Question

A buld is rated at 200v-40w. What is its resistance? 5 such bulbs are lighted for 5 hours. Calculate electrical energy consumed. Find the cost if the rate is Rs 5.10 per KWH

Answer 1

V= 200V
P=40W
P=V^2/R
R=V^2/P
=200×200/40= 1000ohm
TOT P=5×40=200W =0.2kW
E=0.2×5=1kWh
Cost=Rs5.10

Answer 2

Given, V = 200 V, P = 40 W,

As per ohm’s law, P = VI

$\bf\huge I = \frac{P}{V}$

= $\bf\huge\frac{40}{200}$

= $\bf\huge\frac{1}{5}A$

$\bf\huge R = \frac{V}{I}$ = $\bf\huge\frac{200V}{(Y5)A}$

= 200 × 5 Ω

= 1000 Ω

Total power = 40W × 5 = 200W

Time = 5 hours

Electrical energy = 200W × 5 hours

= 1000 Wh

= 1 kWh

Cost of 1 kWh = 5.10 Rs