ECE 211 · Circuit Analysis · Interactive Notes

Lecture 5: Equivalents with Dependent Sources and Maximum Power Transfer

This page extends the Thevenin and Norton constructions to circuits that contain dependent sources, where the looking-in resistance must be measured with a test source rather than read off by series and parallel combination, and then develops the maximum power transfer theorem with its matched-load condition. The material is reconstructed from the Lecture 5 board notes as interactive figures.

Teal indicates linked content. Hovering over (or tabbing to) a boxed term such as V1V_1 highlights the corresponding circuit element; the linkage also operates in the reverse direction, from the circuit to the equations.

§1 · The port experiments, revisited

Dependent sources change the resistance experiment

Lecture 4 established that any linear two-terminal circuit collapses to a Thevenin or a Norton equivalent, and that three port experiments extract the parameters: the open-circuit voltage, the short-circuit current, and the looking-in resistance with all independent sources off. Any two of the experiments determine both equivalents; the third serves as a check.

VT = Voc IN = Isc RT = RN VT = IN · RT

The complication of this lecture is the dependent source. A dependent source is never turned off, so the source-off circuit is no longer a bare resistor network, and RTR_T cannot be read off by series and parallel combination. The resistance experiment must instead be performed literally: the port is driven by a known source, and the response is computed.

The test-source recipe (from the board notes)

With all independent sources OFF (dependent sources remain active):

  1. The circuit is driven at the port with a 1 V or a 1 A source.
  2. The resulting port current or port voltage is solved for.
  3. Ohm's law delivers the resistance: RT=V/IR_T = V / I.
Dependent sources stay in

Only independent sources are turned off (voltage sources become shorts, current sources become opens). A dependent source is part of the circuit's internal behavior, and it participates fully in every experiment, including the resistance measurement. Turning it off produces a wrong RTR_T.

One circuit is carried through the next four sections. All three experiments are performed on it, which is one more than necessary, precisely so that the redundant experiment can confirm the other two.

§2 · Experiment 1: the open circuit

VTV_T from the open port

The problem

Find the Thevenin and Norton equivalents at the terminals a and b. Given: 2 Ω, 5 Ω, and 10 Ω resistors, a 10 V source, and the dependent source 2V12V_1, a voltage-controlled voltage source whose controlling voltage V1V_1 sits across the 2 Ω resistor.

The open port kills the controlling voltage

With the port open, the 2 Ω resistor carries 0 A, so V1=0V_1 = 0 and no voltage drops between terminal a and the middle node. The middle node therefore carries the open-circuit voltage and is labeled VTV_T directly.

The dependent source pins the right node

The dependent source holds the right node at 2V1=02V_1 = 0 V. The source is not removed; its value merely happens to be zero in this particular experiment, because its controlling voltage vanished.

One KCL equation delivers VTV_T

VT−1010+VT−05=0  ⟹  VT−10+2VT=0  ⟹  VT=103 V\tm{oc.r10 oc.vs10}{\frac{V_T - 10}{10}} + \tm{oc.r5 oc.nr}{\frac{V_T - 0}{5}} = 0 \;\Longrightarrow\; V_T - 10 + 2V_T = 0 \;\Longrightarrow\; \boxed{V_T = \tfrac{10}{3}\ \text{V}}

The first term leaves through the 10 Ω resistor toward the 10 V node, and the second leaves through the 5 Ω resistor toward the right node at 0 V. No current leaves through the open port.

§3 · Experiment 2: the short circuit

INI_N from the shorted port

Short the terminals

A short is placed across a-b, and the current INI_N through it, directed from a to b, is sought. The short changes the circuit: the 2 Ω resistor now carries current, so V1V_1 is alive and the dependent source acts.

Label the nodes with the fewest unknowns

Terminal a is pinned at 0 V by the short. The definition of V1V_1 (positive mark at the a side of the 2 Ω resistor) then forces the middle node to −V1-V_1, and the dependent source holds the right node at 2V12V_1. One symbol covers every node.

