Question

in a tv of rating 150 W is operated for 5hours per day.Find the number of this unit consumed in any leap year​

Answer 1

Step-by-step explanation:

ANSWER

It is given that the total power of the TV is 100 W and it is operated 6 hr/day.

The total number of days in a leap year is 366 days. Hence, the total time taken in a leap year is given as 6hours×366days=2196hours.

The electric power in watts associated with a complete electric circuit or a circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other form, e.g., heat, mechanical energy, or energy stored in electric fields or magnetic fields. The power is given by the product of applied voltage and the electric current. That is, P=VI.

The heat produced or the energy dissipated by an appliance is given by the product of power and time taken in seconds. That is, Q=Pt.

So, Q=P×t=100W×2196hrs=219600Wh=219.6kWh.

1 kWh = 1 unit.

Hence, the number of units consumed by the TV in any leap year is 219.6 units.

Answer 2

Answer:

Given: Power (P) = 100 W

P = 100

1000

kW = 1

10

kW

Time (t) = 6 × 366 = 2196 hrs

To find: Unit consumed = ?

Formula: H = P × t

Solution: H = P × t

    H = 1

10

× 2196

H = 219.6 kWh

Ans. Number of units consumed is 219.6

Step-by-step explanation: