1.

A parallel-plate capacitor is charged to 50 μC at 150 V. It is then connected to another capacitor of capacitance 4 times the capacitance of the first capacitor. Find the loss of energy.

Answer»

C1 = 50/150 = 1/3 μF ; C2 = 4 × 1/3 = 4/3 μF 

Before Joining

E1 = 1/2C1V12 = 1/2 x (1/3) x 10-6 x 1502

= 37.5 x 10-4J;

E2 = 0

Total energy = 37.5 × 10− 4 J

After Joining 

When the two capacitors are connected in parallel, the charge of 50 μ C gets redistributed and the two capacitors come to a common potential V.

V = total charge/total capacitance

= 50μC/([(1/3) + (4/3)]μF) = 30V

E1 = 1/2 x (1/3) x 10-6 x 302 = 1.5 x 10-4J

E2 = 18/2 x (4/3) x 10-6 x 302 = 6.0 x 10-4J

Total energy = 7.5 x 10-4J;

Loss of energy = (37.5 − 7.5) × 10− 4 = 3 × 10− 2 J 

The energy is wasted away as heat in the conductor connecting the two capacitors.


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