KCL at the middle node

−V1−02+−V1−1010+−V1−2V15=0\tm{sc.r2}{\frac{-V_1 - 0}{2}} + \tm{sc.r10 sc.vs10}{\frac{-V_1 - 10}{10}} + \tm{sc.r5 sc.nr}{\frac{-V_1 - 2V_1}{5}} = 0

Multiplying by 10 gives −5V1−V1−10−6V1=0-5V_1 - V_1 - 10 - 6V_1 = 0, so V1=−56V_1 = -\tfrac{5}{6} V. The negative sign records that the actual 2 Ω current flows from the middle node toward terminal a.

Ohm's law at the 2 Ω resistor recovers INI_N

IN=−V1−02=5/62  ⟹  IN=512 AI_N = \tm{sc.r2}{\frac{-V_1 - 0}{2}} = \frac{5/6}{2} \;\Longrightarrow\; \boxed{I_N = \tfrac{5}{12}\ \text{A}}

The short carries exactly the current delivered into terminal a by the 2 Ω resistor, which is the Norton current.

§4 · Experiment 3: the test source

RTR_T by driving the dead circuit

Turn off the independent source, then drive the port

The 10 V source is replaced by a short; the dependent source remains. A 1 A test source is connected at the port, and the resulting port voltage VxV_x is sought, because RT=Vx/1 AR_T = V_x / 1\ \text{A}.

The test current labels every node

Terminal a connects only to the test source and the 2 Ω resistor, so the entire 1 A crosses the 2 Ω resistor and V1=1⋅2=2V_1 = 1 \cdot 2 = 2 V. With the port voltage named VxV_x, the middle node sits at Vx−2V_x - 2, and the dependent source pins the right node at 2V1=42V_1 = 4 V.

KCL at the middle node

(Vx−2)−Vx2+(Vx−2)−010+(Vx−2)−45=0\tm{rt.r2}{\frac{(V_x - 2) - V_x}{2}} + \tm{rt.r10 rt.off}{\frac{(V_x - 2) - 0}{10}} + \tm{rt.r5 rt.nr}{\frac{(V_x - 2) - 4}{5}} = 0

The first term is simply −1-1 A (the test current arriving). Multiplying by 10: −10+Vx−2+2Vx−12=0-10 + V_x - 2 + 2V_x - 12 = 0, so 3Vx=243V_x = 24 and Vx=8V_x = 8 V.

Ohm's law delivers the resistance

RT=Vx1 A=8 V1 A  ⟹  RT=8 ΩR_T = \frac{V_x}{1\ \text{A}} = \frac{8\ \text{V}}{1\ \text{A}} \;\Longrightarrow\; \boxed{R_T = 8\ \Omega}

It is worth noting what a resistor-only reduction would have produced with the dependent source wrongly removed: 2+10∥5=5132 + 10 \parallel 5 = 5\tfrac{1}{3} Ω, which is not the port resistance of this circuit.

Alternate route: the same resistance from a 1 V drive

The test source may equally be a voltage source. With 1 V applied at the port, terminal a sits at 1 V and the middle node at 1−V11 - V_1; the dependent source holds the right node at 2V12V_1. KCL at the middle node:

(1−V1)−12+(1−V1)−010+(1−V1)−2V15=0\frac{(1 - V_1) - 1}{2} + \frac{(1 - V_1) - 0}{10} + \frac{(1 - V_1) - 2V_1}{5} = 0

Multiplying by 10 gives −5V1+1−V1+2−6V1=0-5V_1 + 1 - V_1 + 2 - 6V_1 = 0, so V1=14V_1 = \tfrac{1}{4} V. The current drawn from the test source equals the 2 Ω current, Ix=V1/2=18I_x = V_1 / 2 = \tfrac{1}{8} A, and

RT=1 V1/8 A=8 Ω ✓R_T = \frac{1\ \text{V}}{1/8\ \text{A}} = 8\ \Omega\ \checkmark

Either drive is legitimate; the choice is a matter of which arithmetic looks cleaner.

§5 · The equivalents, and the check

Two circuits, three experiments, one consistency test

The three experiments assemble both equivalents. Because all three were performed, the relation VT=INRTV_T = I_N R_T is available as a genuine check rather than as a definition:

VT = 10/3 V IN = 5/12 A RT = 8 Ω 10/3 = (5/12) · 8 ✓

Thevenin: [10/3 V]{eq.vt} in series with [8 Ω]{eq.rt}.

Norton: [5/12 A]{no.inr} in parallel with [8 Ω]{no.rn}.

