In the circuit shown,the potential difference | vedclass

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(D) In the given circuit,the points $X$ and $Y$ are open. Because the circuit is open,no current flows through the resistors $(I = 0 \ A)$. According

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$120$

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(D) In the given circuit,the points $X$ and $Y$ are open. Because the circuit is open,no current flows through the resistors $(I = 0 \ A)$. According to Ohm's law,the potential drop across the resistors is $V_R = I 	imes R = 0 	imes R = 0 \ V$. Therefore,the entire potential difference of the battery appears across the open terminals $X$ and $Y$. Thus,the potential difference between $X$ and $Y$ is equal to the $EMF$ of the battery,which is $120 \ V$.

In the circuit shown,the potential difference between $X$ and $Y$ will be ............. $V$.

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