JEE Main 2022 · 27 Jul · Shift 1 · Physics Q41

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PhysicsSingle correct+4 / -1Question 41 of 90

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 $r_2$ $(r_1>r_2)$ . If the potential difference across the source of internal resistance $r_1$ is zero, then the value of R will be :

r_1-r_2

\dfrac{r_1r_2}{r_1+r_2}

\dfrac{r_1+r_2}2

r_2-r_1

Answer and worked solution

Official answer: A

Current $I=\frac{2E}{R+r_1+r_2}$ . Zero terminal voltage means $E=Ir_1$ .

So $R+r_1+r_2=2r_1$ and $R=r_1-r_2$ . (A)

Topics:EMF and internal resistance

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JEE Main 2022 · 27 Jul · Shift 1, Physics question 41