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
- Ar 1 - r 2
- Br 1 r 2 r 1 + r 2
- Cr 1 + r 2 2
- 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