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

Last updated date: 16th Jul 2024
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Hint: 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, ${{r}_{1}}$ and ${{r}_{2}}$ 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
${{R}_{1}}={{r}_{1}}+{{r}_{2}}$
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
$\dfrac{1}{{{R}_{2}}}=\dfrac{1}{{{r}_{1}}}+\dfrac{1}{{{r}_{2}}}$
Simplifying the above equation and taking the inverse gives us the equivalent resistance in parallel connection of resistors as
${{R}_{2}}=\dfrac{{{r}_{1}}{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}}$
Now, dividing ${{R}_{1}}$ by ${{R}_{2}}$ gives us
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{r_1} + {r_2}}}{{\dfrac{{{r_1}{r_2}}}{{{r_1} + {r_2}}}}}$
Simplifying the above equation gives us
$\Rightarrow \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{{{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}}{{{r}_{1}}{{r}_{2}}}$

Therefore, option (C ) is the correct answer.

To derive the equivalent resistance equation in series and parallel connection of resistors.
Ohm’s law states that $R=\dfrac{V}{I}$ .
For a system of $2$ 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,
$\dfrac{V}{I}=\dfrac{{{V}_{1}}}{I}+\dfrac{{{V}_{2}}}{I}$
$\Rightarrow R={{R}_{1}}+{{R}_{2}}$
In a parallel connection of resistors, the voltage will remain constant but the current will be different for different resistors. Mathematically,
$\dfrac{V}{I}=\dfrac{V}{{{I}_{1}}}+\dfrac{V}{{{I}_{2}}}$
$\Rightarrow \dfrac{1}{R}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}$

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