Question

A lamp consumes 500 J of electrical energy in 20 seconds. What is the power of the lamp?

Also find the electrical energy consumed in units if the lamp operates for 2 hours daily for 15 days.​

Answer 1

• Electrical energy consumed, E = 500 J
• Time, t = 20 seconds.

We know that,

Power = Energy / Time

$P = \dfrac{500}{20} \\ \\ P = \dfrac{50}{2} \\ \\ P = 25 \: watt$

The power of the lamp is 25 watt.

If the lamp operates for 2 hours daily for 15 days.

Let's calculate the time in hours.

One day = 2 hours.

15 days = 15 × 2 hours.

• Time, t = 30 hours.

$1 \: hour \: = 60 \: minutes \\ \\ 30 \: hours \: =30 \times 60 \: minutes \\ \\ t = 1800 \: minutes \\ \\ 1 \: minute \: = 60 \: seconds \\ \\ 1800 \: minutes = 1800 \times 60 \: seconds \\ \\ t = 108000 \: seconds$

• Power, p = 25 watt.

Power = Energy /Time

Energy = Power × Time

$E= 25 \times 108000 \\ \\ = 2700000 \: joules \\ \\ E = 2.7 \times {10}^{6} joules \\ \\ 1 \: unit \: of \: energy \: = 3.6 \times {10}^{6} \: joules \\ \\ x \: unit \: of \: energy \: = 2.7 \times {10}^{6} \: joules \\ \\ On \: dividing \: both \: we \: get , \\ \\ \dfrac{1}{x} = \dfrac{3.6 \times {10}^{6} }{2.7 \times {10}^{6} } \\ \\ \dfrac{1}{x} = \dfrac{3.6}{2.7} \\ \\ x = \dfrac{2.7 \times 10}{3.6 \times 10} \\ \\ x = \dfrac{27}{36} \\ \\ x = \dfrac{3}{4} \\ \\ x = 0.75 \: units$

Electrical energy consumed in units is 0.75 units.

Answer 2

Given:-

$\bigstar$Consumption of electric energy by lamp = 500 J.

$\bigstar$Time taken to consumption of electric energy = 20 S.

To Find:-

$\bigstar$Power of the lamp.

$\bigstar$Consumption of electric energy by lamp for 2 hours daily of 15 days.

Solution:-

$\bigstar$To Find power using formula.

${ \underline{ \boxed{ \boxed{ \mathsf{power= \left( { \frac{energy \: consumed}{duration \: of \: consumption} } \right) }}}}}$

$\\ \longmapsto \sf \: power = \frac{500}{20}$

$\longmapsto \bf \: power = 25 \: watts$

To find the energy consumed in 2 hours of 15 days.

${ \underline{ \boxed{ \boxed{ \mathsf{energy \: consumed \: in \: 15 \: days = \left(25 \times { \frac{2}{15} } \right) = 750\:WH}}}}}$

$\\ \longmapsto \bf \: energy \:consumed \: in \: 15 \: days = 0.75\:units$