Question

A network of four capacitors, $5\mu F$ each, are connected to a $240V$ supply. Determine the equivalent capacitance of the network (b) the charge on each capacitor.

Answer 1

Capacitor $C_1,C_2,C_3$ are in series and $C_1=C_2=C_3=C=5\mu F$

Resultant capacitance of combination $C_s=\frac{C}{3}=\frac{5}{3}=1.67\mu F$

now $C_{s}isparallelto{C}_4$

Equivalent capacitance $C_p=C_s+C_4=1.67+5=6.67\mu F$

$b)Chargeoneachcapacitor$

Total charge $Q=C_{p}V=6.67\mu F\times 240V=1600.8\mu C$

For parallel combination of $C_{s}and{C}_4$ charge is distributed while the potential difference is same

$\frac{Q_s}{C_s}=\frac{Q_4}{C_4}$

$\therefore \frac{Q_s}{Q_4}=\frac{C_s}{C_4}=\frac{1.67}{5}=0.334\mu C$

also $Q_s+Q_4=Q=1600.8\mu C$

$\therefore Q_s=1200\mu Cand{Q}_4=400.8\mu C$

since $C_1,C_2,C_3$ are in series charge on the three capacitors are equal

$Q_1=Q_2=Q_3=Q_s=1200\mu C$