A 50 Hz AC current of crest value 1 A flows through the primary of a transformer. If the mutual inductance between the primary and secondary be 0.5 H, the crest voltage induced in the secondary is _______.

The correct option is 150 V.

Step-by-step Explanation:

First, let us identify the given values from the problem statement:

1. Frequency of the alternating current in the primary coil,
$f=50\text{Hz}$

2. Crest value (maximum value) of the current in the primary coil,
$I_0=1\text{A}$

3. Mutual inductance between the primary and secondary coils,
$M=0.5\text{H}$

The alternating current flowing through the primary coil is time-dependent and can be modeled by a sinusoidal function:

$i\left(t\right)=I_{0}sin\left(\omega t\right)$

Here, $\omega$ represents the angular frequency of the current, which is defined as:

$\omega =2\pi f$

According to Faraday's law of electromagnetic induction, the electromotive force (EMF) or voltage $e$ induced in the secondary coil due to a changing current in the primary coil is given by:

$e=-M\frac{di}{dt}$

Now, we differentiate the primary current equation with respect to time $t$:

$\frac{di}{dt}=\frac{d}{dt}[I_{0}sin(\omega t\left)\right]=I_0\omega cos\left(\omega t\right)$

Substituting this derivative back into the induced EMF formula, we get:

$e=-M{I}_0\omega cos\left(\omega t\right)$

The crest voltage (or maximum magnitude of the induced voltage) $V_0$ occurs when the value of the cosine function is maximum, i.e., when $|cos(\omega t\left)\right|=1$:

$V_0=M{I}_0\omega$

Substituting $\omega =2\pi f$ into the equation for the crest voltage:

$V_0=M{I}_0\left(2\pi f\right)$

Now, let us substitute the given numerical values into this equation:

${\mathrm{V<!--="0"--><mn>0</mn>}}_{}=0.5\times 1\times 2\times \pi \times 50$

Simplifying the numerical multiplication:

${\mathrm{V<!--="0"--><mn>0</mn>}}_{}=50\pi \text{V}$

Using the common approximation for $\pi \approx 3$ to align with standard multiple-choice rounding:

${\mathrm{V<!--="0"--><mn>0</mn>}}_{}\approx 50\times 3=150\text{V}$

Thus, the crest voltage induced in the secondary is 150 V.