Thévenin equivalent with a dependent source, maximum power: 22-Elec-A1 Circuits, May 2016, Q2

Question 2: Thévenin equivalent with a dependent source, maximum power [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. A dependent voltage source $v/3$ (controlled by $v$, the voltage across the $15\,\Omega$ resistor) sits on the far-left branch; the top path runs through $6\,\Omega$ and a $150\text{ V}$ independent source ($-$ left, $+$ right) to node 3, where the $15\,\Omega$ resistor, a $5\text{ A}$ current source (arrow down) and a $30\,\Omega$ resistor to terminal a all connect. The bottom rail is the reference (terminal b).

Find. $V_{th}$ and $R_{th}$ at a–b, the matched load $R_L$, and $P_{L,\max}$.

[Figure not reproduced: Figure 2 — redrawn. Control variable $v$ is the node-3 voltage (across $15\,\Omega$); the dependent source sets the far-left node to $v/3$. See the official exam paper.]

Approach. With the load removed, no current flows in the $30\,\Omega$ so $V_{th}$ equals the node-3 voltage $v$; find $v$ by one KCL/branch equation. Because a dependent source is present, get $R_{th}$ from $V_{th}/I_{sc}$ (a test source would work equally).

Label node voltages. Reference = bottom rail. The dependent source fixes $V_1=v/3$; across the $150\text{ V}$ source $V_3 = V_2+150$, so $V_2 = v-150$ (writing $V_3=v$).
Series current in the left path. The $v/3$ source, $6\,\Omega$ and $150\text{ V}$ source form one series path carrying current $I$ into node 3: $$I=\frac{V_1-V_2}{6}=\frac{v/3-(v-150)}{6}=\frac{-\tfrac{2}{3}v+150}{6}.$$
KCL at node 3 (open circuit, $30\,\Omega$ carries no current). Current in equals the $15\,\Omega$ current plus the $5\text{ A}$ drawn by the source: $$I=\frac{v}{15}+5.$$ Equating and clearing denominators: $600=\dfrac{16}{3}v\Rightarrow v=112.5\text{ V}.$ $$\boxed{V_{th}=112.5\text{ V}}$$ Check: $I=(37.5-(-37.5))/6=12.5\text{ A}$ and $v/15+5=7.5+5=12.5\text{ A}$ ✓.
Short-circuit current. Short a–b ($V_a=0$); the $30\,\Omega$ now carries $I_{sc}=V_3/30$. KCL at node 3 becomes $$\frac{-\tfrac23 V_3+150}{6}=\frac{V_3}{15}+5+\frac{V_3}{30}\;\Rightarrow\;600=\frac{19}{3}V_3\;\Rightarrow\;V_3=\frac{1800}{19}=94.74\text{ V}.$$ Hence $I_{sc}=94.74/30=3.158\text{ A}.$
Thévenin resistance. $$R_{th}=\frac{V_{th}}{I_{sc}}=\frac{112.5}{3.158}=\boxed{35.625\,\Omega}.$$ (A 1 A test source with the independents killed gives the same $R_{th}=\tfrac{19}{3}\cdot\tfrac{45}{8}=35.625\,\Omega$.)
(b) Matched load. Maximum power transfer requires $$\boxed{R_L=R_{th}=35.625\,\Omega}.$$
(c) Maximum power. With $R_L=R_{th}$ the load sees $V_{th}/2$, so $$P_{L,\max}=\frac{V_{th}^{2}}{4R_{th}}=\frac{112.5^{2}}{4(35.625)}=\frac{12656.25}{142.5}=\boxed{88.82\text{ W}}.$$

Quantity	Value
$V_{th}$ at a–b	$112.5\text{ V}$
$R_{th}$ at a–b	$35.625\,\Omega$
Matched load $R_L$	$35.625\,\Omega$
Maximum power $P_{L,\max}$	$88.82\text{ W}$

Topic: Thévenin equivalent of a circuit containing a dependent source; maximum-power transfer.

Key relations: $R_{th}=V_{oc}/I_{sc}$ (valid even with dependent sources, which must stay active); $P_{\max}=V_{th}^2/(4R_{th})$ at $R_L=R_{th}$.

Why this works: A dependent source cannot be zeroed, so the simple “kill the sources and look in” shortcut fails; either drive the terminals with a test source (independents off, dependent on) or take the ratio of open-circuit voltage to short-circuit current. Both encode the same linear two-terminal behaviour $v_{ab}=V_{th}-R_{th}\,i_{ab}$.

Common pitfall: deactivating the dependent source when finding $R_{th}$, or forgetting that the $5\text{ A}$ source and $30\,\Omega$ change the node-3 equation between the open- and short-circuit cases.

Source: Sadiku, §4.5–4.6 (Thévenin with dependent sources) and §4.8 (maximum power transfer).

Question 2 of 6: Thévenin equivalent with a dependent source, maximum power

Question 2: Thévenin equivalent with a dependent source, maximum power [10 + 4 + 6]