123-ME · Node-Voltage & Mesh-Current Analysis

Node-Voltage & Mesh-Current Analysis

Slides 2.4–2.5 · two systematic methods that never run out of road

Series/parallel reduction and the dividers are pattern-matching: they work when the circuit happens to have the pattern. The moment a bridge, a second source, or a dependent source appears, the patterns run out. Node-voltage and mesh-current analysis are the two methods that never run out — they turn any resistive circuit into a linear system you crank through mechanically. Everything in the rest of this course (Thévenin, superposition, phasor AC analysis) is built on top of one of these two. Budget about 90 minutes here; it is the single densest scoring block on the exam.

The one-sentence version

Node voltage picks unknowns that automatically satisfy KVL, then imposes KCL. Mesh current picks unknowns that automatically satisfy KCL, then imposes KVL. Each method gets one law for free and pays for the other. That symmetry is the whole chapter.

1 · Node voltage — where it comes from

Start from the thing we actually want: every element's current and voltage. That's a lot of unknowns. The trick is to notice that voltage is a property of a node, not of a wire. Every point connected by perfect wire sits at the same potential. So if a circuit has $N$ nodes, there are only $N$ distinct potentials in the whole thing — and since only differences matter, we can nail one of them to zero (the reference node, or ground) and be left with $N-1$ unknowns.

Why does this get KVL for free? KVL says the voltage drops around any closed loop sum to zero. But if we define every element's voltage as a difference of node potentials, $v_{ab} = v_a - v_b$ , then going around a loop the potentials telescope: $(v_a - v_b) + (v_b - v_c) + (v_c - v_a) = 0$ identically. KVL can't be violated by construction. That leaves KCL as the only equation we still have to write.

Branch current from node voltages (Ohm's law, rewritten)

i_{a \to b} = \frac{v_a - v_b}{R}

Read it as: current leaves node

a

through

R

toward node

b

. The numerator is always "my node minus the other node."

The course's sign convention — memorize this one

ME 123 writes KCL as all currents leaving the node sum to zero. Every resistor term is therefore $\dfrac{v_{\text{this node}} - v_{\text{other node}}}{R}$ with this node first, and a current source pointing into the node enters the equation with a minus sign. Pick this convention once and never mix it — sign errors here are the single most common way to lose an otherwise perfect exam answer.

The procedure

Choose the reference node. Pick the node with the most connections, or the negative terminal of the main source. Mark it with the ground symbol; it is $0$ V by definition.

A good choice makes several node voltages known immediately.

Label the remaining nodes $v_1, v_2, \ldots$ — one unknown each.

Harvest the free constraints. Any node tied to ground through a voltage source is known: $v_1 = v_s$ . Do not write KCL there — you'd be introducing the source's unknown current for nothing.

Write KCL at every remaining node, all currents leaving, each resistor term as $\dfrac{v_{\text{here}} - v_{\text{there}}}{R}$ .

Solve, then back out whatever the question actually asked for (a current, a power) from the node voltages.

The node voltages are almost never the final answer — they're the intermediate variable.

Worked example 1. Four nodes, one of them ground, one of them pinned by the source — so only

v_2

and

v_3

are genuinely unknown. The three clockwise mesh currents (in red) are there for the cross-check at the end; ignore them on the first pass.

Worked example 1 — plain node analysis2 unknowns

Find $v_2$ and $v_3$ in the circuit above.

The source sits between $v_1$ and ground, so $v_1 = 24$ V for free — no KCL at that node. Two unknowns remain.

KCL at $v_2$ (all currents leaving: through the 4 Ω back toward $v_1$ , down the 12 Ω to ground, right through the 6 Ω toward $v_3$ ):

\frac{v_2 - 24}{4} + \frac{v_2}{12} + \frac{v_2 - v_3}{6} = 0 \quad\Longrightarrow\quad 3v_2 - v_3 = 36

KCL at $v_3$ (left through the 6 Ω, down the 3 Ω, and the 2 A source pushes current into the node so it carries a minus sign):

\frac{v_3 - v_2}{6} + \frac{v_3}{3} - 2 = 0 \quad\Longrightarrow\quad -v_2 + 3v_3 = 12

Substituting $v_3 = 3v_2 - 36$ into the second equation gives $8v_2 = 120$ .

v_2 = 15

v_3 = 9

Cross-check by mesh (three clockwise meshes, the 2 A branch forcing $i_3 = -2$ A): $i_1 = 2.25$ A, $i_2 = 1$ A. Then $v_2 = 12(i_1 - i_2) = 12(1.25) = 15$ V and $v_3 = 3(i_2 - i_3) = 3(3) = 9$ V. ✓ Both methods agree.