Had only two experiments been run, the third parameter would follow from VT=INRTV_T = I_N R_T; for example, RT=Voc/Isc=(10/3)/(5/12)=8R_T = V_{oc} / I_{sc} = (10/3)/(5/12) = 8 Ω reproduces the test-source result without the test source.

§6 · Maximum power transfer

The matched load extracts the most power

A load resistor RLR_L connected at the port receives PL=VLIL=IL2RL=VL2/RLP_L = V_L I_L = I_L^2 R_L = V_L^2 / R_L. Both extremes deliver nothing: a short carries current but holds no voltage, and an open holds voltage but carries no current. A maximum therefore lies between.

RL = 0 Ω ⇒ PL = 0 W RL → ∞ ⇒ PL = 0 W the peak sits in between

Replacing the source network by its Thevenin equivalent makes the load current a one-loop computation, IL=VT/(RT+RL)I_L = V_T / (R_T + R_L), so the load power is

PL=[VTRL+RT]2RL=VT2 RL(RL+RT)2P_L = \left[\frac{V_T}{R_L + R_T}\right]^2 R_L = V_T^2\,\frac{R_L}{(R_L + R_T)^2}

The sliders below drive the Section 5 equivalent (VT=10/3V_T = 10/3 V, RT=8R_T = 8 Ω). The moving dots represent the load current, and the marker rides the power curve.

The equivalent drives [R_L]{mp.rl}; the current [I_L]{mp.il} and the voltage [V_L]{mp.vl} both depend on R_L.

P_L against R_L. The peak sits exactly at R_L = R_T = 8 Ω.

IL = 10/3 V8 Ω + 4.0 Ω = 0.278 A    PL = IL² · RL = 0.309 W  (88.9 % of the maximum)

Derivation: where the peak sits

The maximum is located by setting the derivative to zero. By the quotient rule,

dPLdRL=VT2 (RL+RT)2−RL⋅2(RL+RT)(RL+RT)4=VT2 RT−RL(RL+RT)3=0  ⟹  RL=RT\frac{dP_L}{dR_L} = V_T^2\,\frac{(R_L + R_T)^2 - R_L \cdot 2 (R_L + R_T)}{(R_L + R_T)^4} = V_T^2\,\frac{R_T - R_L}{(R_L + R_T)^3} = 0 \;\Longrightarrow\; \boxed{R_L = R_T}

The derivative is positive below RTR_T and negative above it, so the stationary point is indeed the maximum. Substituting RL=RTR_L = R_T into the power expression gives the ceiling:

PL,max⁡=VT24RTP_{L,\max} = \frac{V_T^2}{4 R_T}

The most this circuit could deliver

For the Section 5 equivalent, the matched load is 8 Ω, and both forms of the equivalent predict the same ceiling, which is a final consistency check:

VL=RLRL+RT VT=816⋅103=53 VPL=VL2RL=(5/3)28=2572 W≈0.347 WV_L = \frac{R_L}{R_L + R_T}\,V_T = \frac{8}{16} \cdot \frac{10}{3} = \frac{5}{3}\ \text{V} \qquad P_L = \frac{V_L^2}{R_L} = \frac{(5/3)^2}{8} = \boxed{\tfrac{25}{72}\ \text{W} \approx 0.347\ \text{W}}

At the matched load the voltage divider splits evenly, so the load sees half of VTV_T.

IL=RNRN+RL IN=12⋅512=524 APL=IL2RL=(524)2⋅8=2572 W ✓I_L = \frac{R_N}{R_N + R_L}\,I_N = \frac{1}{2} \cdot \frac{5}{12} = \frac{5}{24}\ \text{A} \qquad P_L = I_L^2 R_L = \left(\tfrac{5}{24}\right)^2 \cdot 8 = \boxed{\tfrac{25}{72}\ \text{W}}\ \checkmark

At the matched load the current divider also splits evenly, so the load takes half of INI_N.

Efficiency is a different question

At the matched load, the internal resistance dissipates exactly as much as the load receives, so the efficiency is only 50 %. Maximum power transfer is the right objective for communication and sensing, where the signal is scarce; it is the wrong objective for power delivery, where RT≪RLR_T \ll R_L is preferred.

