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Question: The equivalent resistance of \({{r}_{1}}\) and \({{r}_{2}}\) when connected in series is \({{R}_{1}}...

The equivalent resistance of r1{{r}_{1}} and r2{{r}_{2}} when connected in series is R1{{R}_{1}} and that when they are connected in parallel is R2{{R}_{2}}. Then the ratio R1R2\dfrac{{{R}_{1}}}{{{R}_{2}}} is
(A) r1r2\dfrac{{{r}_{1}}}{{{r}_{2}}}
(B) r1+r2r1r2\dfrac{{{r}_{1}}+{{r}_{2}}}{{{r}_{1}}{{r}_{2}}}
(C) (r1+r2)2r1r2\dfrac{{{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}}{{{r}_{1}}{{r}_{2}}}
(D) r1r22r1+r2\dfrac{{{r}_{1}}{{r}_{2}}}{2{{r}_{1}}+{{r}_{2}}}

Explanation

Solution

Two resistors connected in series will result in the voltage drop across each, but the current remains the same. Two resistors connected in parallel will result in the current across each resistor to be different but the voltage will be equal. So the equivalent resistance in series combination will be the sum of individual resistors and that in parallel will be the inverse of the sum of the inverse of the resistors.

Complete Step by step solution
In the first case, r1{{r}_{1}} and r2{{r}_{2}} are connected in series with each other. In this case, the equivalent resistance can be written as the sum of individual resistors. Mathematically, it can be written as
R1=r1+r2{{R}_{1}}={{r}_{1}}+{{r}_{2}}
In the second case, when they are connected in parallel, the equivalent resistance will be the inverse of the sum of the inverse of individual resistors. Mathematically, it can be written as
1R2=1r1+1r2\dfrac{1}{{{R}_{2}}}=\dfrac{1}{{{r}_{1}}}+\dfrac{1}{{{r}_{2}}}
Simplifying the above equation and taking the inverse gives us the equivalent resistance in parallel connection of resistors as
R2=r1r2r1+r2{{R}_{2}}=\dfrac{{{r}_{1}}{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}}
Now, dividing R1{{R}_{1}} by R2{{R}_{2}} gives us
R1R2=r1+r2r1r2r1+r2\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{r_1} + {r_2}}}{{\dfrac{{{r_1}{r_2}}}{{{r_1} + {r_2}}}}}
Simplifying the above equation gives us
R1R2=(r1+r2)2r1r2\Rightarrow \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{{{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}}{{{r}_{1}}{{r}_{2}}}

Therefore, option (C ) is the correct answer.

Additional information
To derive the equivalent resistance equation in series and parallel connection of resistors.
Ohm’s law states that R=VIR=\dfrac{V}{I} .
For a system of 22 resistors,
In a series connection of resistance, the voltage will be dropped at each resistor but the current through all the resistors will be the same and hence,
VI=V1I+V2I\dfrac{V}{I}=\dfrac{{{V}_{1}}}{I}+\dfrac{{{V}_{2}}}{I}
R=R1+R2\Rightarrow R={{R}_{1}}+{{R}_{2}}
In a parallel connection of resistors, the voltage will remain constant but the current will be different for different resistors. Mathematically,
VI=VI1+VI2\dfrac{V}{I}=\dfrac{V}{{{I}_{1}}}+\dfrac{V}{{{I}_{2}}}
1R=1R1+1R2\Rightarrow \dfrac{1}{R}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}

Note:
When nn equal resistors of value RR are connected in series, then the equivalent resistance will be nRnR , and when nn equal resistors of value RR are connected in parallel, then the equivalent resistance will be Rn\dfrac{R}{n} .