ECE 211 · Circuit Analysis · Interactive Notes

Lecture 3: Nodal Analysis and the Supernode

This page presents the node-voltage method: a five-step procedure in which KCL, not KVL, carries the workload. The supernode technique for voltage sources, the handling of dependent sources, and a head-to-head comparison with the branch-current method are developed through four worked examples. The material is reconstructed from the Lecture 3 board notes as interactive figures.

Teal indicates linked content. Hovering over (or tabbing to) a boxed term such as I1I_1 highlights the corresponding circuit element; the linkage also operates in the reverse direction, from the circuit to the equations.

§1 · The nodal recipe

Node voltages do the bookkeeping

Nodal analysis assigns one voltage to every node and lets KCL generate the equations. The procedure from the board notes has five steps:

  1. Select a reference node. Its voltage is defined as 0 V; every other node voltage is measured against it.
  2. Assign node voltages, using the fewest unknowns possible. Every voltage source ties two node voltages together, so each one removes an unknown: $#UNK=#Nodes−1ref−#VS\#\text{UNK} = \#\text{Nodes} - 1_{\text{ref}} - \#\text{VS}$
  3. Apply KCL. Exactly as many KCL equations are written as there are unknowns: #KCL=#UNK\#\text{KCL} = \#\text{UNK}.
  4. Solve for the node voltages.
  5. Answer the question that was asked.

Hovering over the nodes and sources demonstrates the count. Node a is fixed at VsV_s by the source, so only b and c remain unknown; the current source removes nothing, because its voltage is not constrained.

4 nodes − 1 reference − 1 voltage source = 2 unknowns: Vb, Vc
Vocabulary

A supernode (written S.N. in the board notes) is a closed KCL surface drawn around a voltage source and both of its terminals. The current through a voltage source is not given by any element law, so KCL is applied to the surface, where that current never appears. Every example below uses one.

The two methods studied so far are duals of each other, and both end with the same two housekeeping steps:

Unknowns are node voltages; equations come from KCL. Voltage sources are the freebies: each one ties two node voltages together and removes an unknown. Current sources simply appear as known terms inside the KCL sums.

Unknowns are branch currents; equations come from KVL. Current sources are the freebies: #UNK=#BY−#CS\#\text{UNK} = \#\text{BY} - \#\text{CS}, the number of backyards (window panes) minus the number of current sources. Voltage sources simply appear as known terms inside the KVL sums.

§2 · Worked example A: two sources, one unknown

Solve for I1I_1

Two voltage sources and four resistors surround a single unknown node voltage. The requested quantity is the current through a voltage source, which is exactly the current that no element law provides; the supernode makes its first appearance.

The problem

Find I1I_1, the current flowing through the floating 5 V source toward the right. Given: a grounded 10 V source and 2 Ω, 4 Ω, 6 Ω, and 8 Ω resistors.

Step 1: Select a reference node

The bottom rail is chosen as the reference and marked 0 V. Any node may serve, but the node with the most connections usually produces the simplest equations.

Step 2: Assign node voltages

The grounded source pins the left node at 10 V. The middle node is named VxV_x; the floating 5 V source then forces the right node to Vx−5V_x - 5, one symbol covering both nodes.

4 nodes − 1 reference − 2 voltage sources = 1 unknown: Vₓ

Why plain node KCL is not enough

KCL at each source terminal must include the unknown source current I1I_1:

@Vx:Vx−102+Vx−08+I1=0\text{@}V_x:\quad \tm{sa.r2}{\frac{V_x - 10}{2}} + \tm{sa.r8}{\frac{V_x - 0}{8}} + \tm{sa.i1}{I_1} = 0
@Vx−5:(Vx−5)−104+(Vx−5)−06−I1=0\text{@}V_x - 5:\quad \tm{sa.r4}{\frac{(V_x - 5) - 10}{4}} + \tm{sa.r6}{\frac{(V_x - 5) - 0}{6}} - \tm{sa.i1}{I_1} = 0

Two equations now carry two unknowns, VxV_x and I1I_1. The system is solvable, but a cleaner route exists.

Step 3: Apply KCL to the supernode

Adding the two equations cancels I1I_1. Equivalently, KCL is applied directly to a closed surface around the 5 V source and both terminals: only currents that cross the dashed boundary appear.

