Question

A bulb is rated at 330V- 110W. What do you think is its resistance? Three such bulbs burn for

5hrs at a stretch. What is the energy consumed? Calculate the cost in rupees if the rate is 70

paise per unit?​

Answer 1

Answer:

Explanation:

we know that for an electric appliance

P = V2/R

or resistance will be given as R = V2 / P  

So,

P = 110w

V = 330 volts

R = 3302 / 110

= 108,900/110 = 990

thus, resistance of the bulb is 990

now, electric energy consumed will be

E = number of units x power of each unit x time  = n x P x t

n = 3

P = 110 W

t = 5 hours

so,

E = 3 x 110 × 5

thus,

E = 1650Wh

..

the total cost of electricity = total unit of energy consumed x cost per unit

= 1650 x 0.7 Rs

thus,

total cost= 1155rs.

and

 monthly cost = total daily cost x number of days in month =

1155rs. × 30 = 34,650rs.

Answer 2

Answer:

Explanation:

Well here we're given with potential difference and power, and the question asks us to find resistance, hence we must use the formula

$P = V^{2}/R$

Now by substituting the given values,

$110W = (330)^2/R$

$R = \frac{330 * 330}{110}$

$R = 990ohms$

Now, energy is heat expended or power times time; therefore,

$H = Pt\\H = 110 * 5 \\H = 550Wh$

Therefore energy consumed by 3 bulbs will be,

$H = 3*550\\H= 1650Wh\\H = 1.650kWh$

Since 1 unit = 1 kWh

Therefore,

Cost = 1.65*0.7

∴Cost = ₹1.155