After closing the key in the given circuit: ( | vedclass

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(D) When the key is closed,the circuit consists of two parallel branches connected to a $DC$ voltage source $V$. In the branch containing bulb $(2)$,t

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(D) When the key is closed,the circuit consists of two parallel branches connected to a $DC$ voltage source $V$. In the branch containing bulb $(2)$,the current is $I_2 = V/R_2$,where $R_2$ is the resistance of bulb $(2)$. This current is constant,so bulb $(2)$ glows with constant brightness. In the branch containing bulb $(1)$ and capacitor $C$,the capacitor starts charging. Initially,it acts as a short circuit,allowing maximum current to flow. As the capacitor charges,the potential difference across it increases,opposing the source voltage. At steady state,the capacitor is fully charged and acts as an open circuit (infinite resistance),so the current in this branch becomes zero. Thus,bulb $(1)$ gradually dims and eventually goes dark. Therefore,both statements $(a)$ and $(b)$ are correct.

After closing the key in the given circuit:
$(a)$ Bulb $(2)$ will glow and maintain its brightness.
$(b)$ Brightness of bulb $(1)$ will gradually decrease and at steady state it goes completely dark.

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After closing the key in the given circuit: $(a)$ Bulb $(2)$ will glow and maintain its brightness. $(b)$ Brightness of bulb $(1)$ will gradually decrease and at steady state it goes completely dark.