91Ó°ÊÓ

A transformer is a device that transfers electric energy from one alternating-current circuit to one or more other circuits, either increasing (stepping up) or reducing (stepping down) the voltage.

$\frac{{\mathbf{N}}_{\mathbf{2}}}{{\mathbf{N}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{V}}_{\mathbf{1}}}$

The average power equals the root mean square of current times the voltage ${\mathbf{P}}_{\mathbf{av}}\mathbf{=}{\mathbf{I}}_{\mathbf{2}}{\mathbf{V}}_{\mathbf{2}}$,where${\mathbf{I}}_{\mathbf{2}}$ is the rms current in the secondary circuit and${\mathit{V}}_{\mathbf{2}}$ is the voltage in the secondary circuit

Since power in primary circuit is equal to power in secondary circuit

${\mathbf{P}}_{\mathbf{av}}\mathbf{=}{\mathbf{I}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}$

So we get, ${\mathbf{I}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{av}}}{{\mathbf{V}}_{\mathbf{1}}}$, where${\mathbf{I}}_{\mathbf{1}}$ is the rms current in the primary circuit and${\mathbf{V}}_{\mathbf{1}}$ is the voltage in the primary circuit.

Plug the values of${\mathbf{V}}_{\mathbf{2}}$and${\mathbf{V}}_{\mathbf{1}}$to get the ratio of the secondary turns to primary turns

$$\begin{gathered} \frac{{\mathrm{N}}_2}{{\mathrm{N}}_1}=\frac{13000\mathrm{V}}{120\mathrm{V}} \\ =108 \end{gathered}$$

Now plug the values of secondary rms current and voltage

$$\begin{gathered} {\mathrm{P}}_{\mathrm{av}}=\left(8.50×{10}^{-3}\mathrm{A}\right)\left(13000\mathrm{V}\right) \\ =100.5\mathrm{W} \end{gathered}$$

$$\begin{gathered} {\mathrm{I}}_1=\frac{110\mathrm{W}}{120\mathrm{V}} \\ =0.197\mathrm{A} \end{gathered}$$

Therefore, the ratio of secondary to primary turns of the transformer is 108.The power supplied to the transformer is 110.5W. The current rating of fuse in primary circuit is 0.917A.