How many 176 ohm resistors (in parallel) are requiered to carry 5A on a 220 V line?
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Answered by
3
For Resistors in Parallel,
1/R(total) = 1/R₁ + 1/R₂ + 1/Rₓ
From Ohm's Law
V = IR
220 = (5)R
R = 44 Ω
Therefore
1/44 = No. of 176Ω Resistors * 1/176
Let no. of 176Ω Resistors be x
Then 1/44 = x/176
x = 4
Therefore, 4 resistors of 176Ω ( in parallel) are required to carry 5 A on a 220 V line.
1/R(total) = 1/R₁ + 1/R₂ + 1/Rₓ
From Ohm's Law
V = IR
220 = (5)R
R = 44 Ω
Therefore
1/44 = No. of 176Ω Resistors * 1/176
Let no. of 176Ω Resistors be x
Then 1/44 = x/176
x = 4
Therefore, 4 resistors of 176Ω ( in parallel) are required to carry 5 A on a 220 V line.
Answered by
3
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Let the total no. of resistors = x
Given,
I = 5 A
V = 220 V
According to Ohm’s law,
V = IR
R = V/ I = 220/5 = 44Ω
Now for x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is 44 Ω.
1/44 = 1/176 + 1/176 + 1/176 .....χ times
1/44 = x / 176
=> x = 176/ 44
=> x = 4
Therefore, 4 resistors of 176 Ω are required to draw the given amount of current.
I hope, this will help you___❤❤
Thank you
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