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Answer :
Answer:
= 0.1 KWh
Step by step:
We have given two lamps in such a way :
Power of 1st lamp, P₁ = 40W , voltage of 1st lamp, V₁ = 220V
power of 2nd lamp , P₂ = 60W , voltage of 2nd lamp ,V₂ = 220V
we know, one things ,
Power = V²/R [ when potential difference is same then consider P = V²/R ]
R = V²/P
so, resistance of 1st lamp , R₁ = V₁²/P₁ = (220)²/60 = 4840/6 = 2420/3Ω
resistance of 2nd lamp , R₂ = V₂²/P₂ = (220)²/40 = 48400/40 = 1210Ω
Now, question said ,
(a) Both the lamps are in parallel ,
So, 1/Req = 1/R₁ + 1/R₂
1/Req = 3/2420 + 1/1210 = 5/2420 = 1/484
Req = 484Ω
Now, Current drawn from electrical supply ,i = potential difference/Req
= 220V/484 = 20/44 = 5/11 A
(b) energy consumed by lamps in one hour = Energy consumed by 1st lamps in one hour + energy consumed by 2nd lamps in one hour
= P₁ × 1hour + P₂ × 1 hour
=(P₁ + P₂) × 1 hour
= (60W + 40W) × 1 hour
= 100 Wh
= 100/1000 KWh [ ∵ 1 KWh = 10²Wh ]
= 0.1 KWh
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Rewritten by : Barada
We have given two lamps in such a way :
Power of 1st lamp, P₁ = 40W , voltage of 1st lamp, V₁ = 220V
power of 2nd lamp , P₂ = 60W , voltage of 2nd lamp ,V₂ = 220V
we know, one things ,
Power = V²/R [ when potential difference is same then consider P = V²/R ]
R = V²/P
so, resistance of 1st lamp , R₁ = V₁²/P₁ = (220)²/60 = 4840/6 = 2420/3Ω
resistance of 2nd lamp , R₂ = V₂²/P₂ = (220)²/40 = 48400/40 = 1210Ω
Now, question said ,
(a) Both the lamps are in parallel ,
So, 1/Req = 1/R₁ + 1/R₂
1/Req = 3/2420 + 1/1210 = 5/2420 = 1/484
Req = 484Ω
Now, Current drawn from electrical supply ,i = potential difference/Req
= 220V/484 = 20/44 = 5/11 A
(b) energy consumed by lamps in one hour = Energy consumed by 1st lamps in one hour + energy consumed by 2nd lamps in one hour
= P₁ × 1hour + P₂ × 1 hour
=(P₁ + P₂) × 1 hour
= (60W + 40W) × 1 hour
= 100 Wh
= 100/1000 KWh [ ∵ 1 KWh = 10²Wh ]
= 0.1 KWh
