A transformer connected to a 120-V (rms) ac line is to supply 13,000 V (rms) for a neon sign. To reduce shock hazard, a fuse is to be inserted in the primary circuit; the fuse is to blow when the rms current in the secondary circuit exceeds 8.50 mA. (a) What is the ratio of secondary to primary turns of the transformer? (b) What power must be supplied to the transformer when the rms secondary current is 8.50 mA? (c) What current rating should the fuse in the primary circuit have?

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

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

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

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.