The net current flowing in the given circuit | vedclass

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(A) In a steady state,the capacitor acts as an open circuit. Therefore,no current flows through the branch containing the capacitor.Looking at the cir

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(A) In a steady state,the capacitor acts as an open circuit. Therefore,no current flows through the branch containing the capacitor.Looking at the circuit,the $2 	ext{V}$ battery is connected in parallel with the $4 \Omega$ resistor.The current through the $4 \Omega$ resistor is $I_1 = \frac{2 	ext{V}}{4 \Omega} = 0.5 	ext{A}$.The rest of the circuit (the branch with $2 \Omega$,$3 \Omega$,$6 \Omega$ resistors) is also connected in parallel to the battery.The equivalent resistance of the $3 \Omega$ and $6 \Omega$ resistors in parallel is $R_p = \frac{3 	imes 6}{3 + 6} = 2 \Omega$.This $R_p$ is in series with the $2 \Omega$ resistor,so the total resistance of this branch is $R_{branch} = 2 \Omega + 2 \Omega = 4 \Omega$.The current through this branch is $I_2 = \frac{2 	ext{V}}{4 \Omega} = 0.5 	ext{A}$.The total current flowing from the battery is $I = I_1 + I_2 = 0.5 	ext{A} + 0.5 	ext{A} = 1 	ext{A}$.

The net current flowing in the given circuit is . . . . . . $\text{A}$.

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