Question

The sinusoid corresponding to the phasor V2 = 6 + j8 V and ω = 31 rad/s is v2(t) =__________ V. Please report your answer so the magnitude is positive and all angles are in the range of negative 180 degrees to positive 180 degrees.

Answer 1

Answer:

$v_2(t)=10sin[31t+53.13^{\circ}]\ V$

Explanation:

Given in the question

$\omega$ = Angular frequency = 31 rad/s

$V_2=(6+j8)V$

$V_2=\sqrt{6^2+8^2}tan^{-1}\dfrac{8}{6}\\\Rightarrow V_2=10, 53.13^{\circ}$

Now,

$v_2(t)=rsin[\omega t+\theta]\\\Rightarrow v_2(t)=10sin[31t+53.13^{\circ}]\ V$

The required function is

$\mathbf{v_2(t)=10sin[31t+53.13^{\circ}]\ V}$