In the given circuit, what are the potential | vedclass

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(C) $1$. Analyze the circuit: The capacitors are connected in a way that simplifies to two equivalent capacitors in series between points $A$ and $C$.

Vedclass · Accepted Answer

$V_{AB} = 25 \, V, V_{BC} = 75 \, V$

Vedclass · Answer

(C) $1$. Analyze the circuit: The capacitors are connected in a way that simplifies to two equivalent capacitors in series between points $A$ and $C$. $2$. Simplify the circuit: The parallel combinations of $3 \, \mu F$ and $3 \, \mu F$ result in $6 \, \mu F$, and $1 \, \mu F$ and $1 \, \mu F$ result in $2 \, \mu F$. $3$. The circuit effectively becomes two capacitors of $C_1 = 6 \, \mu F$ and $C_2 = 2 \, \mu F$ in series across a $100 \, V$ source. $4$. For capacitors in series, the potential difference across each is inversely proportional to its capacitance: $V_1 = V_{AB} = \left( \frac{C_2}{C_1 + C_2} ight) V$ and $V_2 = V_{BC} = \left( \frac{C_1}{C_1 + C_2} ight) V$. $5$. Substituting the values: $V_{AB} = \left( \frac{2}{6 + 2} ight) 	imes 100 = \left( \frac{2}{8} ight) 	imes 100 = 25 \, V$. $6$. Similarly, $V_{BC} = \left( \frac{6}{6 + 2} ight) 	imes 100 = \left( \frac{6}{8} ight) 	imes 100 = 75 \, V$. $7$. Thus, $V_{AB} = 25 \, V$ and $V_{BC} = 75 \, V$.

In the given circuit, what are the potential differences between points $A$ and $B$ $(V_{AB})$ and between points $B$ and $C$ $(V_{BC})$?

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