The ratio of turns in primary to that in secondary for a transformer is 5 : 1 and power delivered by it is 500 W. What is the current in primary coil if a current of 12.5 A is flowing in the secondary coil?

The transformer in question is ideal and hence it has no power losses. This gives us

${\mathrm{P}}_{\text{in }}={\mathrm{P}}_{\text{out }}$

$\Rightarrow {\mathrm{V}}_1{\mathrm{I}}_1={\mathrm{V}}_2{\mathrm{I}}_2\dots \text{ (i) }$

$\text{ Given that }{\mathrm{P}}_{\text{out }}=500\mathrm{W}\text{ and }$

I₂ = 12.5 A.

But, P_out = V₂I₂

$\Rightarrow {\mathrm{V}}_2=\frac{{\mathrm{P}}_2}{{\mathrm{I}}_2}=\frac{500}{12.5}=40\mathrm{V}$

Also, given $\frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}=\frac{5}{1}\text{ (Turns ratio) }$

For an ideal transformer, $\frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}=\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}$

$\Rightarrow \frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{5}{1}$

$\frac{{\mathrm{V}}_1}{40}=\frac{5}{1}\left(\because {\mathrm{V}}_2=40\mathrm{V}\right)$

$\Rightarrow {\mathrm{V}}_1=200\mathrm{V}$

Form (i), we have

$\Rightarrow {\mathrm{V}}_1{\mathrm{I}}_1={\mathrm{V}}_2{\mathrm{I}}_2$

$\Rightarrow {\mathrm{I}}_1=\frac{{\mathrm{V}}_2{\mathrm{I}}_2}{{\mathrm{V}}_1}=\frac{40(12.5)}{200}=2.5\mathrm{A}$