What will be the current through the 200 Omeg | vedclass

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(C) In a steady state, a capacitor acts as an open circuit, meaning no direct current flows through the branches containing capacitors.Therefore, we c

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$10 mA$

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(C) In a steady state, a capacitor acts as an open circuit, meaning no direct current flows through the branches containing capacitors.Therefore, we can neglect the branches with the $1 \mu F$ and $2 \mu F$ capacitors.The circuit simplifies to a single series loop containing the $6 V$ battery, the $200 \Omega$ resistor, and the $400 \Omega$ resistor.The total resistance of the circuit is $R_{	ext{net}} = 200 \Omega + 400 \Omega = 600 \Omega$.Using Ohm's law, the current $I$ in the circuit is:$I = \frac{V}{R_{	ext{net}}} = \frac{6 V}{600 \Omega} = 0.01 A$.Converting to milliamperes, $I = 0.01 	imes 1000 mA = 10 mA$.

What will be the current through the $200 \Omega$ resistor in the given circuit, a long time after the switch $K$ is closed?

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