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22-Elec-A1 Circuits · May 2015

Question 5 of 6: Thévenin equivalent and maximum power transfer

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — mesh/supermesh and nodal analysis, first-order RC transients, AC steady-state phasors, complex power, Thévenin’s theorem and maximum-power transfer, and Laplace-domain circuit analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of initial-condition source models in the s-domain.

Closed-book national examination, 3 hours, six questions of equal value; any five constitute a complete paper. All six are solved here. A Laplace-transform table is supplied with the paper. Phasor results are quoted as magnitude∡angle with angles in degrees; sources marked rms are treated as rms.

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.

ElementValuePlacement
$V_s$$5\angle 0^\circ\text{ V (rms)}$left branch
Series $j2+3$$3+j2\,\Omega$$V_s$ to node P
Shunt $j4$$j4\,\Omega$P to ground
$1\,\Omega$$1\,\Omega$P to terminal a
$I_s$$10\angle 30^\circ\text{ A (rms)}$terminal b (node Q) to ground
$2-j5$$2-j5\,\Omega$Q to ground

Find. $V_{th}$ and $Z_{th}$ across $a$–$b$, the matched load $Z_L$, and the maximum power delivered to it.

+−Vs = 5∠0° Vj2 Ω3 Ωj4 Ω1 ΩaZLbIs = 10∠30° A2 Ω−j5 Ω
Figure 5 — the two sub-networks meet only at the open terminals $a$–$b$ (and the shared ground), so the Thévenin looking into $a$–$b$ is the left branch in series with the right branch.

Approach. With $Z_L$ removed the two halves are independent: find each terminal’s open-circuit voltage to ground and dead-network impedance, then $V_{th}=V_a-V_b$ and $Z_{th}=Z_a+Z_b$.

  1. Left open-circuit voltage. With $a$ open no current flows in the $1\,\Omega$, so $V_a$ equals the node-P divider:$$V_a=V_s\frac{j4}{3+j2+j4}=5\frac{j4}{3+j6}=2.98\angle 26.57^\circ\text{ V}.$$
  2. Left dead-network impedance. Short $V_s$; looking into $a$: $1\,\Omega$ in series with $j4\parallel(3+j2)$:$$Z_a=1+\frac{j4(3+j2)}{3+j6}=2.07+j1.87\,\Omega.$$
  3. Right open-circuit voltage. With $b$ open the whole $I_s$ flows through $2-j5$:$$V_b=I_s(2-j5)=(10\angle 30^\circ)(5.39\angle{-68.2^\circ})=53.85\angle{-38.2^\circ}\text{ V}.$$
  4. Right dead-network impedance. Open the current source: $Z_b=2-j5\,\Omega$.
  5. Thévenin quantities.$$V_{th}=V_a-V_b=-39.7+j34.6=\boxed{\,52.65\angle 138.9^\circ\text{ V}\,}$$$$Z_{th}=Z_a+Z_b=\boxed{\,4.07-j3.13\,\Omega\,}$$
  6. Matched load and maximum power. Maximum transfer needs $Z_L=Z_{th}^{*}=4.07+j3.13\,\Omega$, and then$$P_{\max}=\frac{|V_{th}|^2}{4\,R_{th}}=\frac{(52.65)^2}{4(4.07)}=\boxed{\,170.4\text{ W}\,}$$ (rms phasors, so no extra $\tfrac12$).
QuantityValue
$V_{th}$$52.65\angle 138.9^\circ\text{ V}$
$Z_{th}$$4.07-j3.13\,\Omega$
$Z_L$ (matched)$4.07+j3.13\,\Omega$
$P_{\max}$$170.4\text{ W}$