Self-check 1 · sign convention

Node $v_2$ connects to node $v_1$ through a 6 Ω resistor, to ground through a 3 Ω resistor, and a 2 A current source pushes current into $v_2$ from ground. Using the course's "all currents leaving" convention, which is the correct KCL equation at $v_2$ ?

2 · Supernodes — when a voltage source gets in the way

Node analysis needs to express every branch current in terms of node voltages. A resistor obliges: $i = \Delta v / R$ . A voltage source does not — it fixes its voltage and lets its current be whatever the rest of the circuit demands. If that source sits between ground and a node, no problem: the node voltage is simply known. But if it floats between two non-reference nodes, we can't write KCL at either one without inventing an unknown source current.

The fix comes straight from KCL's real statement. KCL isn't about nodes specifically — it's about any closed surface: charge doesn't pile up anywhere, so the total current crossing any boundary is zero. So draw a boundary around both nodes and the source between them. The source's mystery current is now entirely inside the boundary and never crosses it — it cancels out before we ever have to name it. That enclosure is a supernode.

One enclosure gives one KCL equation but covers two unknowns, so we're one equation short. The missing one is free: the source itself is a KVL statement about those two nodes.

The supernode pair

\underbrace{\sum i_{\text{leaving the enclosure}} = 0}_{\text{KCL over the boundary}} \qquad\text{and}\qquad \underbrace{v_{+} - v_{-} = v_s}_{\text{KVL across the source}}

Two equations, two node voltages. Note the constraint's direction: the node at the source's + terminal is the higher one.

The 10 V source floats between two non-reference nodes. Enclose it: the dashed boundary is where KCL now gets applied.

Worked example 2 — supernode3 unknowns

Find $v_1$ , $v_2$ , $v_3$ above.

KCL over the enclosure. Four things cross the dashed boundary: the 4 A source (entering, so $-4$ ), the 5 Ω to ground, the 20 Ω to ground, and the 4 Ω heading off to $v_3$ :

-4 + \frac{v_1}{5} + \frac{v_2}{20} + \frac{v_2 - v_3}{4} = 0

Notice what is not in that equation: any current through the 10 V source. It never crossed the boundary.

KVL constraint (the + terminal is at $v_2$ ):

v_2 - v_1 = 10 \quad\Longrightarrow\quad v_1 = v_2 - 10

KCL at $v_3$ — an ordinary node, nothing special:

\frac{v_3 - v_2}{4} + \frac{v_3}{8} = 0 \quad\Longrightarrow\quad v_3 = \tfrac{2}{3}v_2

Substituting both into the enclosure equation and multiplying by 20: $4(v_2 - 10) + v_2 + 5\!\left(v_2 - \tfrac{2}{3}v_2\right) = 80$ , i.e. $\tfrac{20}{3}v_2 = 120$ .

v_1 = 8

v_2 = 18

v_3 = 12

Cross-check by mesh (three clockwise meshes; the 4 A branch forces $i_1 = 4$ A): solving gives $i_2 = 2.4$ A and $i_3 = 1.5$ A. Then $v_1 = 5(i_1 - i_2) = 5(1.6) = 8$ V, $v_2 = 20(i_2 - i_3) = 20(0.9) = 18$ V, and $v_3 = 8i_3 = 12$ V. ✓ Both methods agree.

3 · Step through it yourself

The supernode below is a smaller cousin of the one above — the same structure the Spring 2025 midterm leaned on. Click through the steps and watch the equations accumulate; the last step re-solves the identical circuit by mesh analysis so you can see the two methods land on the same numbers.

Supernode / mesh stepper step 1 of 6

Walkthrough

Self-check 2 · supernode, end to end

A 3 A source pushes current into node $v_1$ from ground. A 4 Ω resistor runs from $v_1$ to ground. A 6 V source sits between $v_1$ and $v_2$ with its + terminal at $v_2$ . A 4 Ω resistor runs from $v_2$ to ground. What is $v_1$ ?

