Question

The voltage v=12cos(60t+45) is applied to a 0.1-H inductor. What will be the the current through the inductor? ​

Answer 1

Answer:

answer in attachment

Answer 2

Answer:

The current through the inductor is $2cos(60t-45\textdegree)$.

Explanation:

The waveform in the question is of the form,

$V=Acos(\omega t+\theta)$      (1)

By comparing the wave in the question with equation (1) we get;

A=12volt

ω=60

θ=45°

The resistance of an inductor is given as,

$X_L=\omega L$        (2)

By substituting the values in equation (1) we get;

$X_L=60\times 0.1=6\Omega$     (3)

And the peak current through the inductor is given as,

$I=\frac{A}{X_L}$     (4)

By substituting the values in equation (4) we get;

$I=\frac{12}{6}=2A$     (5)

Since, the current through the inductor lags behind the voltage by $\pi/2$ then the current through the inductor is given as,

$I=2cos(60t-45\textdegree)$       (6)           (minus sign in phase as the current is lagging)

Hence,  the current through the inductor is $2cos(60t-45\textdegree)$.