=\frac{1}{2}+\frac{1}{5}+\frac{1}{10}$

$C=\frac{5+2+1}{10}\mu F=\frac{8}{10}\mu F$

$C=0.8\mu F$

Charge

$Q=CV=0.8\times12\mu F=9.6\mu F$(B) same capacitor connected parallel combination

$C=C_1+C_2+C_3$

$C=(2+5+10)\mu F=17\mu F$

$V=\frac{Q}{C}=\frac{50\times10^{-6}}{17\times10^{-6}}=2.95V$