Consider the circuit shown. If all the cells | vedclass

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(B) In a steady state, the capacitor acts as an open circuit, meaning no current flows through the branch containing the capacitor.Let the current in

Vedclass · Accepted Answer

$0.29$

Vedclass · Answer

(B) In a steady state, the capacitor acts as an open circuit, meaning no current flows through the branch containing the capacitor.Let the current in the left loop be $I$. The circuit simplifies to a single loop consisting of the $2 \ V$ cell, the $2 \ V$ cell, the $4 \ \Omega$ resistor, the $2 \ \Omega$ resistor, and the $8 \ \Omega$ resistor in series.Applying Kirchhoff's Voltage Law $(KVL)$ to the loop:$2 \ V + 2 \ V - I(4 \ \Omega + 2 \ \Omega + 8 \ \Omega) = 0$$4 \ V = I(14 \ \Omega)$$I = \frac{4 \ V}{14 \ \Omega} = \frac{2}{7} \ A \approx 0.2857 \ A \approx 0.29 \ A$.Thus, the current through the $2 \ \Omega$ resistor is approximately $0.29 \ A$.

Consider the circuit shown. If all the cells have negligible internal resistance, what will be the current through the $2 \Omega$ resistor when steady state is reached (in $A$)?

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