In the figure below, what is the potential di | vedclass

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(C) In steady state, capacitors act as open circuits. The current flows only through the resistors. The total resistance in the circuit is $R_{eq} = 2

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) In steady state, capacitors act as open circuits. The current flows only through the resistors. The total resistance in the circuit is $R_{eq} = 20\, \Omega + 10\, \Omega = 30\, \Omega$. The current in the circuit is $I = \frac{V}{R_{eq}} = \frac{100\, V}{30\, \Omega} = \frac{10}{3}\, A$. The potential at point $A$ is $V_A$ and at point $C$ is $V_C$. The potential difference $V_{AC} = V_A - V_C = 100\, V$. Let the potential at point $B$ be $V_B$. The capacitors form a bridge-like structure. The equivalent capacitance between $A$ and $B$ is $C_{AB} = 3\, \mu F + 3\, \mu F = 6\, \mu F$ (parallel combination). The equivalent capacitance between $B$ and $C$ is $C_{BC} = 1\, \mu F + 1\, \mu F = 2\, \mu F$ (parallel combination). Using the principle of voltage division for capacitors in series (since the charge $Q$ on the branches $AB$ and $BC$ must be equal in steady state), we have $V_{AB} = V_{total} \cdot \frac{C_{BC}}{C_{AB} + C_{BC}} = 100 \cdot \frac{2}{6+2} = 100 \cdot \frac{2}{8} = 25\, V$. Similarly, $V_{BC} = V_{total} \cdot \frac{C_{AB}}{C_{AB} + C_{BC}} = 100 \cdot \frac{6}{6+2} = 100 \cdot \frac{6}{8} = 75\, V$. Thus, $V_{AB} = 25\, V$ and $V_{BC} = 75\, V$.

In the figure below, what is the potential difference between the points $A$ and $B$ and between $B$ and $C$ respectively in steady state?

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