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: (a) r₁ - r₂ (b) (r₁ r₂)/(r₁+r₂) (c) (r₁+r₂)/(2) (d) r₂ - r₁

Answer: (a). I = (2E)/(R + r₁ + r₂) P.d. across source r₁ is zero: E - Ir₁ = 0 ⇒ I = E/r₁ R + r₁ + r₂ = 2r₁ ⇒ R = r₁ - r₂. So the answer is (a).

Two sources of equal emfs are connected in series. This combination… | JEE Main 2022 (27 Jul, Shift 1)

Q20JEE Main 2022 · 27 JUL · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations

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:

Show answer and solution

Answer: (a)

$I = \frac{2E}{R + r_1 + r_2}$
P.d. across source $r_1$ is zero: $E - Ir_1 = 0 \Rightarrow I = \frac{E}{r_1}$
$R + r_1 + r_2 = 2r_1 \Rightarrow R = r_1 - r_2$ . So the answer is (a).

Report a mistake

Cells: EMF, Internal Resistance & Combinations question from JEE Main 2022 (27 Jul, Shift 1)

More questions on Cells: EMF, Internal Resistance & Combinations