Two cells of internal resistance r_{1} and r_ | vedclass

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(C) Since the two cells are connected in series,the total emf is $E_{total} = E + E = 2E$,where $E$ is the emf of each cell.The total resistance of th

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$R=r_{1}-r_{2}$

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(C) Since the two cells are connected in series,the total emf is $E_{total} = E + E = 2E$,where $E$ is the emf of each cell.The total resistance of the circuit is $R_{total} = R + r_{1} + r_{2}$.The current $I$ flowing through the circuit is given by $I = \frac{2E}{R + r_{1} + r_{2}}$.The terminal potential difference $V_{1}$ across the cell with internal resistance $r_{1}$ is given by $V_{1} = E - Ir_{1}$.Given that $V_{1} = 0$,we have $E - Ir_{1} = 0$,which implies $E = Ir_{1}$.Substituting the value of $I$,we get $E = \left( \frac{2E}{R + r_{1} + r_{2}} \right) r_{1}$.Dividing both sides by $E$,we get $1 = \frac{2r_{1}}{R + r_{1} + r_{2}}$.Rearranging the terms,$R + r_{1} + r_{2} = 2r_{1}$.Therefore,$R = 2r_{1} - r_{1} - r_{2} = r_{1} - r_{2}$.

Two cells of internal resistance $r_{1}$ and $r_{2}$ and of the same emf are connected in series across a resistor of resistance $R$. If the terminal potential difference across the cell of internal resistance $r_{1}$ is zero,then the value of $R$ is

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