4 · Dependent sources

A dependent source's value is set by some other voltage or current in the circuit — the control variable. It is not a new unknown, and it is not a known number either. The rule is mechanical: write the node equations exactly as usual, treating the dependent source's value as a symbol, then add one extra equation expressing the control variable in terms of node voltages and substitute it away. The system stays square.

The trap

Never "turn off" a dependent source, and never leave the control variable in the final system. A dependent source is not an independent source — it does not get zeroed in superposition, and it does not get deactivated when finding $R_{Th}$ . It just sits there obeying its controller.

A current-controlled current source (drawn as a diamond, the standard dependent-source symbol). Its strength equals

i_x

, the current in the 5 Ω branch it is nowhere near.

Worked example 3 — dependent sourceCCCS

Find $v_1$ , $v_2$ and $i_x$ above.

Control-variable equation — $i_x$ is the current in the 5 Ω branch, defined flowing from $v_1$ toward $v_2$ :

i_x = \frac{v_1 - v_2}{5}

KCL at $v_1$ (6 A entering ⇒ $-6$ ; out through the vertical 5 Ω and the horizontal 5 Ω):

-6 + \frac{v_1}{5} + \frac{v_1 - v_2}{5} = 0 \quad\Longrightarrow\quad 2v_1 - v_2 = 30

KCL at $v_2$ (out through the 5 Ω back toward $v_1$ and down the 10 Ω; the dependent source drives $i_x$ into the node, so $-i_x$ ):

\frac{v_2 - v_1}{5} + \frac{v_2}{10} - i_x = 0

Substitute the control equation, $i_x = (v_1-v_2)/5$ , and multiply by 10:

2(v_2 - v_1) + v_2 - 2(v_1 - v_2) = 0 \quad\Longrightarrow\quad 5v_2 = 4v_1

Then $2v_1 - 0.8v_1 = 30$ , so $1.2v_1 = 30$ .

v_1 = 25

v_2 = 20

i_x = 1

A (so the dependent source is worth 1 A).

Cross-check by element bookkeeping (no node equations at all): everything arriving at $v_2$ leaves through the 10 Ω, and $i_x$ arrives twice — once from the 5 Ω, once from the dependent source — so $v_2 = 10(2i_x) = 20i_x$ . Walking back through the horizontal 5 Ω, $v_1 = v_2 + 5i_x = 25i_x$ . KCL at $v_1$ : $6 = v_1/5 + i_x = 5i_x + i_x$ , so $i_x = 1$ A and $v_1 = 25$ V, $v_2 = 20$ V. ✓ Agrees.

5 · Mesh current — the mirror image

Now flip the whole idea around. Instead of assigning a potential to each node, assign a circulating current to each mesh — a mesh being a loop with nothing inside it, like a single windowpane. The physical current in any branch is then the sum of the mesh currents flowing through it. ME 123 draws every mesh current clockwise, always; the consistency is what makes the sign bookkeeping automatic.

Why does this get KCL for free? Because a mesh current is, by definition, a loop — it enters every node on its path exactly as much as it leaves. Sum any number of loops at a node and you still get zero net accumulation. KCL is satisfied identically, which leaves KVL as the only equation left to write. Exactly the mirror of node analysis.

The shared-resistor rule — the one place mesh sign errors live

v_R \big|_{\text{mesh 1's KVL}} = \big(i_1 - i_2\big)R

Writing mesh 1's equation, its own mesh current comes first and every neighbouring mesh current is subtracted. Writing mesh 2's equation for the same resistor, you'd write

(i_2 - i_1)R

. A resistor touched by only one mesh contributes plain

i_1R

The procedure

Draw a clockwise current in every windowpane and label them $i_1, i_2, \ldots$

Harvest the free constraints. A current source touched by only one mesh fixes that mesh current outright — write $i_k = \pm I_s$ (sign from whether the source's arrow agrees with the clockwise direction) and skip that mesh's KVL.

Walk each remaining mesh clockwise, summing voltage drops to zero. Through a resistor in the direction of your walk: $+(i_{\text{mine}} - i_{\text{neighbour}})R$ . Through a source from $+$ to $-$ : a drop of $+v_s$ ; from $-$ to $+$ : $-v_s$ .

Walking with the current means you're going downhill — that's a drop.

Solve, then recombine. A branch's real current is the mesh current, or the difference of two of them.

