Let's go through each question step-by-step.

Question 1

You need to find the phase difference between the two voltages: $v_1(t) = -\sin(t - 135^\circ)$ and $v_2(t) = 2 \cos(t + 120^\circ).$

1. Convert Both Equations to the Same Trigonometric Function: It is easier to compare phase angles if both functions use either sine or cosine.

   • We know that $\cos(t + 120^\circ) = \sin(t + 120^\circ + 90^\circ) = \sin(t + 210^\circ)$.
   • Thus, $v_2(t) = 2 \sin(t + 210^\circ)$.

2. Rewrite $v_1(t)$ in Positive Form:

   • $-\sin(t - 135^\circ) = \sin(t - 135^\circ - 180^\circ) = \sin(t - 315^\circ)$.

3. Compare Phases:

   • The phase of $v_1(t)$ is $-315^\circ$ (or equivalently $+45^\circ$).
   • The phase of $v_2(t)$ is $+210^\circ$.

4. Find the Phase Difference:

   • Phase difference = $|45^\circ - 210^\circ| = 165^\circ$.

So, the phase difference between $v_1(t)$ and $v_2(t)$ is 165 degrees.

Question 2

Find the resultant of the complex number $\frac{1 + j2}{2 - j}$ in both Cartesian and Polar forms.

1. Multiply by the Complex Conjugate: $\frac{1 + j2}{2 - j} \cdot \frac{2 + j}{2 + j} = \frac{(1 + j2)(2 + j)}{(2 - j)(2 + j)}.$

2. Calculate the Numerator: [ (1 + j2)(2 + j) = 1 \cdot 2 + 1 \cdot j + j2 \cdot 2 + j2 \cdot j = 2 + j + j4 - 2 = 4j + j =