Question

A network of four capacitors, each of capacitance $30\mu F$ is connected across a battery of 60 V, as shown in the figure. Find the net capacitance and the energy stored in each capacitor

Answer 1

Here $C_1,C_2,C_3$ are in series and the equivalent resistance for these is $C_{123}=30/3=10\mu F$

Now $C_{123},C_4$ are in parallel. So the net capacitance $C_{net}=C_{123}+C_4=10+30=40\mu F$

Now potential across $C_{123}$ and $C_4$ is same i.e, $60V$

So energy stored in $C_4$ is $U_4=\frac{1}{2}{C}_4{V}^2=0.5\times 30\times {10}^{-6}\times {60}^2=0.054J=54mJ$

As $C_1,C_2,C_3$ are in series so they have same charge is equal to $Q_{123}=C_{123}V=10\times 60=600\mu C$

Stored energy, $U_1=\frac{Q_{123}^2}{2C_1}=\frac{(600\times {10}^{-6}{)}^2}{2\times 30\times {10}^{-6}}=6mJ$

As $C_1,C_2,C_3$ have same charge and same capacitance so they store same energy.

Thus, $U_1=U_2=U_3=6mJ$