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

The Correct Option is A

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} $