Both mesh currents clockwise. The 6 Ω in the middle is shared — that's the only place the two equations talk to each other.

Worked example 4 — two-mesh KVL2 unknowns

Find $i_1$ and $i_2$ above.

Mesh 1, walking clockwise from the bottom-left: up through the 24 V source ( $-$ to $+$ , a rise, so $-24$ ), right through the 3 Ω with $i_1$ , then down the shared 6 Ω with $(i_1 - i_2)$ :

-24 + 3i_1 + 6(i_1 - i_2) = 0 \quad\Longrightarrow\quad 3i_1 - 2i_2 = 8

Mesh 2, clockwise: up through the shared 6 Ω — now it's my current first, $(i_2 - i_1)$ — right through the 6 Ω, then down through the 12 V source from $+$ to $-$ , a drop of $+12$ :

6(i_2 - i_1) + 6i_2 + 12 = 0 \quad\Longrightarrow\quad -i_1 + 2i_2 = -2

Adding the two equations: $2i_1 = 6$ .

i_1 = 3

i_2 = 0.5

A. The shared 6 Ω carries

i_1 - i_2 = 2.5

A downward.

Cross-check by node analysis. With the bottom rail as reference, the top-left node is $24$ V, the top-right is $12$ V, and the middle node $v_a$ is the only unknown: $\frac{v_a-24}{3} + \frac{v_a}{6} + \frac{v_a-12}{6} = 0 \Rightarrow 4v_a = 60 \Rightarrow v_a = 15$ V. Then the 3 Ω carries $(24-15)/3 = 3$ A $= i_1$ , the middle 6 Ω carries $15/6 = 2.5$ A, and the right 6 Ω carries $(15-12)/6 = 0.5$ A $= i_2$ . ✓ Both methods agree.

Self-check 3 · the shared-resistor term

Meshes 1 and 2 share a resistor $R$ , and both mesh currents are drawn clockwise as the course requires. You are traversing mesh 1 clockwise and summing voltage drops. Which term does $R$ contribute?

6 · Supermesh — the mirror of the supernode

Mesh analysis has the exact dual of the supernode problem. A current source refuses to tell you its voltage. If it's on the outer edge of one mesh, fine — it fixes that mesh current and you skip the KVL. But if it sits on a branch shared by two meshes, you can't write either mesh's KVL without inventing the source's unknown voltage.

Same fix, dualized: walk a loop that goes around the outside of both meshes, avoiding the offending branch entirely. The source's unknown voltage is never traversed, so it never enters the equation. That loop is a supermesh, and it gives one equation for two unknowns — with the source itself supplying the missing constraint, since it dictates the difference of the two mesh currents.

The supermesh pair

\underbrace{\sum v_{\text{drops around the outer loop}} = 0}_{\text{KVL skipping the source branch}} \qquad\text{and}\qquad \underbrace{i_{\text{with the arrow}} - i_{\text{against}} = I_s}_{\text{KCL at the source}}

The 2 A source is shared, so neither mesh alone can be walked. The dashed path is the loop we walk instead — note the gap where the source branch would be.

Worked example 5 — supermeshshared current source

Find $i_1$ and $i_2$ above.

The constraint. Clockwise $i_1$ runs down the shared branch; clockwise $i_2$ runs up it. So the upward current there is $i_2 - i_1$ , and the source's arrow points up at 2 A:

i_2 - i_1 = 2

The supermesh KVL, walking the dashed loop clockwise from the bottom-left: up through the 32 V source (a rise, $-32$ ), right through the 4 Ω carrying $i_1$ , jump the gap, right through the 2 Ω carrying $i_2$ , down the 4 Ω carrying $i_2$ :

-32 + 4i_1 + 2i_2 + 4i_2 = 0 \quad\Longrightarrow\quad 4i_1 + 6i_2 = 32

Substituting $i_2 = i_1 + 2$ : $4i_1 + 6i_1 + 12 = 32$ , so $10i_1 = 20$ .

i_1 = 2

i_2 = 4

Cross-check by node analysis. Call the top-middle node $v_a$ and the top-right node $v_b$ ; the top-left is $32$ V. KCL at $v_b$ gives $v_b = \frac{2}{3}v_a$ , and KCL at $v_a$ (2 A entering) gives:

\frac{v_a-32}{4} + \frac{v_a-v_b}{2} - 2 = 0 \;\Longrightarrow\; \frac{5}{3}v_a = 40

So $v_a = 24$ V and $v_b = 16$ V. Then the 4 Ω on the left carries $(32-24)/4 = 2$ A $= i_1$ and the 2 Ω carries $(24-16)/2 = 4$ A $= i_2$ . ✓ Both methods agree.

