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  4. Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are r1 and r2(r1>r2). If the potential difference across the source of internal resistance r1 is zero then the value of R will be

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Q. Two sources of equal emfs are connected in series. This combination is connected to an external resistance $R$. The internal resistances of the two sources are $r_1$ and $r_2\left(r_1>r_2\right)$. If the potential difference across the source of internal resistance $r_1$ is zero then the value of $R$ will be

JEE MainJEE Main 2022Current Electricity

Solution:

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$ I =\frac{2 E }{ r _1+ r _2+ R } $
$ IR = E - Ir _2 $
$ I \left( R + r _2\right)= E $
$I =\frac{ E }{ R + r _2}$
$ \frac{2 E }{ r _1+ r _2+ R }=\frac{ E }{ R + r _2}$
$ 2 R +2 r _2= r _1+ r _2+ R $
$ R = r _1- r _2$