Question

When two unknown resistors are connected in series with a battery, $225W$ power is delivered to the combination with a total current $5A$. For the same current, $50W$ is delivered when the resistors are connected in parallel. Find the ratio of the two resistances.
$\begin{aligned} & A)2 \\\ & B)5 \\\ & C)4 \\\ & D)10 \\\ \end{aligned}$

Answer 1

Here, we are required to consider the formulas of power in relation with resistances connected in parallel as well as series combination. Power delivered is equal to the product of square of current flowing through the resistors and the equivalent resistance of the resistors, connected in series as well as parallel combinations. The ratio of resistances can easily be determined once we take the ratio of powers in both cases.

Formula used:
$\begin{aligned} & 1){{P}_{1}}={{I}^{2}}({{R}_{1}}+{{R}_{2}}) \\\ & 2){{P}_{2}}={{I}^{2}}\left( \dfrac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right) \\\ \end{aligned}$

Complete answer:
We know that power delivered in a circuit containing a resistor is equal to the product of a square of current flowing through the resistor and the resistance of the resistor. Also do we know that equivalent resistance of resistors connected in series is the sum of resistances of these resistors. At the same time, reciprocal of equivalent resistance of series connected in parallel is equal to the sum of reciprocals of resistances of resistors connected in parallel.
Relating the above explanation with our question, we can write:
$\begin{aligned} & {{P}_{1}}={{I}^{2}}({{R}_{1}}+{{R}_{2}}) \\\ & {{P}_{2}}={{I}^{2}}(\dfrac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}) \\\ \end{aligned}$
where
${{P}_{1}}$ and ${{P}_{2}}$ are the powers delivered in series combination and parallel combination of resistors, respectively
${{R}_{1}}$ and ${{R}_{2}}$ are the two unknown resistances, whose ratio needs to be determined
$I$ is the identical current flowing through these resistors in both cases
Let this be equation 1.
Substituting the given values from the question in equation 1, we have

& 225={{5}^{2}}({{R}_{1}}+{{R}_{2}}) \\\ & 50={{5}^{2}}\left( \dfrac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right) \\\ \end{aligned}$$ where ${{P}_{1}}=225W;{{P}_{2}}=50W;I=5A$ Let this be equation 2 and equation 3 respectively. Now, dividing equation 2 by equation 3, we have $\dfrac{{{({{R}_{1}}+{{R}_{2}})}^{2}}}{{{R}_{1}}{{R}_{2}}}=\dfrac{225}{50}=\dfrac{9}{2}\Rightarrow \dfrac{{{R}_{1}}^{2}{{\left( 1+\dfrac{{{R}_{2}}}{{{R}_{1}}} \right)}^{2}}}{{{R}_{1}}{{R}_{2}}}=\dfrac{9}{2}$ Now, let us take $\dfrac{{{R}_{1}}}{{{R}_{2}}}=x$, to make our calculations easier. Substituting the same in the above expression, we have $x{{\left( 1+\dfrac{1}{x} \right)}^{2}}=\dfrac{9}{2}\Rightarrow {{x}^{2}}+2x+1=\dfrac{9x}{2}\Rightarrow 2{{x}^{2}}-5x+2=0$ Taking roots of this equation, we have $x=2\Rightarrow \dfrac{{{R}_{1}}}{{{R}_{2}}}=2$ Let this be equation 4. From equation 4, we can conclude that the ratio of resistances is $2$ . **Therefore, the correct answer is option $A$.** **Note:** We could also have proceeded by not substituting $\dfrac{{{R}_{1}}}{{{R}_{2}}}$ with $x$, but this method can be tedious. Therefore, it is always advisable that ratios in big expressions be substituted by any variables to easily determine the same. Using Ohm’s law, power delivered can also be written as $P={{I}^{2}}R=\dfrac{{{V}^{2}}}{R}$ since $I=\dfrac{V}{R}$ using Ohm’s law. This expression would have come in handy if voltages across the resistors were given instead of the identical current.