First slide

combination of resistance

Question

The resistance of the series combination of two resistance is S. When they are joined in parallel the total resistance is P. If S = nP, then the minimum possible value of n is

Easy

A

4
B

3
C

2
D

1

Solution

If two resistances are R₁ and R₂, then
$S = R_{1} + R_{2} and P = \frac{R_{1} R_{2}}{(R_{1} + R_{2})}$
From given condition, S = nP
$or (R_{1} + R_{2}) = n (\frac{R_{1} R_{2}}{R_{1} + R_{2}})$
$\begin{array}{l} \Rightarrow {(R_{1} + R_{2})}^{2} = {nR}_{1} R_{2} \Rightarrow {(R_{1} - R_{2})}^{2} + 4 R_{1} R_{2} = {nR}_{1} R_{2} \\ So n = 4 + \frac{{(R_{1} - R_{2})}^{2}}{R_{1} R_{2}} . \end{array}$
Hence, minimum value of n is 4.

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