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(C) Since the capacitors $C_1$ and $C_2$ are connected in series,the charge $Q$ on both capacitors must be the same.Let $V_D$ be the potential at poin

Vedclass · Accepted Answer

$\frac{C_1V_1 + C_2V_2}{C_1 + C_2}$

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(C) Since the capacitors $C_1$ and $C_2$ are connected in series,the charge $Q$ on both capacitors must be the same.Let $V_D$ be the potential at point $D$.The charge on capacitor $C_1$ is $Q = C_1(V_A - V_D) = C_1(V_1 - V_D)$.The charge on capacitor $C_2$ is $Q = C_2(V_D - V_B) = C_2(V_D - V_2)$.Equating the charges: $C_1(V_1 - V_D) = C_2(V_D - V_2)$.Expanding the terms: $C_1V_1 - C_1V_D = C_2V_D - C_2V_2$.Rearranging to solve for $V_D$: $C_1V_1 + C_2V_2 = V_D(C_1 + C_2)$.Therefore,$V_D = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}$.

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$A$. Tranquilizers	$I$. Anti blood clotting
$B$. Aspirin	$II$. Salvarsan
$C$. Antibiotic	$III$. Antidepressant drugs
$D$. Antiseptic	$IV$. Soframicine

Two capacitors $C_1$ and $C_2$ in a circuit are joined as shown in the figure. The potential of point $A$ is $V_1$ and that of $B$ is $V_2$. The potential of point $D$ will be

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