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- Hint: In this question use the property that efficiency of a transformer is the ratio of power received at the load end to the power supplied by the source end so use this concept to reach the solution of the question.

Given data:

Supply voltage = 2200 V

Load voltage = 220 V

Number of turns in primary coil = 500

Efficiency of transformer = 90% = (90/100)

Power transmitted i.e. power received by the load end = 8 KW

Now as we know that efficiency ($\eta $) of a transformer is the ratio of power received (${P_r}$) at the load end to the power supplied (${P_s}$) by the source end.

$ \Rightarrow \eta = \dfrac{{{P_r}}}{{{P_s}}}$

Now substitute the values we have,

$ \Rightarrow \dfrac{{90}}{{100}} = \dfrac{{8{\text{ KW}}}}{{{P_s}}}$

$ \Rightarrow {P_s} = \dfrac{{8{\text{ KW}}}}{{\dfrac{{90}}{{100}}}} = \dfrac{{800}}{{90}}{\text{ = 8}}{\text{.89 KW}}$

So the power supplied by the transformer is 8.89 KW.

So this is the required answer.

Note – An interesting fact is that for an ideal transformer the power supplied by the transformer is equal to the power received as the efficiency of the ideal transformer is 100%, if not than it is not an ideal transformer it is a practical transformer and the loss in the power while transferring the power from source to load (i.e. the difference of power) is called as core loss and copper loss of the transformer.

__Complete step-by-step solution__-Given data:

Supply voltage = 2200 V

Load voltage = 220 V

Number of turns in primary coil = 500

Efficiency of transformer = 90% = (90/100)

Power transmitted i.e. power received by the load end = 8 KW

Now as we know that efficiency ($\eta $) of a transformer is the ratio of power received (${P_r}$) at the load end to the power supplied (${P_s}$) by the source end.

$ \Rightarrow \eta = \dfrac{{{P_r}}}{{{P_s}}}$

Now substitute the values we have,

$ \Rightarrow \dfrac{{90}}{{100}} = \dfrac{{8{\text{ KW}}}}{{{P_s}}}$

$ \Rightarrow {P_s} = \dfrac{{8{\text{ KW}}}}{{\dfrac{{90}}{{100}}}} = \dfrac{{800}}{{90}}{\text{ = 8}}{\text{.89 KW}}$

So the power supplied by the transformer is 8.89 KW.

So this is the required answer.

__Hence option (B) is the correct answer.__Note – An interesting fact is that for an ideal transformer the power supplied by the transformer is equal to the power received as the efficiency of the ideal transformer is 100%, if not than it is not an ideal transformer it is a practical transformer and the loss in the power while transferring the power from source to load (i.e. the difference of power) is called as core loss and copper loss of the transformer.

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