Let the emf of the battery be 'V' and resistance be r
In position 1
V-2r=0
thus, V=2r
In position 2,
E-ix+V-i:)
- Prankster piyush * E-ix+V-ir=0 10-2x+2r-2r = 0 thus, x = 5 ohm :)
- Akash Anand Good work
Let the emf of the battery be 'V' and resistance be r
In position 1
V-2r=0
thus, V=2r
In position 2,
E-ix+V-i:)