Question

The ratio of the energies stored in capacitors \(A\), \(B\) and \(C\) when connected in parallel to a dc supply is \(1:2:3\). When they are connected in series to the same dc supply, if the energy stored in capacitor \(C\) is \(75\,\text{mJ}\), then the energy stored in capacitor \(A\) is

• A. \(150\,\text{mJ}\)
• B. \(25\,\text{mJ}\)
• C. \(225\,\text{mJ}\)
• D. \(75\,\text{mJ}\)

Hint:

Remember, \[ \boxed{ U=\frac12CV^2 } \] for capacitors connected across the same voltage, and \[ \boxed{ U=\frac{Q^2}{2C} } \] for capacitors carrying the same charge (series connection).

Answer 1

The Correct Option is C

Step 1: Determine the ratio of capacitances. When capacitors are connected in parallel, all have the same potential difference. The energy stored is \[ U=\frac12CV^2. \] Hence, \[ U_A:U_B:U_C = C_A:C_B:C_C = 1:2:3. \] Let \[ C_A=C,\qquad C_B=2C,\qquad C_C=3C. \]

Step 2: Use the relation for capacitors in series. In series connection, the charge on each capacitor is the same. The energy stored is \[ U=\frac{Q^2}{2C}. \] Therefore, \[ U\propto\frac1C. \] Hence, \[ U_A:U_B:U_C = \frac1C:\frac1{2C}:\frac1{3C} = 6:3:2. \]

Step 3: Find the energy stored in capacitor \(A\). Given, \[ U_C=75\,\text{mJ}. \] Since \[ U_A:U_C=6:2=3:1, \] we get \[ U_A=3\times75=225\,\text{mJ}. \] Thus, \[ \boxed{U_A=225\,\text{mJ}.} \] Therefore, the correct option is \(\boxed{(C)}\).