Two condensers of capacities 1,mu F and 2,mu | vedclass

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(C) When capacitors are connected in series,the charge $(Q)$ on each capacitor is the same.Let $C_1 = 1\,\mu F$ and $C_2 = 2\,\mu F$.The total potenti

Vedclass · Accepted Answer

$80$

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(C) When capacitors are connected in series,the charge $(Q)$ on each capacitor is the same.Let $C_1 = 1\,\mu F$ and $C_2 = 2\,\mu F$.The total potential difference is $V = V_1 + V_2 = 120\,V$.Since $Q = C_1 V_1 = C_2 V_2$,we have $1 	imes V_1 = 2 	imes V_2$,which implies $V_1 = 2V_2$.Substituting this into the total voltage equation: $2V_2 + V_2 = 120\,V$.$3V_2 = 120\,V$,so $V_2 = 40\,V$.Therefore,$V_1 = 2 	imes 40\,V = 80\,V$.The potential difference across the $1\,\mu F$ capacitor is $80\,V$.

Two condensers of capacities $1\,\mu F$ and $2\,\mu F$ are connected in series and the system is charged to $120\,V$. Then the potential difference $(P.D.)$ across the $1\,\mu F$ capacitor (in $volts$) will be

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