Question

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?

Short answer

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.

Step-by-step solution

Step-1: Formulas used  

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.

Step-2: Calculations for ratio of secondary to primary turns of the transformer

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

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

Step-3: Calculations for power supplied to the transformer when the rms secondary current is 8.50mA

Now plug the values of secondary rms current and voltage

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

Step-4: Calculations for current rating of the fuse in the primary circuit 

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

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.