
Current through the wire will be,
I=R+r1+r22ϵ.
Now potential difference across cell 1 will be, ϵ−Ir1=0
⇒ϵ=R+r1+r22ϵ×r1⇒R+r1+r2=2r1⇒R=r1−r2
JEE Main 2022 — Physics Electromagnetism
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
Held on 27 Jul 2022 · Verified 6 Jul 2026.
r1−r2
r1+r2r1r2
2r1+r2
r2−r1
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