Three condensers of capacities C_1,C_2,and C_ | vedclass

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(D) When capacitors are connected in series,the charge $Q$ stored on each capacitor is the same.Using the relation $Q = CV$,we can write the potential

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 $Q$ stored on each capacitor is the same.Using the relation $Q = CV$,we can write the potential difference across each capacitor as $V_i = \frac{Q}{C_i}$.Since $Q$ is constant for all capacitors in a series combination,the potential difference $V_i$ is inversely proportional to the capacitance $C_i$.Therefore,the ratio of potentials is $V_1 : V_2 : V_3 = \frac{Q}{C_1} : \frac{Q}{C_2} : \frac{Q}{C_3}$.This simplifies to $V_1 : V_2 : V_3 = \frac{1}{C_1} : \frac{1}{C_2} : \frac{1}{C_3}$.

Three condensers of capacities $C_1$,$C_2$,and $C_3$ are connected in series with a source of e.m.f. $V$. The potentials across the three condensers are in the ratio:

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