Question

The ratio of potential difference that must be applied across parallel and series combination of two capacitors $C_{1}$ and $C_{2}$ with their capacitance in the ratio 1: 2 so that energy stored in these two cases becomes same is

• A. $3:\sqrt{2}$
• B. $\sqrt{2}:3$
• C. 2:9
• D. 9:2

Hint:

Logic Tip: Since parallel capacitance is always greater than series, a smaller voltage is required in the parallel case to store the same amount of energy.

Answer 1

The Correct Option is B

Concept: Energy stored in a capacitor is: \[ E = \frac{1}{2} C V^2 \] For the same stored energy in two different configurations: \[ \frac{1}{2} C_P V_P^2 = \frac{1}{2} C_S V_S^2 \]
Step 1: Equivalent capacitances Given: \[ C_1 : C_2 = 1 : 2 \Rightarrow C_2 = 2C_1 \] Parallel combination: \[ C_P = C_1 + C_2 = C_1 + 2C_1 = 3C_1 \] Series combination: \[ C_S = \frac{C_1 C_2}{C_1 + C_2} = \frac{C_1 \cdot 2C_1}{3C_1} = \frac{2}{3} C_1 \]
Step 2: Equate energies \[ C_P V_P^2 = C_S V_S^2 \] \[ \frac{V_P^2}{V_S^2} = \frac{C_S}{C_P} = \frac{\frac{2}{3}C_1}{3C_1} = \frac{2}{9} \]
Step 3: Final ratio \[ \frac{V_P}{V_S} = \sqrt{\frac{2}{9 = \frac{\sqrt{2{3} \] Final Answer: \[ \boxed{\frac{V_P}{V_S} = \frac{\sqrt{2{3 \]