7 · Which method should you actually use?

On an exam the question usually names the method. When it doesn't, count before you write:

Node	Fewer non-reference nodes than meshes ⇒ fewer equations. Circuit is dominated by current sources (each one just drops into a KCL with a sign — no extra work at all). You are asked for a voltage. Ground is obvious and several nodes hang off it through voltage sources (each is a free known).
Mesh	Fewer meshes than non-reference nodes. Circuit is dominated by voltage sources (each drops into a KVL with a sign). You are asked for a current or a power in a specific branch. The circuit is planar and looks like a ladder or a grid of windowpanes.
Either	Whichever you pick, the other one is your cross-check. Two independent methods landing on the same number is the only real proof you didn't flip a sign — and it's how every answer in this chapter was verified.

Duality, in one table

Node ↔ mesh · KCL ↔ KVL · node voltage ↔ mesh current · voltage source is the awkward one ↔ current source is the awkward one · supernode ↔ supermesh · "all currents leaving" ↔ "all drops clockwise". Learn one method properly and the other is the same method with the words swapped.

8 · Practice

Every answer below was solved twice, by node analysis and by mesh analysis, and shipped only because both agreed. Do the same: solve it one way, then check it the other.

Problem 1 — resistive laddernode · ~6 min

Find $v_2$ and $v_3$ , and the total power delivered by the 28 V source.

Node analysis. The source pins the left node at 28 V. Two unknowns.

\frac{v_2 - 28}{4} + \frac{v_2}{8} + \frac{v_2 - v_3}{8} = 0 \;\Longrightarrow\; 7v_2 - 2v_3 = 112

\frac{v_3 - v_2}{8} + \frac{v_3}{8} = 0 \;\Longrightarrow\; v_3 = \tfrac{1}{2}v_2

So $7v_2 - v_2 = 112$ , giving $v_2 = 16$ V and $v_3 = 8$ V. The source current is the current in the 4 Ω: $(28-16)/4 = 3$ A, so $P = 28 \times 3 = 84$ W delivered.

Cross-check by mesh. Two clockwise meshes: $-28 + 4i_1 + 8(i_1-i_2) = 0$ and $8(i_2-i_1) + 8i_2 + 8i_2 = 0$ . The second gives $i_1 = 3i_2$ ; substituting, $28i_2 = 28$ , so $i_2 = 1$ A and $i_1 = 3$ A. Then $v_2 = 8(i_1-i_2) = 16$ V, $v_3 = 8i_2 = 8$ V, and $P = 28i_1 = 84$ W. ✓

Third sanity check — power balance. $3^2(4) + 2^2(8) + 1^2(8) + 1^2(8) = 36 + 32 + 8 + 8 = 84$ W dissipated, exactly what the source delivers.

v_2 = 16

v_3 = 8

P_{\text{source}} = 84

W delivered.

Problem 2 — supernodesupernode · ~8 min

Find $v_1$ and $v_2$ , then the power delivered by the 18 V source. Finally, verify that total power delivered equals total power dissipated.

Node analysis. The 18 V source floats between two non-reference nodes, so enclose it. Crossing the boundary: the 6 A source (entering, $-6$ ), the 12 Ω, the 6 Ω.

-6 + \frac{v_1}{12} + \frac{v_2}{6} = 0, \qquad v_2 - v_1 = 18

Substituting $v_2 = v_1 + 18$ and multiplying by 12: $-72 + v_1 + 2v_1 + 36 = 0$ , so $3v_1 = 36$ and $v_1 = 12$ V, $v_2 = 30$ V.

Source power. Everything arriving at $v_2$ comes through the source, and it all leaves down the 6 Ω: that current is $30/6 = 5$ A, flowing from $v_1$ through the source to $v_2$ — i.e. entering the $-$ terminal and leaving the $+$ terminal. That's delivery: $P = 18 \times 5 = 90$ W delivered.