Vx−102+Vx−08+Vx−5−104+Vx−5−06=0\tm{sa.r2}{\frac{V_x - 10}{2}} + \tm{sa.r8}{\frac{V_x - 0}{8}} + \tm{sa.r4}{\frac{V_x - 5 - 10}{4}} + \tm{sa.r6}{\frac{V_x - 5 - 0}{6}} = 0

Step 4: Solve for VxV_x

Multiplying through by 24 clears every denominator:

12(Vx−10)+3Vx+6(Vx−15)+4(Vx−5)=0  ⟹  25Vx=230  ⟹  Vx=9.2 V12(V_x - 10) + 3V_x + 6(V_x - 15) + 4(V_x - 5) = 0 \;\Longrightarrow\; 25 V_x = 230 \;\Longrightarrow\; \boxed{V_x = 9.2\ \text{V}}

The right node follows immediately: Vx−5=4.2V_x - 5 = 4.2 V.

Step 5: Answer the question

I1I_1 is recovered from the ordinary KCL equation at VxV_x:

I1=−[9.2−102+9.28]=−[−0.4+1.15]  ⟹  I1=−34 AI_1 = -\left[\tm{sa.r2}{\frac{9.2 - 10}{2}} + \tm{sa.r8}{\frac{9.2}{8}}\right] = -\left[-0.4 + 1.15\right] \;\Longrightarrow\; \boxed{I_1 = -\tfrac{3}{4}\ \text{A}}

The negative sign means the current actually flows leftward. The equation at Vx−5V_x - 5 confirms it: 4.2−104+4.26=−1.45+0.7=−0.75\frac{4.2 - 10}{4} + \frac{4.2}{6} = -1.45 + 0.7 = -0.75 A. ✓

What if the floating source changes?

The floating source value EE (5 V above) shifts both of its nodes together. The slider drives the closed-form solution of the same supernode equation:

Vx = 180 + 10E25 = 9.20 V     Vx − E = 4.20 V     I1 = −0.75 A

Where the closed form comes from

The supernode equation with a general source value EE reads Vx−102+Vx8+Vx−E−104+Vx−E6=0\frac{V_x - 10}{2} + \frac{V_x}{8} + \frac{V_x - E - 10}{4} + \frac{V_x - E}{6} = 0. Multiplying by 24 gives 25Vx=180+10E25 V_x = 180 + 10E. At E=2E = 2 V the recovered current I1I_1 equals zero: the source then carries no current at all, yet it still separates the two node voltages by 2 V.

§3 · Worked example B: a dependent voltage source

Find the power across the current source

The diamond symbol is a current-controlled voltage source: it forces the voltage 5Ix5I_x between its terminals, where IxI_x is the current in the 20 Ω resistor. Three supply terminals (+20 V, +10 V, −15 V) are drawn as open circles; each behaves as a voltage source to ground.

The problem

Find the power absorbed by the 0.7 A current source. Given: supply terminals at +20 V, +10 V, and −15 V; the dependent source 5Ix5I_x; and the resistors shown.

Steps 1 and 2: Reference and node voltages

With the bottom rail at 0 V, the three supply terminals are already known node voltages. The right-hand source node is named VxV_x, and the dependent source ties the left terminal to Vx+5IxV_x + 5I_x. The controlling current is not a new unknown, because Ohm's law on the 20 Ω resistor gives

Ix=10−Vx20  ⟹  Vx=10−20Ix\tm{sb.ix}{I_x} = \frac{10 - V_x}{20} \;\Longrightarrow\; V_x = 10 - 20 I_x
6 nodes − 1 reference − 4 voltage sources = 1 unknown

Step 3: KCL at the supernode

A dependent voltage source needs a supernode exactly as an independent one does. Five currents cross the dashed surface:

Vx−1020  −  710+Vx−(−15)25+Vx+5Ix−2025+Vx+5Ix−050=0\tm{sb.r20}{\frac{V_x - 10}{20}} \tm{sb.cs}{\;-\; \frac{7}{10}} + \tm{sb.r25b}{\frac{V_x - (-15)}{25}} + \tm{sb.r25a}{\frac{V_x + 5I_x - 20}{25}} + \tm{sb.r50}{\frac{V_x + 5I_x - 0}{50}} = 0

Simplify

Multiplying by 100 and collecting terms:

5Vx−50−70+4Vx+60+4Vx+20Ix−80+2Vx+10Ix=0  ⟹  15Vx+30Ix=1405V_x - 50 - 70 + 4V_x + 60 + 4V_x + 20I_x - 80 + 2V_x + 10I_x = 0 \;\Longrightarrow\; 15 V_x + 30 I_x = 140

Step 4: Eliminate the controlling variable and solve

Substituting Vx=10−20IxV_x = 10 - 20 I_x:

