Question 545: Electrostatics & Capacitance

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

Accepted Answer

Given: $$\mathrm{C_1: C_2=1: 2}$$
$$\begin{aligned}
	herefore \quad \mathrm{C}_2 & =2 \mathrm{C}_1 \
\mathrm{C}_{\mathrm{P}} & =\mathrm{C}_1+\mathrm{C}_2=\mathrm{C}_1+2 \mathrm{C}_1=3 \mathrm{C}_1 \
\mathrm{C}_{\mathrm{S}} & =\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{2 \mathrm{C}_1^2}{3 \mathrm{C}_1}=\frac{2}{3} \mathrm{C}_1
\end{aligned}$$
Let $$V_P$$ and $$V_S$$ be the potentials applied across the parallel and series combinations respectively, then using $$\mathrm{E}=\frac{1}{2} \mathrm{CV}^2$$ we can write,
$$\begin{array}{ll} 
& \frac{1}{2} C_P V_P^2=\frac{1}{2} C_S V_S^2 \
	herefore \quad & \frac{V_P^2}{V_S^2}=\frac{C_S}{C_P} \
& \frac{V_P}{V_S}=\sqrt{\frac{\frac{2}{3} C_1}{3 C_1}}=\frac{\sqrt{2}}{3}
\end{array}$$

Answer

$$3: \sqrt{2}$$

Answer

$$\sqrt{2}: 3$$

Answer

$$2: 9$$

Answer

$$9: 2$$

Electrostatics & Capacitance

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

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