Question 5 of 6: Thévenin equivalent and maximum power transfer

Question 5: Thévenin equivalent and maximum power transfer [10 + 4 + 6]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $V_s=15\angle 45^\circ\text{ V rms}$ in series with $j5\,\Omega$; a $5\angle 0^\circ\text{ A rms}$ current source injecting into node $M$; a $-j2\,\Omega$ capacitor from $M$ to node $N$; and a $10\,\Omega$ resistor from $N$ (terminal A) to ground (terminal B).

Find. $V_{th}$, $Z_{th}$, the matched $Z_L$, and $P_{max}$.

[Figure not reproduced: Figure 5 — redrawn. Node $M$ takes the $j5\,\Omega$ branch and the $5\text{ A}$ source; $-j2\,\Omega$ links $M$ to node $N$ (terminal A), which carries the $10\,\Omega$ shunt. See the official exam paper.]

Approach. Find $Z_{th}$ by deactivating both sources (short the voltage source, open the current source) and reducing the network at A–B; find $V_{th}$ as the open-circuit voltage $V_N$ by nodal analysis; then the matched load is $Z_L=Z_{th}^{*}$ and $P_{max}=|V_{th}|^2/(4R_{th})$.

1. Thévenin impedance, part (i). Short $V_s$ (its node becomes ground) and open the $5\text{ A}$ source. Looking into A–B: the $10\,\Omega$ shunt is in parallel with the series path $-j2$ (to $M$) then $+j5$ (to ground):$$Z_{th}=10\parallel(-j2+j5)=10\parallel j3=\frac{10(j3)}{10+j3}=\boxed{\,0.826+j2.752\ \Omega\,}$$
2. Open-circuit node equations. With A–B open, let the source node be $V_P=15\angle45^\circ$. KCL at $M$ and $N$:$$\frac{V_P-V_M}{j5}+5=\frac{V_M-V_N}{-j2},\qquad \frac{V_M-V_N}{-j2}=\frac{V_N}{10}.$$
3. Thévenin voltage, part (i). Solving the two complex equations gives the open-circuit terminal voltage $V_{th}=V_N$:$$\boxed{\,V_{th}=35.59\angle 56.71^\circ\text{ V}\,}\quad(=19.55+j29.75\text{ V}).$$
4. Matched load, part (ii). Maximum power transfer to a complex source requires the conjugate match:$$\boxed{\,Z_L=Z_{th}^{*}=0.826-j2.752\ \Omega\,}$$ The load reactance cancels $X_{th}$ and $R_L=R_{th}$.
5. Maximum power, part (iii). At the conjugate match the source impedance sums to $2R_{th}$, so with rms $V_{th}$:$$P_{max}=\frac{|V_{th}|^2}{4R_{th}}=\frac{(35.59)^2}{4(0.826)}=\boxed{\,383\ \text{W}\,}$$