Question

A 12.5 V Battery operates s 14.2 ohms flashlight intermittently for a cumulative

time of 65.3 min. (a) What is the average current? (b) What quantity of electricity

is furnished by the battery?

Answer 1

(1)We know that

$I=\frac{V}{R}=\frac{12.5}{14.2}=0.88A$

(2)

Power

$P=i^2R$

$P=\frac{12.5^2}{14.2}=11.003W$

Energy

$E=P\times t$

$$E=11.003\times65.3\times60=43111.79J$$