Two lamps, one rated 100W at 220 V and other 40 W at 220 V are conected in parallel to 220 V main supply. Calculate the current drawn from the supply line.
Answers
Explanation:
From Joule's Law of heating,
From Joule's Law of heating,P=V2/R
From Joule's Law of heating,P=V2/RFor first lamp,
From Joule's Law of heating,P=V2/RFor first lamp, R1=2202/100
From Joule's Law of heating,P=V2/RFor first lamp, R1=2202/100For second lamp,
From Joule's Law of heating,P=V2/RFor first lamp, R1=2202/100For second lamp, R2=2202/60
From Joule's Law of heating,P=V2/RFor first lamp, R1=2202/100For second lamp, R2=2202/60When they are connected in parallel,
From Joule's Law of heating,P=V2/RFor first lamp, R1=2202/100For second lamp, R2=2202/60When they are connected in parallel,Req1=R11+R21
From Joule's Law of heating,P=V2/RFor first lamp, R1=2202/100For second lamp, R2=2202/60When they are connected in parallel,Req1=R11+R21=2202/1001+2202/601
From Joule's Law of heating,P=V2/RFor first lamp, R1=2202/100For second lamp, R2=2202/60When they are connected in parallel,Req1=R11+R21=2202/1001+2202/601⟹Req=2202/160
From Joule's Law of heating,P=V2/RFor first lamp, R1=2202/100For second lamp, R2=2202/60When they are connected in parallel,Req1=R11+R21=2202/1001+2202/601⟹Req=2202/160From Ohm's Law,
From Joule's Law of heating,P=V2/RFor first lamp, R1=2202/100For second lamp, R2=2202/60When they are connected in parallel,Req1=R11+R21=2202/1001+2202/601⟹Req=2202/160From Ohm's Law,V=IReq
From Joule's Law of heating,P=V2/RFor first lamp, R1=2202/100For second lamp, R2=2202/60When they are connected in parallel,Req1=R11+R21=2202/1001+2202/601⟹Req=2202/160From Ohm's Law,V=IReq220=1602202I
From Joule's Law of heating,P=V2/RFor first lamp, R1=2202/100For second lamp, R2=2202/60When they are connected in parallel,Req1=R11+R21=2202/1001+2202/601⟹Req=2202/160From Ohm's Law,V=IReq220=1602202II=0.73 A
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Answer:
0.63 A
Explanation:
we have given two lamps in such a way :
Power of 1st lamp, P₁ = 100W , voltage of 1st lamp, V₁ = 220V
power of 2nd lamp , P₂ = 40W , voltage of 2nd lamp ,V₂ = 220V
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)²/100 = 484Ω
resistance of 2nd lamp , R₂ = V₂²/P₂ = (220)²/40 = 48400/40 = 1210Ω
Both the lamps are in parallel ,
So, 1/Req = 1/R₁ + 1/R₂
1/Req = 1/484 + 1/1210 = 2.5+1/1210 = 3.5/1210
Req = 345.71Ω
Now, Current drawn from electrical supply ,i = potential difference/Req
= 220/345.71 = 0.63 A