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(D) The total $emf$ of the circuit is $E + E = 2E$. The total resistance of the circuit is $R + R_1 + R_2$. Thus,the current in the circuit is $i = \f

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$R = R_2 - R_1$

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(D) The total $emf$ of the circuit is $E + E = 2E$. The total resistance of the circuit is $R + R_1 + R_2$. Thus,the current in the circuit is $i = \frac{2E}{R + R_1 + R_2}$. For the source with internal resistance $R_2$,the potential difference $V$ is given by $V = E - i R_2$. Given that $V = 0$,we have $E - i R_2 = 0$,which implies $E = i R_2$. Substituting the value of $i$,we get $E = \left( \frac{2E}{R + R_1 + R_2} \right) R_2$. Dividing both sides by $E$,we get $1 = \frac{2 R_2}{R + R_1 + R_2}$. Therefore,$R + R_1 + R_2 = 2 R_2$,which simplifies to $R = R_2 - R_1$.

Two sources of equal $emf$ are connected to an external resistance $R$. The internal resistances of the two sources are $R_1$ and $R_2$ $(R_2 > R_1)$. If the potential difference across the source having internal resistance $R_2$ is zero,then:

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