Question 314: AC fundamentals

Question

Alternating current of peak value $\left(\frac{2}{\pi}\right)$ A flows through the primary coil of a transformer. The coefficient of mutual inductance between primary and secondary coils is 1 H . The peak e.m.f. induced in secondary coil (Frequency of a.c. $=50 \mathrm{~Hz}$ )

Accepted Answer

To find the peak e.m.f. induced in the secondary coil of the transformer, we can use Faraday's law of electromagnetic induction. The e.m.f. ($ \mathcal{E} $) induced in the secondary coil can be given by:
$$ \mathcal{E} = -M \frac{dI}{dt} $$
where:
$ M = 1 \, 	ext{H} $ is the mutual inductance,
$ \frac{dI}{dt} $ is the rate of change of current in the primary coil.
The current in the primary coil is an alternating current, expressed as:
$$ I(t) = I_0 \sin(\omega t) $$
where:
$ I_0 = \frac{2}{\pi} \, 	ext{A} $ is the peak current,
$ \omega = 2\pi f $ is the angular frequency, with $ f = 50 \, 	ext{Hz} $.
Calculate $ \omega $:
$$ \omega = 2\pi 	imes 50 = 100\pi \, 	ext{rad/s} $$
The derivative of the current with respect to time is:
$$ \frac{dI}{dt} = I_0 \omega \cos(\omega t) $$
Substitute $ I_0 $ and $ \omega $:
$$ \frac{dI}{dt} = \frac{2}{\pi} 	imes 100\pi \cos(\omega t) = 200 \cos(\omega t) $$
The peak value of $ \cos(\omega t) $ is 1, so the peak rate of change of current is:
$$ \left( \frac{dI}{dt} ight)_{	ext{peak}} = 200 \, 	ext{A/s} $$
Thus, the peak e.m.f. in the secondary coil is:
$$ \mathcal{E}_{	ext{peak}} = M 	imes \left( \frac{dI}{dt} ight)_{	ext{peak}} = 1 	imes 200 = 200 \, 	ext{V} $$
Therefore, the peak e.m.f. induced in the secondary coil is
200 V
.
The correct option is
D: 200 V
.

Answer

50 V

Answer

100 V

Answer

150 V

Answer

200 V

AC fundamentals

Alternating current of peak value \left(\frac{2}{\pi}\right) A flows through the primary coil of a transformer. The coefficient of mutual inductance between primary and secondary coils is 1 H . The peak e.m.f. induced in secondary coil (Frequency of a.c. =50 \mathrm{~Hz} )

Choose the correct answer: