Question

2.2 two ac voltages are given by the expressions $\frac{3}{2}+j(y + 1)$ and $(7 + x)-3j$. if the voltages are equal what are the values of $x$ and $y$? (4)
2.3 evaluate $(2angle30^{circ})^6$ and write the answer in exponential form. (2)
2.4 identify and sketch the following curve: $25x^{2}+16y^{2}=400$ (4)

Explanation:

Step1: Equate real - parts for 2.2

For equal complex numbers, real parts are equal. So, $\frac{3}{2}=7 + x$.
Solve for $x$:
$x=\frac{3}{2}-7=\frac{3 - 14}{2}=-\frac{11}{2}$

Step2: Equate imaginary - parts for 2.2

For equal complex numbers, imaginary parts are equal. So, $y + 1=-3$.
Solve for $y$:
$y=-3 - 1=-4$

Step3: Evaluate $(2\angle30^{\circ})^6$ for 2.3

In polar form, if $z = r\angle\theta$, then $z^n=r^n\angle(n\theta)$. Here, $r = 2$, $\theta=30^{\circ}=\frac{\pi}{6}$ radians, and $n = 6$.
$r^n=2^6 = 64$ and $n\theta=6\times\frac{\pi}{6}=\pi$ radians.
In exponential form $z = r e^{j\theta}$, so the answer is $64e^{j\pi}$

Step4: Identify the curve for 2.4

Rewrite the equation $25x^{2}+16y^{2}=400$ in standard form $\frac{x^{2}}{16}+\frac{y^{2}}{25}=1$. This is an ellipse centered at the origin $(0,0)$ with semi - major axis $a = 5$ along the $y$ - axis and semi - minor axis $b = 4$ along the $x$ - axis. To sketch, mark the points $(0,5),(0, - 5),(4,0),(-4,0)$ and draw an ellipse passing through them.

Answer:

2.2: $x=-\frac{11}{2},y=-4$
2.3: $64e^{j\pi}$
2.4: It is an ellipse centered at the origin with semi - major axis $a = 5$ along the $y$ - axis and semi - minor axis $b = 4$ along the $x$ - axis.