§7 · Worked example B: the 110 V network

Two dependent sources, one supernode

The closing example asks for the maximum power that could be delivered to a load connected at a/b. The plan is fixed by the theorem: find the Thevenin equivalent, match the load to RTR_T, and evaluate VT2/(4RT)V_T^2 / (4R_T).

The open-circuit voltage

The problem

Find the maximum power that could be delivered to a load connected at a and b. Given: a 110 V source, 15 Ω, 5 Ω, and 8 Ω resistors, the dependent current source Vy/10V_y/10, and the dependent voltage source 3Ix3I_x. The controls are VyV_y, across the 15 Ω resistor, and IxI_x, the 5 Ω current defined toward the left.

Label the nodes from the source definitions

The 110 V source pins the left node. The definition of VyV_y places the middle node at 110−Vy110 - V_y. Ohm's law across the 5 Ω resistor places the top-right node at 110−Vy+5Ix110 - V_y + 5I_x, and the dependent voltage source adds 3Ix3I_x on the way down to 110−Vy+8Ix110 - V_y + 8I_x. Two symbols, VyV_y and IxI_x, cover every node, so two KCL equations finish the problem.

KCL at the middle node

−Vy15−Vy10−Ix=0  ⟹  Vy=−6Ix-\tm{bo.r15 bo.vy}{\frac{V_y}{15}} - \tm{bo.dcs}{\frac{V_y}{10}} - \tm{bo.r5 bo.ix}{I_x} = 0 \;\Longrightarrow\; V_y = -6 I_x

Summing currents leaving the node: the 15 Ω term is [(110−Vy)−110]/15=−Vy/15[(110 - V_y) - 110]/15 = -V_y/15; the dependent current source injects Vy/10V_y/10, so −Vy/10-V_y/10 leaves; and IxI_x arrives through the 5 Ω resistor, so −Ix-I_x leaves.

KCL at the supernode

The dependent voltage source and its two nodes form a supernode (dashed). With the port open, only two currents cross its surface:

Ix+110−Vy+8Ix8=0  ⟹  110−Vy+16Ix=0\tm{bo.r5 bo.ix}{I_x} + \tm{bo.r8 bo.np}{\frac{110 - V_y + 8I_x}{8}} = 0 \;\Longrightarrow\; 110 - V_y + 16 I_x = 0

Substituting Vy=−6IxV_y = -6I_x gives 110+22Ix=0110 + 22 I_x = 0, so Ix=−5I_x = -5 A and Vy=30V_y = 30 V.

Read off the open-circuit voltage

VT=110−Vy+5Ix=110−30−25  ⟹  VT=55 VV_T = 110 - V_y + 5I_x = 110 - 30 - 25 \;\Longrightarrow\; \boxed{V_T = 55\ \text{V}}

The port voltage is simply the top-right node voltage, because no current flows through the open terminal a.

Check: the solved values satisfy both equations

At the middle node: −30/15−30/10−(−5)=−2−3+5=0-30/15 - 30/10 - (-5) = -2 - 3 + 5 = 0 ✓. At the supernode: Ix=−5I_x = -5 A enters as +5+5 A, while the 8 Ω branch, at 110−30−40=40110 - 30 - 40 = 40 V, drains 40/8=540/8 = 5 A ✓. Both equations balance.

The resistance, again by test source

Turn off the 110 V source, drive with 1 V

The 110 V source becomes a short; both dependent sources remain. This time the port is driven with a 1 V test source, and the current IaI_a it delivers is sought, because RT=1 V/IaR_T = 1\ \text{V} / I_a.

The same labels, with new anchors

The middle node is now −Vy-V_y (its left neighbor sits at 0 V), and the test source pins the top-right node: −Vy+5Ix=1-V_y + 5I_x = 1 V, so Ohm's law across the 5 Ω gives Ix=(1+Vy)/5I_x = (1 + V_y)/5. Below the dependent source, the supernode's other node sits at 1+3Ix1 + 3I_x. KCL at the middle node is unchanged from the previous stepper: Vy=−6IxV_y = -6I_x.

