Question 557: Electrostatics & Capacitance

Question

Capacitors of capacities $$\mathrm{C}_1, \mathrm{C}_2$$ and $$\mathrm{C}_3$$ are connected in series. If the combination is connected to a supply of '$$\mathrm{V}$$' volt, then potential difference across capacitor '$$\mathrm{C}_1$$' is

Accepted Answer

In series combination, the equivalent capacitance $$\mathrm{C}$$ is given by
$$\begin{aligned}
\frac{1}{C} & =\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \
	herefore \quad \frac{1}{C} & =\frac{C_2 C_3+C_1 C_3+C_1 C_2}{C_1 C_2 C_3} \
	herefore \quad C & =\frac{C_1 C_2 C_3}{C_2 C_3+C_1 C_3+C_1 C_2}
\end{aligned}$$
Charge $$\mathrm{Q}$$ stored by the combinations is given by
$$Q=C V=\frac{C_1 C_2 C_3 V}{\mathrm{C}_2 C_3+C_1 C_3+C_1 C_2}$$
Charge on each capacitor is same. Hence potential difference across $$\mathrm{C_1}$$ is $$\mathrm{V_1=\frac{Q}{C_1}}$$
$$	herefore \quad \mathrm{V_1=\frac{C_2 C_3 V}{C_2 C_3+C_1 C_3+C_1 C_2}}$$

Answer

$$\frac{\mathrm{C}_2 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1 \mathrm{C}_2 V}$$

Answer

$$\frac{\mathrm{C}_2 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1 \mathrm{C}_2 \mathrm{C}_3 \mathrm{V}}$$

Answer

$$\frac{\mathrm{C}_2 \mathrm{C}_3 \mathrm{V}}{\mathrm{C}_2 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_2}$$

Answer

$$\frac{\mathrm{C}_1 \mathrm{C}_2 \mathrm{C}_3 \mathrm{V}}{\mathrm{C}_2 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_3+\mathrm{C}_1 \mathrm{C}_2}$$

Electrostatics & Capacitance

Capacitors of capacities $\mathrm{C}_1, \mathrm{C}_2 and \mathrm{C}_3 are connected in series. If the combination is connected to a supply of '\mathrm{V}' volt, then potential difference across capacitor '\mathrm{C}_1$' is

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