Solved numerical of ch. 8 based on electricity

 1. An electrical appliance is rated as 60 watt - 150 volt.

a) what do you understand by this statement?

b) how much current will flow through the appliance when in use?

Ans. 1a. Electrical appliance is rated as 60 watt - 150 volt.

It means that if the appliance is used on a 150 volt supply the electric power consumed by it is 60 watt.i.e., 60 joule of electrical energy is consumed by the appliance in one second.

1b. Power rating as 60W -  150V.

P=60W

V= 150 V

I=?

P=VI

⇒P/V= I

⇒I=P/V

⇒I=60W / 150 V

⇒I= 0.4 A

2. An electric iron of power 1.5 KW is used for 30 minutes to press clothes. Calculate the electrical energy consumed in

a) kilowatt hour

b) joule

Ans. P=1.5 kW

t=30 minutes

  = 30 × 1/ 60  H

P= W/ t

⇒P × t = W

⇒ w = P × t

 ⇒ w= 1.5 kW × 1/2  

⇒ W = 0.75 kWh

2b) 

W= 0.75 kWh

     = 0.75 × 1 kWh

     = 0.75 × 3.6 × 10⁶ J    (  ∵ 1kWh = 3.6 × 10⁶ J )

     = 2.7 × 10⁶ J

3. Assuming the electric consumption per day to be 12 kilowatt hour and the rate of electricity to be rupees 6.25 per unit, find how much money is to be paid in a month of 30 days?

Ans. Energy consumption in one day= 12kWh

Energy consumption in 30 days = 12 kWh ×30

                                                           = 360 kWh

The rate of electricity is to be paid is 6.25 per unit.

Money to be paid in a month= energy consumed × cost of 1 unit

= 360 kWh × ₹ 6.25 / 1kWh

= ₹360× 6.25

= ₹ 2250.0

4. In a premise 5 bulbs each of 100 watt, 2fans each of 60 watt,2 AC s each of 1.5 kilowatt hour used for 5 hour per day.find:

A. Total power consumed per day

B. Total power consumed in 30 days

C. Total electrical energy consumed in 30 days,

D. The cost of electricity at the rate of rupees 6.25 per unit

Ans. Power of one bulb= 100W

∴ power of five bulbs = 5× 100W 

     =500W

Power of one fan= 60W

Power of 2 fan= 60W × 2 

  = 120W

Power of one AC = 1.5 kW

Power of 2 AC = 1.5kW × 2

  = 3.0 kW

  = 3.0 × 1kW

   = 3.0 × 1000W

   = 3000W

 ∴ total power of 5 bulbs 2 fans and 2 AC =500W + 120W + 3000 W

  = 3620 W

Total power consumed in 30 days = power consumed in one day × number of days

  = 3620W × 30 

   = 108600 W

    = 108.6 × 10³ W

    = 108.6 × 1 k W

    = 108.6 kW

Total electrical energy consumed in 30 days = P ×  t

   = 108.6 × 10³ W × 5 h

   = 543 kWh

The cost of electricity at the rate of 6.25 per unit = electrical energy consumed × cost of 1 unit

  = 543 kWh × ₹ 6.25 / 1kWh

   = 543 × ₹ 6.25

   = ₹ 3393.75