15 [10−20Ix]+30Ix=140  ⟹  −270Ix=−10  ⟹  Ix=127 AVx=25027≈9.26 V15\,[10 - 20 I_x] + 30 I_x = 140 \;\Longrightarrow\; -270 I_x = -10 \;\Longrightarrow\; \boxed{I_x = \tfrac{1}{27}\ \text{A}} \qquad V_x = \tfrac{250}{27} \approx 9.26\ \text{V}

Step 5: Answer with the passive sign convention

The source voltage is VxV_x with the positive mark on top, and the 0.7 A flows out of the positive terminal. The absorbed power is therefore negative:

P=−Vx⋅710=−[10−2027]⋅710=−17527  ⟹  P≈−6.48 WP = -V_x \cdot \tfrac{7}{10} = -\left[10 - \tfrac{20}{27}\right] \cdot \tfrac{7}{10} = -\tfrac{175}{27} \;\Longrightarrow\; \boxed{P \approx -6.48\ \text{W}}

The current source delivers about 6.48 W to the rest of the circuit. (The board notes round this to −6.5 W; the exact value is −175/27-175/27 W.)

§4 · Worked example C: nodal versus branch currents

Solve for VxV_x, two ways

Two current sources, one floating voltage source, and three resistors meet in one circuit. The unknown count decides which method is cheaper, and this example runs both to the same answer, which is a powerful self-check.

The problem

Find VxV_x, the voltage across the 20 Ω resistor with the positive mark on the right.

Count before committing

Nodal: 5 nodes − 1 ref − 1 VS = 3 unknowns Branch: 3 BY − 2 CS = 1 unknown

Hovering over the three backyards and the two current sources demonstrates the branch count. The branch-current method is cheaper here; the nodal route is walked first anyway, because it exercises every rule of this lecture, and the branch route returns at the end as the check.

Assign node voltages

The bottom rail is the reference. Three unknowns are named: V1V_1, V2V_2, and V3V_3. The floating 10 V source pins its left terminal at V3+10V_3 + 10, so no fourth symbol is needed.

KCL at V1V_1

+2+V1−05+V1−[V3+10]20=0\tm{sc.cs2}{+2} + \tm{sc.r5}{\frac{V_1 - 0}{5}} + \tm{sc.r20}{\frac{V_1 - [V_3 + 10]}{20}} = 0

The 2 A source current leaves V1V_1, so it enters the sum with a positive sign.

KCL at V2V_2: a one-line decoupled node

−2  −  3+V210=0  ⟹  V2=50 V\tm{sc.cs2}{-2} \tm{sc.cs3}{\;-\; 3} + \tm{sc.r10}{\frac{V_2}{10}} = 0 \;\Longrightarrow\; \boxed{V_2 = 50\ \text{V}}

Both neighbors of V2V_2 are current sources, so this node decouples from the rest of the circuit entirely: 5 A must flow down the 10 Ω resistor.

KCL at the supernode

The surface encloses the 10 V source together with V3V_3:

V3+10−V120  +  3+V315=0\tm{sc.r20}{\frac{V_3 + 10 - V_1}{20}} \tm{sc.cs3}{\;+\; 3} + \tm{sc.r15}{\frac{V_3}{15}} = 0

Solve and answer

Clearing denominators gives 5V1−V3=−305V_1 - V_3 = -30 and −3V1+7V3=−210-3V_1 + 7V_3 = -210; solving the pair yields V1=−13.125V_1 = -13.125 V and V3=−35.625V_3 = -35.625 V. Then

Vx=[V3+10]−V1=−35.625+10+13.125  ⟹  Vx=−12.5 VV_x = [V_3 + 10] - V_1 = -35.625 + 10 + 13.125 \;\Longrightarrow\; \boxed{V_x = -12.5\ \text{V}}

Check: the branch-current route

With branch current I1I_1 up the 5 Ω resistor, KCL during labeling fills every branch, and one KVL around the outer loop suffices:

5I1+20[I1−2]+10−15[5−I1]=0  ⟹  40I1=105  ⟹  I1=2.625 A\tm{sc.r5}{5I_1} + \tm{sc.r20}{20[I_1 - 2]} + \tm{sc.vs10}{10} - \tm{sc.r15}{15[5 - I_1]} = 0 \;\Longrightarrow\; 40 I_1 = 105 \;\Longrightarrow\; I_1 = 2.625\ \text{A}
Vx=20 [2−I1]=20 [−0.625]=−12.5 V ✓V_x = 20\,[2 - I_1] = 20\,[-0.625] = -12.5\ \text{V}\ \checkmark

Two independent methods, one answer. Agreement between routes is the best self-check available.

