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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# In Fig. 31-35, let the rectangular box on the left represent the (high-impedance) output of an audio amplifier, with ${\mathbit{r}}{\mathbf{=}}{\mathbf{1000}}{\mathbf{}}{\mathbf{ohm}}$ . Let ${\mathbit{R}}{\mathbf{=}}{\mathbf{10}}{\mathbf{}}{\mathbf{ohm}}$ represent the (low-impedance) coil of a loudspeaker. For maximum transfer of energy to the load ${\mathbit{R}}$ we must have ${\mathbit{R}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbit{r}}$ , and that is not true in this case. However, a transformer can be used to “transform” resistances, making them behave electrically as if they were larger or smaller than they actually are. (a) Sketch the primary and secondary coils of a transformer that can be introduced between the amplifier and the speaker in Fig. 31-35 to match the impedances. (b) What must be the turns ratio?

(a) Sketch the primary and secondary coils.

(b) Turns ratio is 10.

See the step by step solution

## Step 1: Listing the given quantities

The resistance $r=1000\mathrm{ohm}$,

The resistance $R=10\mathrm{ohm}$,

## Step 2: Understanding the concepts of transformer

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

Use the relation between primary current, the number of turns in the primary coil, secondary current, and the number of turns in the secondary coil.

Then by using the formula for power,define the ratio of turns.

Formula:

${I}_{s}=\frac{{N}_{p}{I}_{p}}{{N}_{s}}$

$P={I}_{s}^{2}R$

${I}_{p}=\frac{{\epsilon }_{rms}}{{R}_{eq}+r}$

## Step 4: (b) Calculations of the turn’s ratio:

The RMS current in secondary coil is as follows

${I}_{s}=\frac{{N}_{p}{I}_{p}}{{N}_{s}}$

The primary current in circuit is given as,

$\begin{array}{rcl}{I}_{p}& =& \frac{{\epsilon }_{rms}}{{R}_{eq}+r}\\ & =& \frac{{\epsilon }_{rms}}{{\left(\frac{{N}_{p}}{{N}_{s}}\right)}^{2}R+r}\end{array}$

Then power equation is written as follows.

$P={\left(\frac{{\epsilon }_{rms}}{{\left(\frac{{N}_{p}}{{N}_{s}}\right)}^{2}R+r}\right)}^{2}{\left(\frac{{N}_{p}}{{N}_{s}}\right)}^{2}R$

Let’s assume

$x={\left(\frac{{N}_{p}}{{N}_{s}}\right)}^{2}$

${P}_{avg}=\frac{{\epsilon }_{rms}^{2}×R×x}{{\left(xR+r\right)}^{2}}$

Then differentiate with respect to $x$:

$\begin{array}{rcl}\frac{d{P}_{avg}}{dx}& =& \frac{{\epsilon }_{rms}^{2}×R×\left(r-xR\right)}{{\left(xR+r\right)}^{3}}\\ & =& 0\end{array}$

$\frac{{\epsilon }_{rms}^{2}×R×\left(r-xR\right)}{{\left(xR+r\right)}^{3}}=0$

So,

$\begin{array}{rcl}x& =& \frac{r}{R}\\ & =& \frac{1000}{10}\\ & =& 100\mathrm{ohm}\end{array}$

But we know

$\begin{array}{rcl}x& =& {\left(\frac{{N}_{p}}{{N}_{s}}\right)}^{2}\\ {\left(\frac{{N}_{p}}{{N}_{s}}\right)}^{2}& =& 100\\ \frac{{N}_{p}}{{N}_{s}}& =& 10\end{array}$

Hence, number of turns ratio is 10.

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