for the circuit shown:\n(a) (5 marks) determine the magnitude and direction of the total impedance\n(b) (2…

Explanation:

Step1: Analyze the circuit components in complex - form

The impedance of a resistor $R$ is $Z_R = R$, the impedance of a capacitor $C$ is $Z_C=\frac{1}{j\omega C}$ (assuming $\omega = 1$ for simplicity, and here we have $Z_{C1}=-j10$ and $Z_{C2}=-j7$), and the impedance of an inductor $L$ is $Z_L = j\omega L$ (assuming $\omega = 1$, so $Z_{L}=j6$). The resistors are $R_1 = 8$ and $R_2 = 5$. First, consider the right - most part of the circuit. The impedance of the inductor and the resistor in series is $Z_{s1}=5 + j6$. The impedance of the capacitor and the resistor in parallel is $Z_{p1}=\frac{8\times(-j7)}{8 - j7}=\frac{-j56}{8 - j7}=\frac{-j56(8 + j7)}{(8 - j7)(8 + j7)}=\frac{-j448+392}{64 + 49}=\frac{392-j448}{113}\approx3.47 - j3.96$. Then, consider the parallel combination of $Z_{p1}$ and $Z_{C1}=-j10$. [ \begin{align*} Z_{p2}&=\frac{(3.47 - j3.96)\times(-j10)}{(3.47 - j3.96)+(-j10)}\ &=\frac{-j34.7 - 39.6}{3.47 - j13.96}\ &=\frac{39.6 - j34.7}{3.47 - j13.96}\times\frac{3.47 + j13.96}{3.47 + j13.96}\ &=\frac{39.6\times3.47+39.6\times j13.96 - j34.7\times3.47+34.7\times13.96}{3.47^{2}+13.96^{2}}\ &=\frac{137.412 + j553.816 - j120.409+484.412}{12.0409 + 194.8816}\ &=\frac{621.824 + j433.407}{206.9225}\approx3.00+ j2.09 \end{align*} ]

Step2: Calculate the total impedance

The total impedance $Z_{total}$ is the impedance of the parallel combination of the source and the rest of the circuit. Since we have calculated the impedance of the non - source part of the circuit as $Z_{p2}\approx3.00 + j2.09$, the magnitude of the total impedance $|Z_{total}|=\sqrt{3.00^{2}+2.09^{2}}=\sqrt{9 + 4.3681}=\sqrt{13.3681}\approx3.66$. The direction $\theta=\arctan(\frac{2.09}{3.00})\approx34.9^{\circ}$.

Step3: Calculate the voltage drop

Given the current $I = 1.8 - j0.10$, using Ohm's law $V = IZ$. So $V=(1.8 - j0.10)\times(3.00 + j2.09)=1.8\times3.00+1.8\times j2.09 - j0.10\times3.00+0.10\times2.09=(5.4 + 0.209)+j(3.762 - 0.3)=5.609 + j3.462$. The magnitude of the voltage drop $|V|=\sqrt{5.609^{2}+3.462^{2}}=\sqrt{31.460881+11.985444}=\sqrt{43.446325}\approx6.59$

Answer:

(a) Magnitude = $3.66$, Direction = $34.9^{\circ}$ (b) Voltage drop is: $5.609 + j3.462$ (or magnitude $\approx6.59$)