Question
Find the total energy stored in the capacitors in the given network.
✓
Answer
The equivalent capacitance of $C_1$ and $C_2$ in series,$\text{C}'=\frac{\text{C}_1\text{C}_2}{\text{C}_2+\text{C}_2}=\frac{2\times2}{2+2}=1\mu\text{F}$
C′ is in parallel with $C_3$, so equivalent capacitance of $C_1, C_2$ and $C_3$ is:$\text{C}''=1+1=2\mu\text{F}$
C′′ is in series with $C_4;$ their equivalent capacitance,$\text{C}'''=\frac{\text{C}_4\text{C}}{\text{C}_4+\text{C}''}=\frac{2\times2}{2+2}=1\mu\text{F}$
This is in parallel with $C_5$; So equivalent capacitance across AB is $\text{C}_{\text{AB}}=1+1=2\mu\text{F},$ Energy stored $\text{V}'=\frac{1}{2}\text{C}_{\text{AB}}\text{V}^2=\frac{1}{2}\times2\times10^{-6}\times(6)^2=36\times10^{-6}\text{J}$
C′ is in parallel with $C_3$, so equivalent capacitance of $C_1, C_2$ and $C_3$ is:$\text{C}''=1+1=2\mu\text{F}$
C′′ is in series with $C_4;$ their equivalent capacitance,$\text{C}'''=\frac{\text{C}_4\text{C}}{\text{C}_4+\text{C}''}=\frac{2\times2}{2+2}=1\mu\text{F}$
This is in parallel with $C_5$; So equivalent capacitance across AB is $\text{C}_{\text{AB}}=1+1=2\mu\text{F},$ Energy stored $\text{V}'=\frac{1}{2}\text{C}_{\text{AB}}\text{V}^2=\frac{1}{2}\times2\times10^{-6}\times(6)^2=36\times10^{-6}\text{J}$
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