Two identical bulbs are connected in parallel | vedclass

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(C) The source of emf $E$ is ideal,meaning it has zero internal resistance. Therefore,the terminal voltage across the parallel combination of the bulb

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Reading of $A$ will decrease but that of $V$ will remain same

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(C) The source of emf $E$ is ideal,meaning it has zero internal resistance. Therefore,the terminal voltage across the parallel combination of the bulbs is always equal to the emf $E$ of the source.Since the voltmeter $V$ is connected in parallel with the source,it will always read the emf $E$,regardless of whether bulb $B_2$ is fused or not. Thus,the reading of $V$ remains constant.Initially,both bulbs $B_1$ and $B_2$ are in parallel,so the total resistance of the circuit is $R_{eq} = R/2$,where $R$ is the resistance of each bulb. The current measured by the ammeter $A$ is $I = E / (R/2) = 2E/R$.When bulb $B_2$ gets fused,it acts as an open circuit. The circuit now only contains bulb $B_1$. The new total resistance is $R_{eq}' = R$. The new current measured by the ammeter $A$ is $I' = E/R$.Comparing the two,$I' < I$. Therefore,the reading of the ammeter $A$ decreases. Hence,the correct option is $C$.

Two identical bulbs are connected in parallel across an ideal source of emf $E$. The ammeter $A$ and voltmeter $V$ are ideal. If bulb $B_2$ gets fused,then

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