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

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

Step-by-step explanation:

Given in the question

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

`V_2=(6+j8)V`

`V_2=√(6^2+8^2)tan^(-1)(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}`