Question

A current of 0.5 A flows through the filament of a bulb with a resistance of 500 Ω for 100 s.

1. What is the power of bulb?
2. What is magnitude of energy consumed by bulb in watt hours?

Answer 1

Given:

Current (I) = 0.5 A

Resistance (R) = 500 Ω

Time (t) = 100 s

(i) P = I² × R

= (0.5)² × 500

= 0.25 × 500

= 125 W

(ii) Convert time from seconds to hours:

Since there are 3600 seconds in 1 hour:

\[t_{\text{hours}} = \frac{100}{3600}\text{hours} = \frac{1}{36}\text{hours}\]

Calculate energy in Watt-hours (Wh):

\(\text{Energy\ }(E)=\text{Power\ }(P)\times \text{Time\ in\ hours\ }(t_{\text{hours}})\)

\[= 125\text{W} \times \frac{1}{36}\text{hours}\]

\[=\frac{125}{36}\text{\ Wh}\]

\(E\approx {3.47}\text{\ Wh}\)