Question

A transformer rated at $10\,kW$ is used to connect a $5\,kV$ transmission line to a $240\,V$ circuit. The ratio of turn in the winding of a transformer is
A. \[40\]
B. \[110\]
C. \[20.8\]
D. \[45\]

Answer 1

Hint:In order to answer the question, to know about the ratio of turn in the winding of a transformer first we will convert the given voltage of the transmission line from kilovolt to volt, then we will use the ratio formula of primary and secondary voltage which we will make equal to with the ratio of primary and secondary number of turns to find out the desired value.

Formula used:
$\dfrac{{{V_p}}}{{{V_s}}} = \dfrac{{{N_p}}}{{{N_s}}}$
Here, ${V_p}$ and ${V_s}$ are the primary and secondary voltage respectively whereas, ${N_p}$ and ${N_s}$ are the number of turns in primary and secondary respectively.

Complete step by step answer:
Here, we are given that; the voltage in the transmission line is $5kV$ and if we convert it in voltage then it will we
$5kV = 5000V$
Now, It is said that this transmission line is connected with $240V$ circuit. So, we can say that this is the voltage of the secondary of the transformer as we connect the load in the secondary. So, we can say that this is a step down transformer. The output voltages of a step-down transformer are reduced, or in other words, high voltage, low current power is converted to low voltage, high current power. The turn ratio of the primary and secondary coils determines the voltage reduction capabilities of step-down transformers.Therefore, the turn ratio will be;
$\dfrac{{{V_p}}}{{{V_s}}} = \dfrac{{{N_p}}}{{{N_s}}}$
\[\Rightarrow \dfrac{{5000}}{{240}} = \dfrac{{{N_p}}}{{{N_s}}} \\
\therefore \dfrac{{{N_p}}}{{{N_s}}} = 20.8 \]
Thus, the ratio of turn in the winding of a transformer is \[20.8\] .

So, the correct answer is option C.

Note:To act as a "Step-Down Transformer," the number of turns in the secondary winding should always be smaller than the number of turns in the primary winding, i.e. \[Np > Ns\] . Because the secondary winding has fewer turns, the total induced emf will be lower, and the secondary output voltage will be lower than the primary input voltage.