Obtain an expression for the capacitance of a parallel plate (air) capacitor. The given figure shows a network of five capacitors connected to a 100 V supply. Calculate the total charge and energy stored in the network.

Solution

Five capacitors are connected to a 100 V supply.
As given in the figure, the capacitors C₁ and C₂are in parallel combination.

∴ C₁₂ = 1 + 2 = 3

μF

Now, the circuit reduces to

As seen, capacitors C₄ and C₅ are in parallel combinations

∴

C₄₅ = 3 + 3 = 6 μF

Now, the circuit reduces to

and, capacitors C₁₂ and C₄₅ are in series combinations.

∴

\frac{1}{C_{1245}} = \frac{1}{6} + \frac{1}{3}

\Rightarrow

C_{1245} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 μF

[∵ C_{5} = \frac{C_{1} . C_{2}}{C_{1} + C_{2}}]

Now, C₁₂₄₅ and C₃ are in parallel combinations, thus, equivalent capacitance is

C = C_{1245} + C_{3} = 2 + 2 = 4 μF

Therefore,

Total charge, q = CV

= 4 \times 10^{- 6} \times 100 = 4 \times 10^{- 4} C

Energy stored, U =

\frac{1}{2} {CV}^{2}

= \frac{1}{2} \times 4 \times 10^{- 6} \times (100)^{2} = 2 \times 10^{- 6} \times 10^{4} = 2 \times 10^{- 2} J

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