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Question: The resistance of the series combination of two resistances is \(S\) . When they are joined in paral...

The resistance of the series combination of two resistances is SS . When they are joined in parallel the total resistance is PP . If S=nPS = nP , then the minimum possible value of nn is
A. 44
B. 33
C. 1
D. 22

Explanation

Solution

Let the two resistances be R1{R_1} and R2{R_2} . In a series combination, the net resistance is R1+R2=S{R_1} + {R_2} = S , and in parallel combination the net resistance is 1P=1R1+1R2\dfrac{1}{P} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} . Then put these values in the given equation, i.e. S=nPS = nP . For the minimum value of nn , R1=R2=R{R_1} = {R_2} = R use this value to reach the solution.

Complete answer:
Resistance - Resistance is the property of a material due to which it opposes the flow of current through it. Every material has some resistance, conductors (metals, etc.) have a very low resistance, and insulators (wood, rubber,etc.) have very high resistance.
In the problem, we are given two resistances that when connected in series combination have a net resistance of SS and when they are connected in parallel combination have a net resistance of PP .
We are also given that S=nPS = nP .
Let the two resistances be R1{R_1} and R2{R_2} .
In a series combination the same current flows through all of the resistances but in parallel combination, the current flowing through the resistances can be the same or different.
So, when these two resistances are connected in a series combination
R1+R2=S{R_1} + {R_2} = S
When these two resistances are connected in parallel combination
1P=1R1+1R2\dfrac{1}{P} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}
1P=R1+R2R1R2\Rightarrow \dfrac{1}{P} = \dfrac{{{R_1} + {R_2}}}{{{R_1}{R_2}}}
P=R1R2R1+R2\Rightarrow P = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}
So, for the relation S=nPS = nP
R1+R2=nR1R2R1+R2{R_1} + {R_2} = n\dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}
(R1+R2)2=nR1R2\Rightarrow {\left( {{R_1} + {R_2}} \right)^2} = n{R_1}{R_2}
R12+R22+2R1R2=nR1R2\Rightarrow R_1^2 + R_2^2 + 2{R_1}{R_2} = n{R_1}{R_2}
R12+R22R1R2=n2\Rightarrow \dfrac{{R_1^2 + R_2^2}}{{{R_1}{R_2}}} = n - 2
For the minimum value of nn , the value of R1=R2=R{R_1} = {R_2} = R
R2+R2R×R=n2\therefore \dfrac{{{R^2} + {R^2}}}{{R \times R}} = n - 2
n2=2\Rightarrow n - 2 = 2
n=4\Rightarrow n = 4

So, the correct answer is “Option A”.

Note:
Series and parallel combination of resistances are used in the circuit to alter the net resistance of the circuit using the same resistances. Aside from resistances, all electrical parts have some resistance which is to be taken into account when designing a piece of electrical equipment.