Question

The current I in an alternating-current circuit is given by I=I m sinθ, where I m is the maximum current in the circuit. Find I if I m = 0.0246A and θ= 489.1 degrees

Answer 1

$\begin{equation*} 0.01909A \end{equation*}$

1) Given that the formula provided here is:

$\begin{gathered} i_=i_msin(\theta) \\ \end{gathered}$

2) Then let's plug into that formula, the given data:

$\begin{gathered} i=0.0246*sin(489.1) \\ i=0.0246*\sin\left(\frac{3600+1291}{1800}180^{\circ\:}\right) \\ i=0.0246*\sin\left(129.1^{\circ\:}\right) \\ i=0.01909A \end{gathered}$

Note that we needed to rewrite the value of that angle into a simpler form making use of the periodicity of that angle.

So the current is 0.01909 ampere (A)