CBSE Class 12 · 2015 · Delhi · CapacitanceShort Answer

Two capacitors of unknown capacitances

C_{1}

and

C_{2}

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

C_{1}

and

C_{2}

. Also calculate the charge on each capacitor in parallel combination.

Solution

Series combination:

U_{s} = \frac{1}{2} \frac{C _{1} C _{2}}{C _{1} + C _{2}} V^{2} = 0.045 J \Rightarrow \frac{C _{1} C _{2}}{C _{1} + C _{2}} = 9 \times 1 0^{- 6}

Parallel combination:

U_{p} = \frac{1}{2} (C_{1} + C_{2}) V^{2} = 0.25 J \Rightarrow C_{1} + C_{2} = 50 \times 1 0^{- 6}

Hence the product:

C_{1} C_{2} = 9 \times 1 0^{- 6} \times 50 \times 1 0^{- 6} = 4.5 \times 1 0^{- 10}

The difference follows from

(C_{1} - C_{2})^{2} = (C_{1} + C_{2})^{2} - 4 C_{1} C_{2}

(C_{1} - C_{2})^{2} = (5 0^{2} - 4 \times 450) \times 1 0^{- 12} = 7 \times 1 0^{- 10}

C_{1} - C_{2} = 2.64 \times 1 0^{- 5}

Solving the sum and difference:

C_{1} = 35 μ F, C_{2} = 15 μ F

In parallel both have 100 V:

Q_{1} = 35 \times 1 0^{- 6} \times 100 = 35 \times 1 0^{- 4} C, Q_{2} = 15 \times 1 0^{- 6} \times 100 = 15 \times 1 0^{- 4} C

CBSE Class 12 Board · 2015 · Delhi · Subjective

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CBSE Class 12 · 2015 · Delhi · CapacitanceShort Answer

Solution

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