Question

Please solve this question....
Dont spam⚠️⚠️
And the answer must be 1161.22....​

Answer 1

Answer:

Rs 1177.416

Explanation:

Energy consumed by the 20 bulbs in one day= (number of bulbs × power × hours of consumption) = 20 x 100 x 6 = 12,000 Wh

For Fan,

Given I = 0.5 A and V = 220 V

So, Power = Current × voltage = 0.5 × 220 = 110 Wh

Energy consumed by the 10 fans in one day = 10 × 110 × 6 = 6600 Wh

For Electric kettle

Given R = 100 Ohms and V = 220 V Using Ohms Law = I = V/R Amp   therefore, I = 220 V/100 Ohms = 2.2 Amp

So, Power = Current × voltage = 2.2 × 220 = 484 Wh

Energy consumed by the 1 electric kettle in one day = 1 × 484 × 6 = 2904 Wh

Total energy consumption in one day = (12000 + 2904 + 6600)Wh = 21,504 W-h

Total energy consumption in the month of November (30 days) = 30 × 21,504 = 654,120 W-h = 654.12 kW-h = 654.12 units

So, the cost will be 654.12 × 1.8 = Rs 1177.416  

Hope this information will clear your doubts about the topic.

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Answer 2

$\huge\red{\mid{\fbox{\tt{\:QUESTION}}\mid}}$

A house is fitted with 20 bulbs rated 100w each ,10 fans each running at 0.5A current and an electric kettle of 100 ohm resistance electricity is supplied at a rate of rs 1.80 per unit what will be the expendirure in the month of November if all the appliances are used for 6 hours daily.

$\bold{\huge{\fbox{\color{blue}{SOLUTION}}}}$

_________

- GIVEN

I= 0.5A

let V=200

_________

Number of electric bulbs= 20

Time= 6 hours

Energy consumed by bulb= (100×6)

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀=600 wh

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀( wh= watt hour)

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 0.6 kwh( kilo watt hour)

⠀⠀⠀⠀⠀⠀⠀20×0.6= 12kwh

____________

number of fans= 10

time= 6hrs.

power= V×I

= 10×200×$\frac{1}{2}$×6

=6000wh

= 0.6kwh

____________

energy..For kettle= 1×$\frac{V}{R²}$×6

= $\frac{200×200}{100}$ ×6

=2400wh

=2.4 kwh

________

Total energy consumed in 1 daY=(12+0.6+20) kwh

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 15kwh

Energy for month November =30×15 kwh

cost⠀⠀⠀⠀⠀⠀⠀⠀=1.80×30×20.5

$\bold{\huge{\fbox{\color{black}{1107.22approx.}}}}$

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