How many 176(ohm) resistors in are required to carry 5 A on 220v line

a)in parallel

b)in series

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Solution

R = V/I
= 220/5
= 44 Ω
To make 44Ω from 176Ω,
$R_{r e q} = \frac{R}{4}$
$A s w e k n o w, \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}}; ∵ R_{1} = R_{2} = R_{3} = R_{4} = 176$
we need to connect four 176Ω resistors in parallel so that their effective resistance is 176Ω/4 = 44Ω

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