Physics, asked by drover, 9 months ago

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 ichaappu
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

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