Question

a bulb is rated at 200V_40W.what is its resistance?5 such bulbs are lighted for 5 hours. calculate the electrical energy consumed . find the cost if the rate is Rs 5.10perKWH.

Answer 1

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