Maximum power transfer with a dependent source: 22-Elec-A1 Circuits, December 2018, Q4

Question 4: Maximum power transfer with a dependent source [10 + 10]

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.

Figure‑4 — 50 V source, 2Ω and 5Ω with a current-controlled…voltage-dependent current source 0.5V₀ (arrow toward M) in parallel with the 4Ω between node M and terminal a.

Given. A $50\text{ V}$ source in series with $2\,\Omega$ feeds node $M$; a $5\,\Omega$ resistor runs from $M$ to the bottom rail (b), and $V_o$ is its voltage ($+$ at $M$). Between $M$ and terminal a a $4\,\Omega$ resistor is in parallel with a current-dependent current source $0.5V_o$ whose arrow points from a toward $M$.

Find. (a) $R_L=R_{th}$ at a–b; (b) $P_{max}$.

Approach. Because a dependent source is present, find the open-circuit voltage $V_{th}$ and the short-circuit current $I_{sc}$ directly; then $R_{th}=V_{th}/I_{sc}$, $R_L=R_{th}$, and $P_{max}=V_{th}^2/(4R_{th})$.

Open-circuit voltage. With a–b open, KCL at node a (only the $4\,\Omega$ and the source connect) gives $\dfrac{V_a-V_M}{4}+0.5V_o=0$ with $V_o=V_M$, so $V_a=-V_M$. KCL at $M$, $\dfrac{V_M-50}{2}+\dfrac{V_M}{5}+\dfrac{V_M-V_a}{4}-0.5V_M=0$, reduces (the $4\,\Omega$ and source terms cancel) to $\dfrac{V_M-50}{2}+\dfrac{V_M}{5}=0$, giving $V_M=\dfrac{250}{7}=35.71\text{ V}$ and $$\boxed{V_{th}=V_a=-35.71\text{ V}\;(|V_{th}|=35.71\text{ V}).}$$
Short-circuit current. With a–b shorted ($V_a=0$), KCL at $M$ becomes $\dfrac{V_M-50}{2}+\dfrac{V_M}{5}+\dfrac{V_M}{4}-0.5V_M=0\Rightarrow V_M=\dfrac{500}{9}=55.56\text{ V}.$ The short current from a to b is $$I_{sc}=\frac{V_M}{4}-0.5V_M=-0.25V_M=-13.89\text{ A}.$$
Thévenin resistance. $$\boxed{R_{th}=\frac{V_{th}}{I_{sc}}=\frac{-35.71}{-13.89}=\frac{18}{7}=2.571\,\Omega.}$$ For maximum power transfer the load is matched: $\boxed{R_L=R_{th}=2.571\,\Omega.}$
Maximum power. With $R_L=R_{th}$ the load sees half of $V_{th}$: $$\boxed{P_{max}=\frac{V_{th}^{\,2}}{4R_{th}}=\frac{35.71^{2}}{4(2.571)}=124.0\text{ W}.}$$

Note on the source type The figure labels the dependent element $0.5V_o$ with a current-source (diamond) symbol, so it is treated as a voltage-controlled current source of $0.5V_o$ amperes. The sign work above keeps $R_{th}$ positive, which is the physically meaningful check.

Quantity	Result
Thévenin voltage $\|V_{th}\|$	$35.71\text{ V}$
Short-circuit current $\|I_{sc}\|$	$13.89\text{ A}$
Thévenin resistance $R_{th}=R_L$	$2.571\,\Omega$
Maximum power $P_{max}$	$124.0\text{ W}$

Concept & Theory

Topic: Thévenin equivalent and maximum-power transfer with a dependent source.

Key relations: $R_{th}=V_{oc}/I_{sc}$ (needed when a dependent source forbids “resistance by inspection”); matched load $R_L=R_{th}$; $P_{max}=V_{th}^2/(4R_{th})$.

Why this works: a dependent source can only be handled by exciting the port — either open/short as here, or with a test source — because it changes value with the circuit’s own variables and cannot simply be zeroed. Once $V_{th}$ and $R_{th}$ are known the load is a plain resistive-divider power problem.

Common pitfall: deactivating the dependent source to find $R_{th}$ by inspection (illegal), or forgetting that maximum power to a purely resistive load requires $R_L=R_{th}$ (not conjugate matching, which applies to reactive Thévenin impedances).

Source: Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., §4.5–4.6 and §4.8 (Thévenin, maximum power).

Question 4 of 6: Maximum power transfer with a dependent source

Question 4: Maximum power transfer with a dependent source [10 + 10]