Resistance of flashlight is 1.5ohm
It uses 3cells each of emf 1.5v and internal resistance 0.25 what is the net potential difference across the lamp
Answers
Answered by
1
Answer:
The emf of cell is, E=1.5V
The current in circuit is, I=15A
The resistance of ammeter in circuit is, R
A
=0.04Ω
The equation for current in circuit is
I=
r+R
A
E
where, r is the internal resistance of cell.
15=
r+0.04
1.5
r+0.04=
15
1.5
=0.1
r=0.1−0.04=0.06Ω
Similar questions