Cross-check by mesh. Two clockwise meshes; the 6 A branch is touched only by mesh 1, so $i_1 = 6$ A. Mesh 2, walking clockwise: up the 12 Ω (a rise of $12(i_1-i_2)$ ), right through the source $-$ to $+$ (a rise of 18), down the 6 Ω (a drop of $6i_2$ ): $-12(i_1 - i_2) - 18 + 6i_2 = 0 \Rightarrow 18i_2 = 90 \Rightarrow i_2 = 5$ A. Then $v_1 = 12(i_1-i_2) = 12$ V, $v_2 = 6i_2 = 30$ V, and the source current is $i_2 = 5$ A. ✓

Power balance. Delivered: $6 \times 12 = 72$ W by the current source, plus 90 W by the voltage source $= 162$ W. Dissipated: $12^2/12 + 30^2/6 = 12 + 150 = 162$ W. ✓

v_1 = 12

v_2 = 30

V; the 18 V source delivers 90 W.

Problem 3 — mesh with a dependent sourcemesh · CCVS · ~10 min

The dependent voltage source on the right is worth $5i_x$ volts, where $i_x$ is the current in the 5 Ω resistor (rightward, as drawn). Find $i_1$ , $i_2$ and $i_x$ , and say whether the dependent source is absorbing or delivering power.

Control equation first. The 5 Ω sits only in mesh 1, and $i_x$ points the same way clockwise $i_1$ runs along the top, so $i_x = i_1$ .

Mesh 1, clockwise: up through the 30 V source (rise, $-30$ ), right through the 5 Ω, down the shared 20 Ω:

-30 + 5i_1 + 20(i_1 - i_2) = 0 \;\Longrightarrow\; 5i_1 - 4i_2 = 6

Mesh 2, clockwise: up the shared 20 Ω as $(i_2-i_1)$ , right through the 10 Ω, then down through the dependent source from $+$ to $-$ — a drop of $+5i_x = +5i_1$ :

20(i_2 - i_1) + 10i_2 + 5i_1 = 0 \;\Longrightarrow\; -15i_1 + 30i_2 = 0 \;\Longrightarrow\; i_1 = 2i_2

Substituting: $10i_2 - 4i_2 = 6$ , so $i_2 = 1$ A and $i_1 = 2$ A.

Cross-check by node analysis. Bottom rail is reference, top-left is 30 V; call the middle node $v_a$ and the right node $v_b$ . Now $i_x = (30 - v_a)/5$ and the dependent source forces $v_b = 5i_x = 30 - v_a$ . KCL at $v_a$ :

\frac{v_a - 30}{5} + \frac{v_a}{20} + \frac{v_a - v_b}{10} = 0 \;\Longrightarrow\; 4(v_a-30) + v_a + 2(2v_a - 30) = 0 \;\Longrightarrow\; 9v_a = 180

So $v_a = 20$ V, $v_b = 10$ V, $i_x = (30-20)/5 = 2$ A $= i_1$ ✓, the 10 Ω carries $(20-10)/10 = 1$ A $= i_2$ ✓, and the shared 20 Ω carries $20/20 = 1$ A $= i_1 - i_2$ ✓.

Power. The source is worth $5(2) = 10$ V and carries $i_2 = 1$ A flowing downward, i.e. into its $+$ terminal — passive sign convention says it is absorbing $10 \times 1 = 10$ W.

i_1 = i_x = 2

i_2 = 1

A; the dependent source absorbs 10 W.

Problem 4 — supermeshsupermesh · ~8 min

Both mesh currents are clockwise, as always. The 3 A source points downward. Find $i_1$ and $i_2$ , and the voltage across the current source (top terminal positive).

Constraint. Clockwise $i_1$ runs down the shared branch and clockwise $i_2$ runs up it, so the downward current there is $i_1 - i_2$ . The arrow points down at 3 A:

i_1 - i_2 = 3

Supermesh KVL, walking the outer loop clockwise and skipping the source branch: up through the 30 V source ( $-30$ ), right through the 6 Ω with $i_1$ , across the gap, right through the 4 Ω with $i_2$ , down the 2 Ω with $i_2$ :

-30 + 6i_1 + 4i_2 + 2i_2 = 0 \;\Longrightarrow\; 6i_1 + 6i_2 = 30 \;\Longrightarrow\; i_1 + i_2 = 5

With $i_1 - i_2 = 3$ : $i_1 = 4$ A, $i_2 = 1$ A.

