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 $\mathrm{r}_{1}$ is zero then the value of R will be

(B) $\frac{\mathrm{r}_{1} r_{2}}{\mathrm{r}_{1}+r_{2}}$

(C) $\frac{\mathrm{r}_{1}+r_{2}}{2}$

(D) $\mathrm{r}_{2}-\mathrm{r}_{1}$

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Current Electricity

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

Answer: (A)

The current in the circuit is

I = \frac{2 E}{r _{1} + r _{2} + R}

. For the potential difference across the first source to be zero:

E - I r_{1} = 0

, so

I = \frac{E}{r _{1}}

. Also, considering the second source and external resistance:

I R = E - I r_{2}

, giving

I = \frac{E}{R + r _{2}}

. Equating the two expressions for

I

\frac{2 E}{r _{1} + r _{2} + R} = \frac{E}{R + r _{2}}

. Simplifying:

2 (R + r_{2}) = r_{1} + r_{2} + R

, which gives

R = r_{1} - r_{2}

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option