1.

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

Answer»

Solution :Here, I = 5 A, V = 220 V
So, the total resistance of the given circuit is
`R_("total") = V/I =220/5 = 44 Omega`
i.e., when `44Omega`resistance is connected with220 V line, 5 A current would flow through the given circuit.
Now, SUPPOSE .n. resistors, each of resistance R, are required to be connected in parallel, so that the total resistance `R_("total")`becomes `44Omega`.
Hence, `1/(R_("total")) = 1/R + 1/R +...n` times
`=(1 + 1 + ...n" time")/R = n/R`
`:.R_("total") =R/n`
Now, `R_("total") =44 Omega and R = 176 Omega`
So, `44 = 176/n`
`:. n = 176/44 =4`
Thus, 4 resistors each of `176 Omega` connected in parallel will RESULT in total resistance of `44 Omega` causing a current of 5 A to flow when connected to 220 V line.


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