In the given figure,the steady state current | vedclass

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(D) In the steady state condition,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.Therefore

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$1.5$

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(D) In the steady state condition,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.Therefore,the branch with the $C = 0.2 \ \mu F$ capacitor and $4 \ \Omega$ resistor can be ignored.The circuit effectively consists of the $6 \ V$ battery in series with the parallel combination of $2 \ \Omega$ and $3 \ \Omega$ resistors,and the $2.8 \ \Omega$ resistor.First,calculate the equivalent resistance of the parallel combination of $2 \ \Omega$ and $3 \ \Omega$ resistors:$R_p = \frac{2 	imes 3}{2 + 3} = \frac{6}{5} = 1.2 \ \Omega$.Now,the total resistance $R_{eq}$ of the circuit is the sum of $R_p$ and the $2.8 \ \Omega$ resistor:$R_{eq} = 1.2 \ \Omega + 2.8 \ \Omega = 4.0 \ \Omega$.Using Ohm's law,the steady state current $I$ is:$I = \frac{V}{R_{eq}} = \frac{6 \ V}{4 \ \Omega} = 1.5 \ A$.

In the given figure,the steady state current $I$ is ................. $A$.

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