Two sources of equal emfs are connected in series. This combination is connected to an external resistance R…

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 r2r1>r2. If the potential difference across the source of internal resistance r1 is zero then the value of R will be
  1. r1-r2
  2. r1r2r1+r2
  3. r1+r22
  4. r2-r1

Solution

Current through the wire will be,

I=2εR+r1+r2.

Now potential difference across cell 1 will be, ε-Ir1=0

⇒ε=2εR+r1+r2×r1⇒R+r1+r2=2r1⇒R=r1-r2

Asked in: JEE Main 2022 (27 Jul Shift 1)

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