Question

bulb is rated at 200v -400w what is its resistance? 5 such bulbs are lighted for 5 hours. calculate the electrical energy consumed. find the cost if the ratio is rs 5.10 per kwh

Answer 1

p=400w
v=200v

a) calculation of resistance
p=v2×r

400=200×200÷r
r=200×200÷400
r=100ohm

b)calculation of energy consumed
t=5 hour
contverting power in kw
p= 400÷1000
p =0.4kw

e=p×t
e=0.4×5
e=2kwh
energy consumed by 1 bulb=2kwh
energy consumed by 5 bulb=5×2=10kwh

c) calculation of cost of electical energy
cost of 1 unit electricity= rs 5.10
cost of 2 unit electricity = rs10.20

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