Two sources of equal emf E are connected in s | vedclass

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(D) The total $emf$ of the series combination is $E_{net} = E + E = 2E$.The total resistance of the circuit is $R_{total} = R + R_1 + R_2$.The current

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$R = R_2 - R_1$

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(D) The total $emf$ of the series combination is $E_{net} = E + E = 2E$.The total resistance of the circuit is $R_{total} = R + R_1 + R_2$.The current flowing through the circuit is $I = \frac{E_{net}}{R_{total}} = \frac{2E}{R + R_1 + R_2}$.The potential difference across the source with internal resistance $R_2$ is given by $V = E - IR_2$.Given that the potential difference across this source is zero,we have $E - IR_2 = 0$,which implies $E = IR_2$.Substituting the value of $I$ into this equation:$E = \left( \frac{2E}{R + R_1 + R_2} ight) R_2$.Dividing both sides by $E$ (assuming $E 
eq 0$):$1 = \frac{2R_2}{R + R_1 + R_2}$.$R + R_1 + R_2 = 2R_2$.$R = 2R_2 - R_2 - R_1$.$R = R_2 - R_1$.

Two sources of equal $emf$ $E$ are connected in series with 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 of internal resistance $R_2$ is zero,then the value of $R$ is:

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