Two capacitors C_1 = 2 mu F and C_2 = 3 mu | vedclass

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(A) The capacitors are connected in series,so the charge $q$ on each capacitor is the same. Let $V_B$ be the potential at point $B$. The potential dif

Vedclass · Accepted Answer

$22$

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(A) The capacitors are connected in series,so the charge $q$ on each capacitor is the same. Let $V_B$ be the potential at point $B$. The potential difference across $C_1$ is $V_A - V_B = 40 - V_B$. The potential difference across $C_2$ is $V_B - V_C = V_B - 10$. Since the charge $q = CV$ is the same for both,we have $C_1(V_A - V_B) = C_2(V_B - V_C)$. Substituting the given values: $2(40 - V_B) = 3(V_B - 10)$. Expanding this,we get $80 - 2V_B = 3V_B - 30$. Rearranging the terms,$5V_B = 110$,which gives $V_B = 22 \ V$.

Two capacitors $C_1 = 2 \ \mu F$ and $C_2 = 3 \ \mu F$ are connected in series between points $A$ and $C$. The potential at point $A$ is $40 \ V$ and the potential at point $C$ is $10 \ V$. Find the potential at point $B$ (between the two capacitors). (in $V$)

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