Two capacitors of different capacitances are connected first (1) in series…

Two capacitors of different capacitances are connected first (1) in series and then (2) in parallel across a de source of 100 V. If the total energy stored in the combination in the two cases are 40 mJ and 250 mJ respectively, the capacitance of the capacitors.

Ans. Capacitors Connected in Series: When capacitors are connected in series, their total capacitance ��Cs​ is given by the reciprocal of the sum of the reciprocals of individual capacitances:

1��=1�1+1�2Cs​1​=C1​1​+C2​1​

Given that the total energy stored in the combination is 40 mJ, we can use the formula for the energy stored in a capacitor:

�=12���2E=21​Cs​V2

where $E$ is the energy, ��Cs​ is the total capacitance, and $V$ is the voltage.

Given that $V=100$ V, we have:

�=12��(100)2=40×10−3 JE=21​Cs​(100)2=40×10−3J

��=2×40×10−31002 FCs​=10022×40×10−3​F

��=810000 F=8×10−4 FCs​=100008​F=8×10−4F

2. Capacitors Connected in Parallel: When capacitors are connected in parallel, their total capacitance ��Cp​ is the sum of the individual capacitances:

��=�1+�2Cp​=C1​+C2​

Given that the total energy stored in the combination is 250 mJ, we can use the same formula for the energy stored in a capacitor:

�=12���2E=21​Cp​V2

Given that $V=100$ V, we have:

�=12(�1+�2)(100)2=250×10−3 JE=21​(C1​+C2​)(100)2=250×10−3J

�1+�2=2×250×10−31002 FC1​+C2​=10022×250×10−3​F

�1+�2=5010000 F=5×10−3 FC1​+C2​=1000050​F=5×10−3F

Now, we have two equations:

1. 1��=1�1+1�2Cs​1​=C1​1​+C2​1​

2. �1+�2=5×10−3 FC1​+C2​=5×10−3F