(a) Derive an expression for the energy stored in a parallel plate capacitor $C$, charged to a potential difference $V$. (b) Obtain the equivalent capacitance of the network given below ($C_1 = 100\,\text{pF}$ with $200\,\text{pF}$ capacitors on either side, $C_2$ and $C_3$ of $200\,\text{pF}$ in series, and $C_4 = 100\,\text{pF}$). For a supply of 300 V, determine the charge and voltage across $C_4$.

(a) $U = \tfrac12 CV^2 = \dfrac{Q^2}{2C} = \tfrac12 QV$. (b) $C_{eq} = \dfrac{200}{3}\ \text{pF}$; $V_4 = 200\ \text{V}$ and $Q_4 = 2 \times 10^{-8}\ \text{C}$.

CBSE Class 12 · 2008 · CapacitanceLong Answer

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(a) Derive an expression for the energy stored in a parallel plate capacitor

C

, charged to a potential difference

V

.
(b) Obtain the equivalent capacitance of the network given below (

C_{1} = 100 pF

with

200 pF

capacitors on either side,

C_{2}

and

C_{3}

200 pF

in series, and

C_{4} = 100 pF

). For a supply of 300 V, determine the charge and voltage across

C_{4}

Solution

(a) When the charge is

q

the potential is

\frac{q}{C}

, so the work to add

d q

\frac{q}{C} d q

.
Integrating from

0

Q

U = \int_{0}^{Q} \frac{q}{C} d q = \frac{Q ^{2}}{2 C}

U = \frac{Q ^{2}}{2 C} = \frac{1}{2} C V^{2}

(b) Series pair

C_{2}, C_{3}

C^{'} = \frac{200 \times 200}{200 + 200} = 100 pF

In parallel with

C_{1}

gives

200 pF

, in series with the two

200 pF

capacitors:

C_{e q} = \frac{200}{3} pF

With 300 V across the network,

V_{4} = 200 V

Q_{4} = C_{4} V_{4} = 100 \times 1 0^{- 12} \times 200 = 2 \times 1 0^{- 8} C

CBSE Class 12 Board · 2008 · Subjective

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CBSE Class 12 · 2008 · CapacitanceLong Answer

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