Question

Three capacitors of capacitances 25 μF, 30 μF and 45 μF 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 \(\frac{9}{x} E\)  . The value of x is .................

Answer 1

Correct Answer: 86

In parallel combination: Potential difference is the same across all capacitors.

\[ \text{Energy} = \frac{1}{2}(C_1 + C_2 + C_3)V^2 \]

\[ = \frac{1}{2}(25 + 30 + 45) \times (100)^2 \times 10^{-6} = 0.5 = E \]

In series combination: Charge is the same on all.

\[ \frac{1}{C_{\text{equ}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{25} + \frac{1}{30} + \frac{1}{45} \]

\[ \frac{1}{C_{\text{equ}}} = \frac{18 + 15 + 10}{450} = \frac{43}{450} \implies C_{\text{equ}} = \frac{450}{43} \]

Energy:

\[ \text{Energy} = \frac{Q^2}{2C_1} + \frac{Q^2}{2C_2} + \frac{Q^2}{2C_3} \]

\[ = \frac{Q^2}{2} \left[ \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \right] \]

\[ = \frac{(V \times C_{\text{equ}})^2}{2 \times C_{\text{equ}}} \times \frac{1}{C_{\text{equ}}} = \frac{V^2C_{\text{equ}}}{2} \]

\[ = \frac{(100)^2}{2} \times \frac{450}{43} \times 10^{-6} \]

\[ = \frac{4.5}{86} = \frac{9}{x} \times 0.5 \implies x = 86 \]

Answer 2

Step 1: Given data
Capacitances: \( C_1 = 25\,\mu\text{F}, \; C_2 = 30\,\mu\text{F}, \; C_3 = 45\,\mu\text{F} \)
Supply voltage: \( V = 100\,\text{V} \)
Energy in parallel combination: \( E \)
When connected in series to the same supply, the energy becomes \( \dfrac{9}{x}E \). We must find \(x\).

Step 2: Energy in the parallel combination
For capacitors in parallel: \[ C_{\text{p}} = C_1 + C_2 + C_3 = 25 + 30 + 45 = 100\,\mu\text{F}. \] The energy stored: \[ E = \frac{1}{2} C_{\text{p}} V^2 = \frac{1}{2} \times 100 \times 10^{-6} \times (100)^2 = \frac{1}{2} \times 100 \times 10^{-6} \times 10^4 = 0.5\,\text{J}. \] Hence, \(E = 0.5\,\text{J}\) (we will just use the ratio later).

Step 3: Energy in the series combination
For capacitors in series: \[ \frac{1}{C_{\text{s}}} = \frac{1}{25} + \frac{1}{30} + \frac{1}{45}. \] Compute the reciprocal: \[ \frac{1}{C_{\text{s}}} = \frac{(1800/25)(1/1800) + (1800/30)(1/1800) + (1800/45)(1/1800)}{} = \frac{36 + 30 + 20}{900} = \frac{86}{900}. \] Thus, \[ C_{\text{s}} = \frac{900}{86} \approx 10.465\,\mu\text{F}. \] Now, energy in series connection: \[ E_{\text{s}} = \frac{1}{2} C_{\text{s}} V^2 = \frac{1}{2} \times 10.465 \times 10^{-6} \times 10^4 = 0.052325\,\text{J}. \]
Step 4: Ratio of energies
\[ \frac{E_{\text{s}}}{E} = \frac{0.052325}{0.5} = 0.10465. \] Given: \[ E_{\text{s}} = \frac{9}{x}E \quad \Longrightarrow \quad \frac{9}{x} = 0.10465 \quad \Longrightarrow \quad x = \frac{9}{0.10465} \approx 86. \]
Final answer

86