electrostatic-potential-and-capacitance Question 43

Question: Q. 17. Two capacitors of unknown capacitances C1 and C2 are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of C1 and C2. Also calculate the charge on each capacitor in paraller combination.

A. Delhi I, II, III 2015]

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Solution:

Ans. Energy stored in a capacitor

In series combination

(i)0.045=12C1C2C1+C2(100)2

C1C2C1+C2=0.09×104

In parallel combination

(ii)0.25=12(C1+C2)(100)2

C1+C2=0.5×104

On simplifying (i) and (ii)

(iii)C1C2=0.045×108 (C1C2)2=(C1+C2)24C1C2 =(0.5×104)24×0.045×108 =0.25×1080.180×108 (C1C2)2=0.07×108 (C1C2)=2.6×105=0.25×104( iii )

From (ii) and (iii) we have

1/2 Charges on capacitors C1 and C2 in parallel combination

$Q_{1}=\mathrm{C}{1} V=\left(0.38 \times 10^{-4} \times 100\right)=0.38 \times 10^{-2} \mathrm{C} \quad 1 / 2Q{2}=C_{2} V=\left(0.12 \times 10^{-4} \times 100\right)=0.12 \times 10^{-2} \mathrm{C} \quad 1 / 2$ Alternatively,

and

U=12CV2 0.045=12(C1C2C1+C2)(100)2 0.25=12(C1+C2)(100)2

If the student is unable to calculate C1 and C2, award him/her full 2 marks.

Also if the student just writes

Q1=C1V=C1(100) and Q2=C2V=C2(100)

award him/her one mark for this part of the question.

[CBSE Marking Scheme, 2015]



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