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

Solution

Given,
Potential applied across the network, V=300 V
Capacitance on capacitor 1, C₁=100 pF
Capacitance on capacitor 2, C₂=200 pF
Capacitance on capacitor 3, C₃=200 pF
Capacitance on capacitor 4, C₄=100 pF

A similar network is shown below as given in the problem.

Assume, that the series combination of C₂ and C₃ is C₂₃.

∴

\frac{1}{C_{23}} = \frac{1}{C_{2}} + \frac{1}{C_{3}} = \frac{1}{200} + \frac{1}{200} = \frac{1}{100}

C_{23} = 100 pF

Suppose, the parallel combination of C₁ and C₂₃ is C’. Then,

C' = C_{1} + C_{23} = 100 + 100 = 200 pF

Let, the series combination of C₄ and C is C.

∴

\frac{1}{C} = \frac{1}{C_{4}} + \frac{1}{C'} = \frac{1}{100} + \frac{1}{200} = \frac{3}{200}

\Rightarrow

C = \frac{200}{3} pF

Therefore, total charge on all capacitors.

q = CV = \frac{200}{3} \times 10^{- 12} \times 300 = 2 \times 10^{- 8} C ∴ V_{4} = \frac{q_{4}}{C_{4}} = \frac{2 \times 10^{- 8}}{100 \times 10^{- 12}} V_{4} = 2 \times 100 = 200 V V_{1} = [300 - 200] = 100 V q_{1} = C_{1} V_{1} = 100 \times 10^{- 12} \times 100 = 10^{- 8} C V_{2} = V_{3} V_{2} + V_{3} = 100 V 2 V_{2} = 100 V_{2} = 50 V q_{2} = C_{2} V_{2} = 200 \times 10^{- 12} \times 50 = 10000 \times 10^{- 12} = 10^{- 8} C q_{3} = C_{3} V_{3} = 200 \times 10^{12} \times 50 = 10^{- 8} C

Hence, the charge and voltage across each capacitor is as given below :

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Electrostatic Potential And Capacitance

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

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