Cross-check by node analysis. Top-left is 30 V; call the top-middle node $v_a$ and the top-right $v_b$ . KCL at $v_b$ gives $\frac{v_b - v_a}{4} + \frac{v_b}{2} = 0$ , i.e. $v_a = 3v_b$ . Now KCL at $v_a$ — the 3 A source leaves this node going down, so it enters with a $+3$ :

\frac{v_a - 30}{6} + \frac{v_a - v_b}{4} + 3 = 0 \;\Longrightarrow\; 5v_a - 3v_b = 24

Substituting $v_a = 3v_b$ : $12v_b = 24$ , so $v_b = 2$ V and $v_a = 6$ V. Then the 6 Ω carries $(30-6)/6 = 4$ A $= i_1$ ✓ and the 4 Ω carries $(6-2)/4 = 1$ A $= i_2$ ✓.

The source voltage is just the node voltage at its top terminal (its bottom is grounded): $v_a = 6$ V. That was the quantity mesh analysis refused to give us directly — which is exactly why the supermesh existed.

i_1 = 4

i_2 = 1

A; the current source has 6 V across it (top +), delivering

6 \times 3 = 18

Problem 5 — three-mesh laddermesh · 3×3 · ~12 min

All three mesh currents are clockwise. Find $i_1$ , $i_2$ , and $i_3$ . (They will not come out to round numbers — that's fine; carry fractions and check them exactly.)

Mesh 1, clockwise: rise through the 30 V source, drop across the 2 Ω, drop across the shared 4 Ω:

-30 + 2i_1 + 4(i_1-i_2) = 0 \;\Longrightarrow\; 3i_1 - 2i_2 = 15

Mesh 2, clockwise: the shared 4 Ω on the left as $(i_2-i_1)$ , the 2 Ω exclusive to mesh 2, the shared 4 Ω on the right as $(i_2-i_3)$ :

4(i_2-i_1) + 2i_2 + 4(i_2-i_3) = 0 \;\Longrightarrow\; -2i_1 + 5i_2 - 2i_3 = 0

Mesh 3, clockwise: the shared 4 Ω as $(i_3-i_2)$ , the 2 Ω exclusive to mesh 3, then the 10 V source — its $+$ faces mesh 2, so clockwise (top to bottom through it) is a drop:

4(i_3-i_2) + 2i_3 + 10 = 0 \;\Longrightarrow\; -2i_2 + 3i_3 = -5

Solve. From the first equation, $i_1 = 5 + \tfrac{2}{3}i_2$ . Substituting into the second and combining with the third eliminates $i_1$ and $i_3$ down to a single equation in $i_2$ : $7i_2 = 20$ , so $i_2 = \tfrac{20}{7}\text{ A} \approx 2.857\text{ A}$ . Then $i_3 = \tfrac{5}{21}\text{ A} \approx 0.238\text{ A}$ and $i_1 = \tfrac{145}{21}\text{ A} \approx 6.905\text{ A}$ .

Cross-check by node analysis. The 30 V source pins the top-left corner at 30 V, and the 10 V source pins the top-right corner at 10 V — both directly, no resistor in the way of either source's own terminal — so only the two internal rung-top nodes $B$ and $C$ are unknown:

\frac{B-30}{2} + \frac{B}{4} + \frac{B-C}{2} = 0 \;\Longrightarrow\; 5B - 2C = 60

\frac{C-B}{2} + \frac{C}{4} + \frac{C-10}{2} = 0 \;\Longrightarrow\; 5C - 2B = 20

Solving gives $B = \tfrac{340}{21}$ V, $C = \tfrac{220}{21}$ V. Then $i_1 = (30-B)/2 = \tfrac{145}{21}$ A, $i_2 = (B-C)/2 = \tfrac{20}{7}$ A, and $i_3 = (C-10)/2 = \tfrac{5}{21}$ A — all three match the mesh solution exactly. ✓

i_1 = \tfrac{145}{21}\text{ A} \approx 6.90

i_2 = \tfrac{20}{7}\text{ A} \approx 2.86

i_3 = \tfrac{5}{21}\text{ A} \approx 0.24

Problem 6 — mesh with a CCVS, take twomesh · dependent source · ~10 min

The dependent source is worth $4i_1$ volts, oriented $-$ on top / $+$ on bottom as drawn. Both mesh currents are clockwise. Find $i_1$ , $i_2$ , and the voltage delivered by the dependent source.

