Question

An electric bulb of resistance of 500 $\Omega$, draws a current of 0.4 A. Calculate the power of bulb and Potential Difference at its ends.

Answer 1

The resistance of the bulb is R = 500A and it draws a current of 0.4 A

So the power dissipated will be given as

p =$i^{2}R$

= $\frac{4}{10}\times \frac{4}{10}\times$500

=80 w

also, p = VI

or, 80 = V $\times$0.4

V = $\frac{80\times 10}{4}$

V = 200V