91Ó°ÊÓ

Evaluate in polar form (a) \(\frac{16 \angle 75^{\circ}}{2 \angle 15^{\circ}}\) (b) \(\frac{10 \angle \frac{\pi}{4} \times 12 \angle \frac{\pi}{2}}{6 \angle-\frac{\pi}{3}}\) (a) \(\frac{16 \angle 75^{\circ}}{2 \angle 15^{\circ}}=\frac{16}{2} \angle\left(75^{\circ}-15^{\circ}\right)=8 \angle 60^{\circ}\) (b) \(\frac{10 \angle \frac{\pi}{4} \times 12 \angle \frac{\pi}{2}}{6 \angle-\frac{\pi}{3}}=\frac{10 \times 12}{6} \angle\left(\frac{\pi}{4}+\frac{\pi}{2}-\left(-\frac{\pi}{3}\right)\right)\) $$ =20 \angle \frac{13 \pi}{12} \text { or } 20 \angle-\frac{11 \pi}{12} \text { or } $$ \(20 \angle 195^{\circ}\) or \(20 \angle-165^{\circ}\)

Evaluate, in polar form \(2 \angle 30^{\circ}+5 \angle-45^{\circ}-4 \angle 120^{\circ}\) Addition and subtraction in polar form is not possible directly. Each complex number has to be converted into Cartesian form first. $$ \begin{aligned} 2 \angle 30^{\circ} &=2\left(\cos 30^{\circ}+j \sin 30^{\circ}\right) \\ &=2 \cos 30^{\circ}+j 2 \sin 30^{\circ}=1.732+j 1.000 \\ 5 \angle-45^{\circ} &=5\left(\cos \left(-45^{\circ}\right)+j \sin \left(-45^{\circ}\right)\right) \\ &=5 \cos \left(-45^{\circ}\right)+j 5 \sin \left(-45^{\circ}\right) \\ &=3.536-j 3.536 \\ 4 \angle 120^{\circ} &=4\left(\cos 120^{\circ}+j \sin 120^{\circ}\right) \\ &=4 \cos 120^{\circ}+j 4 \sin 120^{\circ} \\ &=-2.000+j 3.464 \end{aligned} $$ Hence \(2 \angle 30^{\circ}+5 \angle-45^{\circ}-4 \angle 120^{\circ}\) $$ \begin{aligned} =(1.732+j 1.000)+(3.536-&j 3.536) \\ -(-2.000+j 3.464) \end{aligned} $$ \(=7.268-j 6.000\), which lies in the fourth quadrant $$ \left.=\sqrt{\left[(7.268)^{2}+(6.000)^{2}\right.}\right] \angle \tan ^{-1}\left(\frac{-6.000}{7.268}\right) $$ $$ =9.425 \angle-39.54^{\circ} $$

Determine the resistance and series inductance (or capacitance) for each of the following impedances, assuming a frequency of \(50 \mathrm{~Hz}\) (a) \((4.0+j 7.0) \Omega\) (b) \(-j 20 \Omega\) (c) \(15 \angle-60^{\circ} \Omega\) (a) Impedance, \(Z=(4.0+j 7.0) \Omega\) hence, resistance \(=4.0 \Omega\) and reactance \(=7.00 \Omega\) Since the imaginary part is positive, the reactance is inductive, i.e. \(X_{L}=7.0 \Omega\) Since \(X_{L}=2 \pi f L\) then inductance, $$ L=\frac{X_{L}}{2 \pi f}=\frac{7.0}{2 \pi(50)}=\mathbf{0 . 0 2 2 3} \mathbf{H} \text { or } \mathbf{2 2 . 3} \mathbf{~ m H} $$ (b) Impedance, \(Z=j 20\), i.e. \(Z=(0-j 20) \Omega\) hence resistance \(=0\) and reactance \(=20 \Omega\), Since the imaginary part is negative, the reactance is capacitive, i.e., \(X_{C}=20 \Omega\) and since \(X_{C}=\frac{1}{2 \pi f C}\) then: $$ \text { capacitance, } \begin{aligned} C &=\frac{1}{2 \pi f X_{C}}=\frac{1}{2 \pi(50)(20)} \mathrm{F} \\ &=\frac{10^{6}}{2 \pi(50)(20)} \mu \mathrm{F}=\mathbf{1 5 9} .2 \mu \mathrm{F} \end{aligned} $$ (c) Impedance, \(Z\) $$ \begin{aligned} &=15 \angle-60^{\circ}=15\left[\cos \left(-60^{\circ}\right)+j \sin \left(-60^{\circ}\right)\right] \\ &=7.50-j 12.99 \Omega \end{aligned} $$ Hence resistance \(=7.50 \Omega\) and capacitive reactance, \(X_{C}=12.99 \Omega\) Since \(X_{C}=\frac{1}{2 \pi f C}\) then capacitance, $$ \begin{aligned} \boldsymbol{C} &=\frac{1}{2 \pi f X_{C}}=\frac{10^{6}}{2 \pi(50)(12.99)} \mu \mathrm{F} \\ &=\mathbf{2 4 5} \mu \mathbf{F} \end{aligned} $$