Question

If a transformer of an audio amplifier has output impedance 8000 0 and the speaker has input impedance of 80, the primary and secondary turns of this transformer connected

• A. 1000:01:00
• B. 100:01:00
• C. 1:32
• D. 32:01:00

Answer 1

Answer: (A)

1000:01:00

Answer 2

From Faraday's law, the induced emf across primary and secondary is
${{e}_{p}}=-{{N}_{p}}\frac{\Delta \phi}{\Delta t}$
${{e}_{s}}=-{{N}_{s}}\frac{\Delta \phi}{\Delta t}$
Also, $e=iR$
$\therefore$ $\frac{{{R}_{p}}}{{{R}_{s}}}=\frac{{{N}_{p}}}{{{N}_{s}}}$
Given, ${{R}_{s}}=8000\,\Omega ,{{R}_{p}}=8\,\Omega$
$\therefore$ $\frac{{{N}_{s}}}{{{N}_{p}}}=\frac{{{R}_{s}}}{{{R}_{p}}}=\frac{8000}{8}=\frac{1000}{1}$