Question 436: Electrostats & Capacitance

Question

The potential difference across the 4 capacitor in the following circuit is

Accepted Answer

$\mathrm{C}_2$ and $\mathrm{C}_3$ capacitors are in parallel. Their equivalent capacitance is given by
$$ C_p=C_2+C_3=2+4=6 \mu F $$
$\mathrm{C}_{\mathrm{p}}$ is in series with $\mathrm{C}_1$
$$ V=\frac{q}{C} $$
As for series combination, the charge on the capacitors is same, therefore, potential across them is inversely proportional to their capacitances.
$$ V_1: V_p=C_p: C_1 $$
$$ 	herefore \quad V_1=\frac{C_p}{C_p+C_1} V=\frac{6}{10} 	imes 9=5.4 V $$
Similarly, voltage across $\mathrm{C}_2$ and $\mathrm{C}_3$ will be
$$ 	herefore \quad V_p=9-5.4 \mathrm{~V}=3.6 \mathrm{~V} $$
The question asks for the voltage across a $4 \mu \mathrm{~F}$ capacitor. However, the circuit contains two $4 \mu \mathrm{~F}$ capacitors. Among the given options, 5.4 V is listed, but 3.6 V is not. Therefore, it is presumed that the voltage across the first $4 \mu \mathrm{~F}$ capacitor ( $\mathrm{C}_1$ ) is intended.

Answer

3.4 V

Answer

4.6 V

Answer

5.4 V

Answer

6.2 V

Electrostats & Capacitance

The potential difference across the 4 capacitor in the following circuit is

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