Question

A Step-Down Transformer. A transformer connected to a 120-V (rms) ac line is to supply 12.0 V (rms) to a portable electronic device. The load resistance in the secondary is 5.00 Ω. (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the 120-V line would draw the same power as the transformer? Show that this is equal to 5.00 Ω times the square of the ratio of primary to secondary turns.

Short answer

a)The ratio of primary to secondary turns of the transformer is $\frac{{\mathrm{N}}_2}{{\mathrm{N}}_1}=0.5$.b)The secondary must supply a rms current of I=2.40A .c)Average power delivered to the load will be 28.8W.d)A$500\Omega$resistance connected directly across the 120-V line would draw the same power as the transformer.

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 expression for average power is

$$\begin{gathered} {\mathbf{P}}_{\mathbf{avg}}\mathbf{=}{\left(I_2\right)}^{\mathbf{2}}\mathbf{R} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\left(V\right)}^{\mathbf{2}}}{\mathbf{R}} \end{gathered}$$

The Ohm’s law relation to get the amplitude voltage V in the circuit.

V=IR

For a transformer,

$$\begin{gathered} {\mathbf{V}}_{\mathbf{1}}\mathbf{,}{\mathbf{I}}_{\mathbf{1}}\mathbf{=}{\mathbf{V}}_{\mathbf{2}}{\mathbf{I}}_{\mathbf{2}} \\ {\mathbf{I}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}{\mathbf{(}{\displaystyle \frac{{\mathbf{N}}_{\mathbf{2}}}{{\mathbf{N}}_{\mathbf{1}}}}\mathbf{}\mathbf{)}}^{\mathbf{2}}}{\mathbf{R}} \end{gathered}$$

Step-2: Calculations for ratio of turns.

$$\begin{gathered} {\mathrm{V}}_1=120\mathrm{V} \\ {\mathrm{V}}_2=12\mathrm{V} \end{gathered}$$

$$\begin{gathered} \frac{{\mathrm{N}}_2}{{\mathrm{N}}_1}=\frac{120\mathrm{V}}{12\mathrm{V}} \\ =10 \end{gathered}$$

Therefore, to operate the dryer, use a step down transformer with $\frac{{\mathrm{N}}_2}{{\mathrm{N}}_1}=10$.

Step-3: Calculations for rms current and power

$$\begin{gathered} \mathrm{R}=5\mathrm{\Omega } \\ {\mathrm{I}}_1=\frac{{\mathrm{V}}_1{({\displaystyle \frac{{\mathrm{N}}_2}{{\mathrm{N}}_1}})}^2}{\mathrm{R}} \\ =\frac{120{\left(0.01\right)}^2\mathrm{V}}{5\mathrm{\Omega }} \\ =0.24\mathrm{A} \end{gathered}$$

$$\begin{gathered} {\mathrm{V}}_1{\mathrm{I}}_1={\mathrm{V}}_2{\mathrm{I}}_2 \\ {\mathrm{I}}_2=\frac{\left(120\mathrm{V}\right)\left(0.24\mathrm{A}\right)}{\left(12\mathrm{A}\right)} \\ =2.4\mathrm{A} \end{gathered}$$

Therefore, the secondary must supply a rms current of I = 2.40A.

$$\begin{gathered} P_{av}={\left(2.4A\right)}^2\left(5.00\Omega \right) \\ =28.8W \end{gathered}$$

Step-4: Calculations for resistance across 120V.

$$\begin{gathered} R=\frac{{\left(V\right)}^2}{P_{avg}} \\ R=\frac{{\left(120V\right)}^2}{28.8W} \\ =500\Omega \end{gathered}$$

Now,

$500\Omega ={(\frac{N_1}{N_2})}^2\left(5\Omega \right){(\frac{N}{}}^{}$₁N₂)2=10

Therefore, The ratio of primary to secondary turns of the transformer is $\frac{N_1}{N_2}=0.5$.b)The secondary must supply a rms current of I=2.40A.c)Average power delivered to the load will be 28.8W.d) A$500\Omega$ resistance connected directly across the 120-V line would draw the same power as the transformer.