Question

What is the turns ratio i.e, transformer ratio ${n_s}:{n_p}$ in an ideal transformer which increases ac voltage from 220 V to 33000 V?

Answer 1

The ideal transformer is free from all sorts of losses. It is an imaginary transformer that has no core loss, no leakage flux, and no ohmic resistance. The ideal transformer gives output power exactly equal to the input power. The turns ratio of a transformer is defined as the number of turns on its secondary divided by the number of turns on its primary coil.

Complete step by step answer: Given:
Primary voltage of the transformer, ${V_P} = 220V$
Secondary voltage of the transformer, ${V_S} = 33000V$
Number of turns in primary coil$= {n_P}$
Number of turns in secondary coil$= {n_S}$
Now, we know that the voltage ratio of an ideal transformer is directly related to the turns ratio which is given as,
$\dfrac{{{n_S}}}{{{n_P}}} = \dfrac{{{V_S}}}{{{V_P}}}$
Now after substituting the values of VP and VS in the above equation, we get,
$\dfrac{{{n_S}}}{{{n_P}}} = \dfrac{{33000}}{{220}}$ $= \dfrac{{150}}{1}$
Therefore, turns ratio i.e., transformer ratio, ${n_s}:{n_p}$ in an ideal transformer which increases ac voltage from 220 V to 33000 V is $150:1$.

Note: In this question you are asked to find the turns ratio in the ideal transformer. We must have knowledge about the concept of an ideal transformer. Here, we apply the property of ideal transformers. After, substitute the values ${V_P}$ and ${V_S}$ in the voltage ratio of an ideal transformer equation we get the number of turns in the primary and secondary coil and the ratio of the number of turns in the ideal transformer.