Two capacitors C_1 = 3 mu F and C_2 = 2 mu F | vedclass

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(B) When capacitors are connected in series,the charge $Q$ on each capacitor is the same.Given: $C_1 = 3 \mu F$,$C_2 = 2 \mu F$.The potential differen

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$3:2$

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(B) When capacitors are connected in series,the charge $Q$ on each capacitor is the same.Given: $C_1 = 3 \mu F$,$C_2 = 2 \mu F$.The potential difference across a capacitor is given by $V = \frac{Q}{C}$.Since $Q$ is constant for both capacitors in series,$V \propto \frac{1}{C}$.Therefore,the ratio of potential across $C_2$ to $C_1$ is:$\frac{V_2}{V_1} = \frac{Q/C_2}{Q/C_1} = \frac{C_1}{C_2}$.Substituting the values:$\frac{V_2}{V_1} = \frac{3 \mu F}{2 \mu F} = \frac{3}{2}$.Thus,the ratio is $3:2$.

Two capacitors $C_1 = 3 \mu F$ and $C_2 = 2 \mu F$ are connected in series across a d.c. source of $100 \ V$. The ratio of the potential across $C_2$ to $C_1$ is

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