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How many `176 Omega` resistors (in parallel) are required to carry 5 A in 220 V line ?

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Here, Potential difference (V) is 220 V and current (I) is 5A.
So, Resistance, `R = (V)/(I)`
`R = (220)/(5)`
`R = 44 Omega`
Thus, the total resistance of the circuit is 44 ohms. Now, all that we have to do is to find out how many 176 ohm resistors should be connected in parallel to obtain a resultant resistance of `44 Omega`. Suppose the number of `176 Omega` resistors required is x.
Then: `(1)/(44) = (1)/(176) xx x` (Resistances in parallel)
`44 x = 176`
`x = (176)/(44)`
`x = 4`
Thus, 4 resistors of `176 Omega` each should be connected in parallel.

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