Network Theorems Practice Questions

$V_{L_1}=\left ( \frac{j4}{3+j4} \right )100\angle 53.13^{\circ}$ $\;\;=80\angle 90^{\circ}V$ $V_{Th}=10V_{L_1}=800\angle 90^{\circ}V$

Using maximum power transfer theorem, $R_L=|Z|=|4-j3|$ $\;\;=\sqrt{4^{2+}3^2}=5\Omega$

$I=\frac{7}{R+2}$ $V=10-2I$ $\;\;=10-\frac{14}{R+2}=\frac{10R+6}{R+2}$ Power transferred from circuit A to circuit B $P=VI$ $\;\;=\frac{10R+6}{R+2} \times \frac{7}{R+2}$ For P to be maximum $\frac{dP}{dR}=0$ $(R+2)^2(10)-(10R+6)\times 2 (R+2)=0$ $5R^{2+}20R+20-10R^{2+}26R+12=0$ $5R^{2+}6R=8$ $\Rightarrow \;R=0.8\Omega$

To find thevenin impedance across node 1 and node 2. Connect a 1 V source and find the current through voltage source. Then, $Z_{Th}=\frac{1}{I_{Th}}$ By applying KCL at node B and A $i_{AB}+99i_{b}=I_{Th}$ $i_b==i_A+i_{AB}$ $\Rightarrow \;\; i_b-i_A+99i_b=I_{Th}$ $\Rightarrow \;\; 100i_b-i_A=I_{Th}\;...(i)$ By applying KVL in outer loop $10 \times 10^3i_b=1$ $i_b=10^{-4}A$ and $10 \times 10^{3}i_b=-100i_A$ $i_A=-100i_b$ $\therefore$ From equation (i), $100i_b+100i_b=I_{Th}$ $\Rightarrow \; I_{Th}=200i_b$ $\;\;=200 \times 10^{-4}=0.02$ $\therefore \;\;Z_{Th}=\frac{1}{I_{Th}}=\frac{1}{0.02}=50\Omega$

For maximum power transfer to the load, the source resistance must be minimum i.e. zero. So, $R=0\Omega$ .

To calculate thevenin's resistance 5 V source is short-circuited and $V_{dc}$ source is connected at terminals A and B $R_{th}=\frac{V_{dc}}{I}$ $V_{AB}=V_{dc}$ and assuming $I \; and \; I_1$ in mA. Current through $1k\Omega$ resistance $I_1=\frac{V_{dc}}{1}=V_{dc}$ $V_{AB}=V_{dc}$ $V_x=2||2\times (I-I_1)$ $V_x=I-I_1=I-V_{dc}\;...(i)$ $V_x=3V_{AB}+I_1$ $\;\;3V_{dc}+V_{dc}=4V_{dc}\;...(ii)$ Comparing equation (i) and (ii), $I-V_{dc}=4V_{dc}$ $R_{th}=\frac{V_{dc}}{I}=\frac{1}{5}=0.2\Omega$

To calculate thevenin's volatage, terminal A-B are kept open. Applying source transformation, voltage sorce is transformed into current source. Applying source transformation current source is transformed into voltage souce. Applying KVL, (I is assumed to be in mA) $\frac{5}{2}-I-3V_{OC}-I=0\;\;...(i)$ $I=\frac{V_{OC}}{I}\;\;...(ii)$ Put value of I in equation (i), $\frac{5}{2}-5V_{OC}=0$ $\Rightarrow \; V_{OC}=0.5 V$

Thevenin's Impedance: $Z_0=2.38-j0.667\Omega$ as real part is not zero, so $Z_0$ has resistor $Img[Z_0]=-j0.667$ CASE-I: $Z_0$ has capacitor (as $Img[Z_0]$ is negative) CASE-II: $Z_0$ has both capacitor and inductor, but inductive reactance $\lt$ capacitive reactance. At, $\omega$ =5 rad/sec For minimal realization case-I is considered. Therefore, $Z_0$ will have a resistor and a capacitor.

To calculate Thevenin's impedance, current-souce is open-circuited $Z_{th}=R+Z_L+Z_c$ $\;\;=1+2j-j$ $\;\;=1+j \Omega$ Open-circuit voltage at terminals X-Y $=I \times Z_{th}$ $\;\;=1\angle 0 \times (1+j)$ $\;\;=\sqrt{2}\angle 15^{\circ} volts$

To calculate $R_{th}$ (seen at terminals P-Q), voltage source is short-circuit $V_{th}$ = open-circuit voltage at terminals P-Q. $V_{th}=\frac{4}{10+10}\times 10=2 V$ Thevenin's equivalent circuit

For obtaining power absorbed by $R_L$ under maximum power transfer condition. We find thevenin's equivalent circuit across $R_L$ . $Z_{th}$ is calculated by short-circuting the voltage source. $Z_{th}=(6+j8)||(6+j8)=3+j4\Omega$ $\frac{V_{th}-110\angle 0^{\circ}}{6+j8}+\frac{V_{th}-90\angle 0^{\circ}}{6+j8}=0$ $V_{th}=100\angle 0^{\circ}V$ For maximum power transfer , $R_L=\sqrt{R_{th}^{2+}X_{th}^2}$ $\;\;=\sqrt{3^{2+}4^2}=5\Omega$ $I=\frac{V_{th}}{(3+j4)+R_L}=\frac{100}{8+j4}$ $\;\;=11.18\angle -26.56^{\circ}A$ Power absorbed by $R_L (max)$ $\; =I^2R_L=11.18^2 \times 5$ $\;\;=625W$

By Thevenin's theorem $Z_{Th}=Z_{X-Y}=Z_1||Z_2+Z_3$ $\;\;=\frac{Z_1 \times Z_2}{Z_1+Z_2}+Z_3$ $\;\;=\frac{10\angle -60 \times 10\angle 60}{10\angle -60 + 10\angle 60}+50\angle 53.13$ $\;\;=56.66\angle 45^{\circ}\Omega$

Currently no explanation available

Currently no explanation available