Question

Obtain the equivalent capacitance of the network shown in the figure. For a 300 V supply, determine the charge and voltage across each capacitor.

Answer 1

$\text{Step 1: Redraw the circuit}$ $\text{[Ref. Fig.]}$

$\text{Step 2: Equivalent Capacitance}$

Equivalent capacitance for the loop A shown in figure:

$C_A=\frac{C_2{C}_3}{C_2+C_3}+C_1$

$\Rightarrow C_A=\frac{200\times 200}{200+200}+100$$=200pF$

Now $C_A$ and $C_4$ are in series

$C_{eq}=\frac{C_A{C}_4}{C_A+C_4}=\frac{200\times 100}{200+100}=\frac{200}{3}pF$

$\text{Step 3: Total charge supplied}$

$Q=C_{eq}V$$=\frac{200}{3}pF\times 300V=2\times {10}^{-8}C$

As, $C_4$ and $C_A$ are in series,

$\therefore Q_4=Q_A=Q=2\times 10{}^{-8}C$

$\text{Step 4: Potential across each capacitor}$

Voltage across capacitor $C_4:$ $V_4=\frac{Q_4}{C_4}=\frac{2\times {10}^{-8}}{100\times {10}^{-12}}=200V$

Voltage across capacitor $C_1:V_1=V-V_4=300-200=100V$

$C_2$ and $C_3$ are parallel with $C_1$. That means $C_2$ and$C_3$ will have a potential difference of $100V$ together.

$C_2\&C_3$, they both are in series and have equal capacitances, so potential will be equal.

$\therefore V_2=V_3=50V$

$\text{Step 5: Charge on each capacitor:}$

$Q_1=C_1{V}_1=100pF\times 100V={10}^{-8}C$

$Q_2=C_2{V}_2=200pF\times 50V={10}^{-8}C$

$Q_3=C_3{V}_3=200pF\times 50V={10}^{-8}C$

$Q_4=2\times {10}^{-8}C$ $\{$ as calculated in step 3$\}$