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Electrostatic Potential And Capacitance

Question
CBSEENPH12039011
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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 parallel combination.

Solution

When the capacitors are connected in parallel, 
Equivalent resistance, straight C subscript straight P space equals space straight C subscript 1 space plus space straight C subscript 2 
Energy stored in capacitors, EP  = 1 half straight C subscript straight P straight V squared 
therefore straight space straight E subscript straight P straight space equals straight space 1 half left parenthesis straight C subscript 1 plus straight C subscript 2 right parenthesis left parenthesis 100 right parenthesis squared straight space equals straight space 0.25 straight space straight J rightwards double arrow straight space straight C subscript 1 straight space plus straight space straight C subscript 2 straight space equals straight space 0.5 straight space cross times straight space 10 to the power of – 4 end exponent straight F space straight C subscript 1 straight space plus straight space straight C subscript 2 straight space equals 50 straight space μF space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis thin space

When the capacitors are connected in series,
Error converting from MathML to accessible text.
Error converting from MathML to accessible text.

Putting this in equation (2), we get 
C2 = 50 – 11.2 = 38.2 μF
The charge on capacitor is given by, Q = CV

Charge on capacitor 1, C1 = straight Q subscript 1 straight space equals straight C subscript 1 straight V straight space equals straight space 35 straight space cross times straight space 10 to the power of negative 6 end exponent cross times 100 straight space equals straight space 35 straight space cross times straight space 10 to the power of negative 4 end exponent straight space straight C
Charge on capacitor 2, C2 = straight Q subscript 2 straight space equals straight space straight C subscript 2 straight V straight space equals straight space 15 straight space cross times 10 to the power of negative 6 end exponent cross times 100 straight space equals straight space 15 cross times straight space 10 to the power of negative 4 end exponent straight space straight C

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