Computer Science, asked by rishu5816, 1 year ago

The bulb in a flashlight draws 1.7A with new batteries producing a potential difference of 3V. What is the current, when the voltage of the batteries drops to 2.5V?

Answers

Answered by rampaul90

4

Answer:

1.42A

Explanation:

resistance of the bulb is R=V/l

=(3/1.7)ohm

current drawn in 2nd case I=V/R

=2.5/(3/1.7)=(2.5×1.7/3)=1.42A

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