Question

you are given 'n' identical wire, each of resistance R . when this are connected in parallel, the equivalent resistance is X. when this will be connected in series, then what will be the equivalent resistance?​

Answer 1

Answer:

n²X

Explanation:

When wires are connected in parallel:

$\frac{1}{R(eq)} = \frac{1}{R} + \frac{1}{R} + ...n \: terms \\ = \frac{n}{2} ( \frac{2}{R} + (n - 1)(0)) \: \\ = \frac{n}{R}$

[Using Sum upto n-terms for A.P., where a = 1/R, d = 0]

So, R(eq) = R/n

or, R = n[R(eq)] = nX ...(1)

Similarly, if we connect wires in series,

R(eq)' = R + R + R + ...upto n-terms

R(eq)' = (n/2)[2R + (n − 1)(0)] = nR = n(nX) [by (1)]

So, the required equivalent resistance when wires will be connected in series = n²X

Hope it helps!

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