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Q66P

Expert-verifiedFound in: Page 939

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 ${\mathit{r}}{\mathbf{=}}{\mathbf{1000}}{\mathbf{}}{\mathbf{ohm}}$** **. Let **${\mathit{R}}{\mathbf{=}}{\mathbf{10}}{\mathbf{}}{\mathbf{ohm}}$** represent the (low-impedance) coil of a loudspeaker. For maximum transfer of energy to the load ${\mathit{R}}$ we must have ${\mathit{R}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathit{r}}$** **, 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.

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

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

**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:**

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

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

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

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

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

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

Then differentiate with respect to $x$:

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

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

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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