(i) Find equivalent capacitance between A ...

CBSE Class 12 Physics 2017 Delhi Set 1 Paper

Question : 17 of 26

Marks: +1, -0

(i) Find equivalent capacitance between

A

and

B

in the combination given below. Each capacitor is of

2 \mu \mathrm{F}

capacitance.

(ii) If a dc source of

7 \mathrm{V}

is connected across

AB

, how much charge is drawn from the source and what is energy stored in the network?

Solution:

(i) Calculation of equivalent capacitance

(ii) Calculation of charge and energy stored

(i) Capacitors

C_2, C_3

and

C_4

are in parallel

\therefore \; C_{234}=C_2+C_3+C_4

\therefore \; C_{234}=6 \mu \mathrm{F}

Capacitors

C_1, C_{234}

and

C_5

are in series

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_{234}} + \frac{1}{C_5}

= \frac{1}{2} + \frac{1}{6} + \frac{1}{2}

= \frac{7}{6} \mu \mathrm{F}

C_{eq} = \frac{6}{7} \mu \mathrm{F}

(ii) Charge drawn from the source

Q = C_{eq} V,

= \frac{6}{7} \times 7 \mu \mathrm{C} = 6 \mu \mathrm{C}

Energy stored

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

= \frac{6 \times 6 \times 10^{-12} \times 7}{2 \times 6 \times 10^{-6}} = 21 \mu \mathrm{J}

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