Combine the two relations

Ix=1+Vy5=1−6Ix5  ⟹  11Ix=1  ⟹  Ix=111 AI_x = \frac{1 + V_y}{5} = \frac{1 - 6I_x}{5} \;\Longrightarrow\; 11 I_x = 1 \;\Longrightarrow\; \boxed{I_x = \tfrac{1}{11}\ \text{A}}

KCL at the supernode recovers IaI_a

Ix−Ia+1+3Ix8=0  ⟹  Ia=111+14/118=8+1488=14 A\tm{bt.r5 bt.ix}{I_x} - \tm{bt.ia}{I_a} + \tm{bt.r8 bt.np}{\frac{1 + 3I_x}{8}} = 0 \;\Longrightarrow\; I_a = \frac{1}{11} + \frac{14/11}{8} = \frac{8 + 14}{88} = \frac{1}{4}\ \text{A}
RT=1 V1/4 A  ⟹  RT=4 ΩR_T = \frac{1\ \text{V}}{1/4\ \text{A}} \;\Longrightarrow\; \boxed{R_T = 4\ \Omega}

The matched load, and the answer

The Thevenin equivalent, 55 V behind 4 Ω, drives the matched 4 Ω load. The divider splits evenly, so the load holds 55/2 V.

PL,max⁡=VT24RT=(55/2)24=302516  ⟹  PL,max⁡≈189.1 WP_{L,\max} = \frac{V_T^2}{4 R_T} = \frac{(55/2)^2}{4} = \frac{3025}{16} \;\Longrightarrow\; \boxed{P_{L,\max} \approx 189.1\ \text{W}}

No load attached at a/b can extract more than about 189 W from this network, and only the matched 4 Ω load reaches that figure.

§8 · Your turn

Practice: a resistance that defies inspection

The circuit below is small, yet it rewards care: the dependent source spans the port, so the looking-in resistance cannot be obtained by combining resistors. The problem should be attempted on paper before the hints are opened, in order.

Find the Thevenin equivalent at the terminals a and b, and the maximum power that a load connected there could extract. Given: an 18 V source in series with 6 Ω, a second 6 Ω resistor carrying VxV_x, and the dependent current source Vx/6V_x/6.

Hint 1: The open-circuit voltage costs one equation

With the port open, KCL at node a involves three terms: the 6 Ω branch to the source, the 6 Ω branch to ground, and the dependent source, whose value is Vx/6=Va/6V_x/6 = V_a/6 flowing into the node. Two of the three terms cancel.

Hint 2: The tempting resistance answer is wrong

With the 18 V source shorted, the two 6 Ω resistors appear to lie in parallel, suggesting RT=3R_T = 3 Ω. That reduction silently discards the dependent source, which remains active and feeds current back into the port. The test-source recipe of Section 1 is required.

Hint 3: The two equations

Open circuit: Va−186+Va6−Va6=0\tm{pr.r1}{\frac{V_a - 18}{6}} + \tm{pr.r2}{\frac{V_a}{6}} - \tm{pr.dep}{\frac{V_a}{6}} = 0. Test source (18 V off, 1 A injected into a): Va6+Va6−Va6=1\tm{pr.r1}{\frac{V_a}{6}} + \tm{pr.r2}{\frac{V_a}{6}} - \tm{pr.dep}{\frac{V_a}{6}} = 1.

Solution

The open-circuit equation collapses to (Va−18)/6=0(V_a - 18)/6 = 0, because the dependent source supplies exactly the current the grounded 6 Ω resistor drains. Hence VT=Voc=18V_T = V_{oc} = 18 V. The test-source equation collapses the same way, to Va/6=1V_a / 6 = 1, so Va=6V_a = 6 V and RT=6 ΩR_T = 6\ \Omega (twice the naive 3 Ω).

The short-circuit experiment provides the independent check: with a-b shorted, Vx=0V_x = 0, the dependent source idles, and the 18 V source drives Isc=18/6=3I_{sc} = 18/6 = 3 A through its 6 Ω resistor. Indeed VT=INRT=3⋅6=18V_T = I_N R_T = 3 \cdot 6 = 18 V ✓.

VT=18 V,RT=6 Ω,IN=3 APL,max⁡=1824⋅6=13.5 W\boxed{V_T = 18\ \text{V}, \quad R_T = 6\ \Omega, \quad I_N = 3\ \text{A}} \qquad\quad P_{L,\max} = \frac{18^2}{4 \cdot 6} = \boxed{13.5\ \text{W}}