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JEE Main202227 Jul 2022Morning ShiftPhysicsCurrent ElectricityActual

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

Options

  1. Ar 1 - r 2
  2. Br 1 r 2 r 1 + r 2
  3. Cr 1 + r 2 2
  4. Dr 2 - r 1

Correct answer

A. r 1 - r 2

Step-by-step solution

Current through the wire will be, I = 2 ε R + r 1 + r 2 . Now potential difference across cell 1 will be, ε - I r 1 = 0 ⇒ ε = 2 ε R + r 1 + r 2 × r 1 ⇒ R + r 1 + r 2 = 2 r 1 ⇒ R = r 1 - r 2

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