Two capacitors A and B with capacitances 2 mu | vedclass

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(B) In a series combination of capacitors,the charge $Q$ on each capacitor is the same. Given $C_A = 2 \mu F$ and $C_B = 3 \mu F$. The total voltage $

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The potential difference across the plates of $A$ is $6 \ V$ and across the plates of $B$ is $4 \ V$.

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(B) In a series combination of capacitors,the charge $Q$ on each capacitor is the same. Given $C_A = 2 \mu F$ and $C_B = 3 \mu F$. The total voltage $V = 10 \ V$ is distributed such that $V_A + V_B = 10 \ V$. Since $Q = C_A V_A = C_B V_B$,we have $V_A / V_B = C_B / C_A = 3 / 2$. From $V_A = (3/2) V_B$,substituting into the sum equation: $(3/2) V_B + V_B = 10 \ V$. $(5/2) V_B = 10 \ V \implies V_B = 4 \ V$. Then $V_A = 10 \ V - 4 \ V = 6 \ V$. Thus,the potential difference across $A$ is $6 \ V$ and across $B$ is $4 \ V$.

Two capacitors $A$ and $B$ with capacitances $2 \mu F$ and $3 \mu F$ respectively are connected in series with a $10 \ V$ battery as shown in the figure. When the switch $S$ is closed and the two capacitors get charged fully,then:

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