Three condensers of capacities C_{1}, C_{2}, | vedclass

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(D) When capacitors are connected in series,the charge $Q$ remains the same on all the capacitors.Given the relation $Q = C V$,the potential differenc

Vedclass · Accepted Answer

$\frac{1}{C_{1}}: \frac{1}{C_{2}}: \frac{1}{C_{3}}$

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(D) When capacitors are connected in series,the charge $Q$ remains the same on all the capacitors.Given the relation $Q = C V$,the potential difference across each capacitor is $V_i = \frac{Q}{C_i}$.Since $Q$ is constant for all capacitors in series,the potential difference $V_i$ is inversely proportional to the capacitance $C_i$.Therefore,the ratio of potentials $V_1 : V_2 : V_3$ is given by:$V_1 : V_2 : V_3 = \frac{Q}{C_1} : \frac{Q}{C_2} : \frac{Q}{C_3}$$V_1 : V_2 : V_3 = \frac{1}{C_1} : \frac{1}{C_2} : \frac{1}{C_3}$

Three condensers of capacities $C_{1}, C_{2}, C_{3}$ are connected in series with a source of e.m.f. $V$. The potentials across the three condensers are in the ratio of

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