Question

Two capacitors are connected as shown in figure below. If the equivalent capacitance of the combination is $4\mu F$
(i) Calculate the value of $C$.
(ii) Calculate the charge on each capacitor.
(iii) What will be the potential drop across each capacitor?

Answer 1

i) Here two capacitors are in series. So the equivalent resistance $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$

or $4=\frac{20C}{20+C}$

or $80+4C=20C$ or $C=80/16=5\mu F$

ii) The equivalent charge $Q_{eq}=C_{eq}V=4\left(12\right)=48\mu C$

As the capacitors are in series so they have same charge is equal to $Q_{eq}$

Thus, $Q_1=Q_2=48\mu C$

iii) Potential across $20\mu F$ is $V_1=Q1/C_1=48/20=2.4V$

and potential across $C=5\mu F$ is $V_2=Q_2/C_2=48/5=9.6V$