In the circuit shown,the potential difference | vedclass

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(B) The circuit consists of three parallel branches connected between points $D$ and $C$. Each branch contains a battery of $EMF$ $E$ and internal res

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$2$

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(B) The circuit consists of three parallel branches connected between points $D$ and $C$. Each branch contains a battery of $EMF$ $E$ and internal resistance $r$. For the parallel combination of batteries,the equivalent $EMF$ $E_{eq}$ and equivalent internal resistance $r_{eq}$ are given by:$E_{eq} = \frac{\sum \frac{E_i}{r_i}}{\sum \frac{1}{r_i}} = \frac{\frac{1}{1} + \frac{2}{1} + \frac{3}{1}}{\frac{1}{1} + \frac{1}{1} + \frac{1}{1}} = \frac{6}{3} = 2 \, V$$r_{eq} = \frac{1}{\sum \frac{1}{r_i}} = \frac{1}{1+1+1} = \frac{1}{3} \, \Omega$Since there is no current flowing through the external resistors ($5 \, \Omega$ and $10 \, \Omega$) because the circuit is open at $A$ and $B$,the potential difference between $A$ and $B$ is equal to the potential difference between $D$ and $C$.Therefore,$V_{AB} = V_{DC} = E_{eq} = 2 \, V$.

In the circuit shown,the potential difference between $A$ and $B$ is ............. $V$.

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