Three capacitors of capacitances 25 μ F, 30 μ F and 45 μ F are connected in parallel to a supply of 100 V. Energy stored in the above combination is E. When these capacitors are connected in series to the same supply, the stored energy is 9/x E. The value of x is _________.

Solution

Physics sir · 4 steps

Same supply both times, so energy goes as capacitance: compare the series and parallel capacitances.

1. Energy ratio. With the same V in both cases, U = 1/2CV² means energy is proportional to the equivalent capacitance. So the series energy divided by E equals Cₛ/Cₚ.
2. Parallel. Capacitances in parallel add: Cₚ = 25 + 30 + 45 = 100 μF.
3. Series. Reciprocals add. Over a common denominator of 450, 1/Cₛ = (18 + 15 + 10)/450 = 43/450, so Cₛ = 450/43 μF.

   Eₛ/E = Cₛ/Cₚ = (450/43)/100 = 4.5/43 = 9/86

4. So. The series energy is 9/86E, so x = 86.

The idea to keep. When the supply voltage stays fixed, stored energy scales directly with equivalent capacitance. Take the ratio of capacitances and skip computing the energies.

Answer from the NTA final answer key. Solution by PrepWiser.