Question

A circuit has section AB as shown in figure. Th~ emf of the source equals $E=10V$, the capacitor capacitances are equal to $C_1=1.0\mu F$ and $C_2=2.0\mu F$, and the potential difference $V_A-V_B=5.0V$. Find the voltage across each capacitor.

Answer 1

Here the capacitors are in series so the charge on each is equal to equivalent charge.

The equivalent capacitance is $C_{eq}=\frac{1\times 2}{1+2}=\frac{2}{3}\mu F$ and the equivalent charge , $Q_{eq}=C_{eq}V=\frac{2}{3}(10+5)=10\mu C$

Thus, $Q_1=Q_2=Q_{eq}=10\mu C$

Potential across $C_1$ is $V_1=Q_1/C_1=10/1=10V$ and Potential across $C_2$ is $V_2=Q_2/C_2=10/2=5V$