A student who is a member of energy club replace forty CFL'S of 20w with 9w LED bulb, which is continusly for 8 hour a day in his cattle farm. What is the amount of energy he saved in a month?
Answers
Given
- A student replaces CFL bulbs with LED bulbs
- CFL = 40 bulbs of 20 W each
- LED = 40 bulbs of 9 W each
- Time = 8 hours
To Find
- Amount of energy saved
Solution
● E = P × t
● First convert all the power to kW and then simple substitution would give us the energy consumed then their difference would be our answer!!
✭ Converting Units :
- CFL = 20 W = 20/1000 = 0.02 kW
- LED = 9 W = 9/1000 = 0.009 kW
✭ In the case of CFL :
→ E = P × T
→ E = 0.02 × 8
→ E = 0.16 kWh
For 40 such bulbs
- 40 × 0.16 = 6.4 kWh
✭ In the case of LED :
→ E = P × t
→ E = 0.009 × 8
→ E = 0.072 kWh
For 40 such bulbs
- 40 × 0.072 = 2.88 kWh
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✭ Energy Saved :
→ Energy Saved = Energy Consumed by CFL - Energy Consumed by LED
→ Energy Saved = 6.4 - 2.88
→ Energy Saved = 3.52 kWh
∴ The energy saved is 3.52 kWh
Answer:
Given:
CFL = 40 bulbs of 20 W
LED = 40 bulbs of 9 W
Time = 8 hours
To Find:
Amount of energy saved in a month
Solution:
First we need to convert the power to KW
★ Converting Units:
CFL = 20 W
= 20/1000
= 0.02 kW
LED = 9 W
= 9/1000
= 0.009 kW
★ Case of CFL:
E = P × T
E = 0.02 × 8
E = 0.16kWh
→ 40 × 0.16 = 6.4 kWh
Energy consumed by CFL = 6.4 kWh
★ Case of LED:
E = P × T
E = 0.009 × 8
E = 0.072 kWh
→ 40 × 0.072 = 2.88 kWh
Energy consumed by LED = 2.88 kWh
_______________________
→ Energy saved = Energy consumed by CFL - Energy consumed by LED
→ Energy saved = 6.4 - 2.88
→ Energy saved = 3.52 kWh
•°• The Amount of energy he saved in a month = 3.52 kWh