In the transformer shown in figure, the load resistor is 50Ω. The turns ratio ${N_1}:{N_2}$ is \[10:2\] and the source voltage is \[200{\rm{ }}V\](r.m.s). If a voltmeter across the load measures \[25{\rm{ }}V\](r.m.s.), then the source resistance \[{R_s}\] is (power loss is negligible)
A $450\;\Omega $
B $750\;\Omega $
C $2000\;\Omega $
D $100\;\Omega $

Hint: This question is based on the transformer. First, we have to know the transformer. The transformer is the device that is responsible for transferring the electrical energy from one to another circuit. We can use the expression for the primary voltage. Then we can find the current in the primary as well as in secondary windings. And last using the relation of Ohm’s law to find the source resistance in the circuit.

Complete step by step answer:
Given data: The load resistance is ${V_2} = 50\;\Omega $, the turn ratio is $\dfrac{{{N_1}}}{{{N_2}}} = \dfrac{{10}}{2}$and the voltage across the load resistance is $25\;{\rm{V}}$.
We know that the transformer is the electrical device that transfers the electricity to different locations. This transformer works on the principle of electromagnetic induction.

We have to find the expression of the primary voltage is given as,
$\dfrac{{{V_1}}}{{{V_2}}} = \left( {\dfrac{{{N_1}}}{{{N_2}}}} \right)$
Here, ${V_1}$ is the primary voltage and ${V_2}$ is the voltage across load resistance.
Now, substitute the values in above equation we get,

$\begin{array}{l}
\dfrac{{{V_1}}}{{25\;{\rm{V}}}} = \left( {\dfrac{{10}}{2}} \right)\\
{V_1} = 125\;{\rm{V}}
\end{array}$
Now, we have to calculate the current in the secondary windings is given as,
${I_2} = \dfrac{{{V_2}}}{{{R_2}}}$
Now, substitute the values in above equation we get,
$\begin{array}{l}
{I_2} = \dfrac{{25\;{\rm{V}}}}{{50\;\Omega }}\\
\Rightarrow {I_2} = 0.5\;{\rm{A}}
\end{array}$
Similarly, the current in the primary windings is given as,

Substitute the values in the above equation we get,
$\begin{array}{l}
{I_1} = \dfrac{{0.5\;{\rm{A}}}}{{\left( {\dfrac{{10}}{2}} \right)}}\\
\Rightarrow {I_1} = 0.1
\end{array}$
Now to get the source resistance, we use Ohm’s law is given as,
\[\begin{array}{l}
V = I{R_s}\\
{R_s} = \dfrac{V}{i}
\end{array}\]
Substitute the values in the above equation we get,
\[\begin{array}{l}
{R_s} = \dfrac{{200\;V}}{{0.1\;{\rm{A}}}}\\
\Rightarrow {R_s} = 2000\;\Omega
\end{array}\]
Thus, the source resistance is $2000\;\Omega $.

So, the correct answer is “Option C”.

Note:
In this question, students must have knowledge of transformer. Also, the source resistance is the as the internal resistance of both primary and secondary windings. The number of turns in primary and secondary windings is different.