two lamps one is rated 100w at 200V and the other 60W at 220V are connected in parallel to a 220V supply. find the current drawn from the supply line
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3
If this is an text qn then ur qn is wrong..
1.P= 100w and v= 220
2.p=60w at 220v
Using p=vi
P/v =I
100/ 220+ 60/220=160/200
Which is 16/22 =0.727...
Therefore current =0.73 A
1.P= 100w and v= 220
2.p=60w at 220v
Using p=vi
P/v =I
100/ 220+ 60/220=160/200
Which is 16/22 =0.727...
Therefore current =0.73 A
Answered by
4
Case I
Given:
Power = P = 100 W
Voltage = V = 220v
Resistance = R
P = V²/R
100 = 220 ×220/R
R = 220 × 220/100
= 484 Ω
Case II
Given:
Power = P = 60 W
Voltage = V = 220v
Resistance = R
P = V²/R
60 = 220 × 220/R
R = 220× 220/60
= 806.7 Ω
As the resistors are connected in parallel ,
total resistance = 1/R
1/R = 1/484 + 1/806.7
= 806.7 + 484/484×806.7
= 1290.7/390442.8
R = 390442.8/ 1290.7
= 302.5 ohms
Total resistance = 302.5 Ω
Then,
I = current
V = IR
220 = I × 302.5
I = 220/302.5
= 0.73 A
The current drawn is 0.73 A
:) Hope this Helps !!!
Given:
Power = P = 100 W
Voltage = V = 220v
Resistance = R
P = V²/R
100 = 220 ×220/R
R = 220 × 220/100
= 484 Ω
Case II
Given:
Power = P = 60 W
Voltage = V = 220v
Resistance = R
P = V²/R
60 = 220 × 220/R
R = 220× 220/60
= 806.7 Ω
As the resistors are connected in parallel ,
total resistance = 1/R
1/R = 1/484 + 1/806.7
= 806.7 + 484/484×806.7
= 1290.7/390442.8
R = 390442.8/ 1290.7
= 302.5 ohms
Total resistance = 302.5 Ω
Then,
I = current
V = IR
220 = I × 302.5
I = 220/302.5
= 0.73 A
The current drawn is 0.73 A
:) Hope this Helps !!!
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