§5 · Worked example D: dependent sources everywhere

Find V1V_1, V0V_0, and IxI_x

The capstone example contains four sources: a 20 V independent source, a current-controlled voltage source (2Ix2I_x), a voltage-controlled current source (0.4V10.4V_1), and a 5 A independent source. Careful labeling reduces the whole circuit to two unknowns before any KCL equation is written.

The problem

Find V1V_1, V0V_0, and IxI_x. Note the 20 V source polarity: its negative terminal faces the top rail.

Label nodes by chained substitution

Each source constraint feeds the next label. The 20 V source pins the left node at −20 V; Ohm's law across the 5 Ω resistor puts the next node at −20 − 5IxI_x; the dependent source lifts its top terminal to 2IxI_x; and V1V_1 across the 10 Ω resistor stacks on top of it: 2IxI_x + V1V_1.

5 nodes − 1 reference − 2 voltage sources = 2 unknowns: Iₓ, V₁

KCL at −20 − 5IxI_x

−Ix+−20−5Ix−020−0.4V1=0-\tm{sd.ix}{I_x} + \tm{sd.r20}{\frac{-20 - 5I_x - 0}{20}} - \tm{sd.dep}{0.4V_1} = 0

IxI_x arrives on the wire and the dependent current source injects 0.4V10.4V_1, so both enter with negative signs.

KCL at 2IxI_x + V1V_1: the shortcut

0.4V1+[2Ix+V1]−2Ix10−5=0  ⟹  0.4V1+V110=5  ⟹  V1=10 V\tm{sd.dep}{0.4V_1} + \tm{sd.r10}{\frac{[2I_x + V_1] - 2I_x}{10}} - \tm{sd.cs5}{5} = 0 \;\Longrightarrow\; 0.4V_1 + \frac{V_1}{10} = 5 \;\Longrightarrow\; \boxed{V_1 = 10\ \text{V}}

The 2Ix2I_x terms cancel inside the resistor current, which collapses this equation to a single unknown.

Back-substitute

With V1=10V_1 = 10 V, the first equation contains only IxI_x:

−Ix+−20−5Ix20−4=0  ⟹  −25Ix=100  ⟹  Ix=−4 A-I_x + \frac{-20 - 5I_x}{20} - 4 = 0 \;\Longrightarrow\; -25 I_x = 100 \;\Longrightarrow\; \boxed{I_x = -4\ \text{A}}

Answer and check

V0=2Ix+V1=−8+10  ⟹  V0=2 VV_0 = 2I_x + V_1 = -8 + 10 \;\Longrightarrow\; \boxed{V_0 = 2\ \text{V}}

A numerical check costs one line: the node −20−5Ix-20 - 5I_x evaluates to −20+20=0-20 + 20 = 0 V, and KCL there reads −(−4)+0/20−0.4(10)=4−4=0-(-4) + 0/20 - 0.4(10) = 4 - 4 = 0. ✓

§6 · Your turn

Practice: find V2V_2 and the source current

The circuit below distills this lecture to its essentials: one current source, one floating voltage source, two resistors. The problem should be attempted on paper before the hints are opened, in order.

Solve for the node voltage V2V_2 and for II, the current flowing through the 6 V source toward the right.

Hint 1: The count

With the bottom rail as reference, the count reads 3 nodes − 1 reference − 1 voltage source = 1 unknown. Naming V2V_2 is enough, because the 6 V source forces V1V_1 =V2+6= V_2 + 6.

Hint 2: The plan

A supernode is drawn around the 6 V source and both of its terminals. The 3 A source injects current into the surface, and the two resistors carry it out; one KCL equation results.

Hint 3: The equation
[V2+6]−04+V2−08=3\tm{pr.r4}{\frac{[V_2 + 6] - 0}{4}} + \tm{pr.r8}{\frac{V_2 - 0}{8}} = \tm{pr.cs3}{3}
Solution

Multiplying by 8 gives 2V2+12+V2=242V_2 + 12 + V_2 = 24, so 3V2=123V_2 = 12 and V2=4 V\boxed{V_2 = 4\ \text{V}} (with V1=10V_1 = 10 V). The source current follows from plain KCL at V2V_2: everything arriving through the source leaves through the 8 Ω resistor, so

I=V28=0.5 AI = \frac{V_2}{8} = \boxed{0.5\ \text{A}}

A check at V1V_1 confirms the balance: 10/4+0.5=310/4 + 0.5 = 3 A, exactly the injected current. ✓ As a final observation, the current enters the source's positive terminal, so the 6 V source absorbs 6×0.5=36 \times 0.5 = 3 W.