Mesh 1, clockwise: rise through the 60 V source, drop across the 10 Ω, drop across the shared 20 Ω:

-60 + 10i_1 + 20(i_1-i_2) = 0 \;\Longrightarrow\; 3i_1 - 2i_2 = 6

Mesh 2, clockwise: the shared 20 Ω as $(i_2-i_1)$ , then the dependent source — clockwise crosses it $-$ to $+$ , a rise, contributing $-4i_1$ :

-4i_1 + 20(i_2-i_1) = 0 \;\Longrightarrow\; -6i_1 + 5i_2 = 0 \;\Longrightarrow\; i_2 = \tfrac{6}{5}i_1

Substituting into the first equation: $3i_1 - \tfrac{12}{5}i_1 = 6 \Rightarrow \tfrac{3}{5}i_1 = 6 \Rightarrow i_1 = 10$ A, so $i_2 = 12$ A.

Cross-check by node analysis. The 60 V source pins the top-left corner at 60 V. Because the dependent source has no series resistor, it pins the middle node $B$ directly: $B = -4i_1$ , and $i_1 = (60-B)/10$ .

B = -4\left(\frac{60-B}{10}\right) \;\Longrightarrow\; 14B = -240 \;\Longrightarrow\; B = -40\text{ V}

Then $i_1 = (60-(-40))/10 = 10$ A ✓. The rung current is $B/20 = -2$ A downward, which is $i_1 - i_2$ , giving $i_2 = i_1 - (-2) = 12$ A ✓ — both match the mesh solution.

Power. The source is worth $4(10) = 40$ V and $B=-40$ V confirms the polarity drawn ( $-$ on top). It carries $i_2 = 12$ A flowing from its $+$ (bottom) terminal out into the external circuit — that's delivery: $P = 40 \times 12 = 480$ W.

i_1 = 10

i_2 = 12

A; the dependent source delivers 480 W.

Problem 7 — chained supermeshsupermesh · 3 meshes · ~12 min

All three mesh currents are clockwise; mesh 2 has no exclusive element of its own. The 4 A source points downward. Find $i_1$ , $i_2$ , $i_3$ , and the voltage across the current source (top terminal positive).

Constraint. Clockwise $i_2$ runs down the shared branch and clockwise $i_3$ runs up it, so the downward current there is $i_2-i_3$ . The arrow points down at 4 A:

i_2 - i_3 = 4

Mesh 1, clockwise (unaffected by the supermesh): rise through the 50 V source, drop across the 5 Ω, drop across the shared 10 Ω:

-50 + 5i_1 + 10(i_1-i_2) = 0 \;\Longrightarrow\; 3i_1 - 2i_2 = 10

Supermesh (meshes 2 and 3 combined), clockwise, skipping the current-source branch: the shared 10 Ω as $(i_2-i_1)$ (mesh 2's own contribution), then straight through to the 8 Ω as $i_3$ (mesh 3's own current — mesh 2 has nothing else to contribute):

10(i_2-i_1) + 8i_3 = 0

Combine all three equations: from the constraint, $i_3 = i_2 - 4$ . Substituting into the supermesh equation: $10i_2 - 10i_1 + 8i_2 - 32 = 0 \Rightarrow 9i_2 - 5i_1 = 16$ . With $i_1 = \tfrac{10+2i_2}{3}$ from mesh 1, this reduces to $17i_2 = 98$ , so $i_2 = \tfrac{98}{17}\text{ A} \approx 5.765$ A, $i_1 = \tfrac{122}{17}\text{ A} \approx 7.176$ A, and $i_3 = \tfrac{30}{17}\text{ A} \approx 1.765$ A.

Cross-check by node analysis. The 50 V source pins the top-left corner at 50 V. Mesh 2's top is a plain wire, so the rung-top node and the current-source-top node are the same node $B$ ; the right edge is grounded directly, so the far-right corner is 0 V. One unknown, $B$ :

\frac{B-50}{5} + \frac{B}{10} + 4 + \frac{B}{8} = 0 \;\Longrightarrow\; 17B = 240 \;\Longrightarrow\; B = \tfrac{240}{17}\text{ V}

Then $i_1 = (50-B)/5 = \tfrac{122}{17}$ A ✓, the rung current $B/10 = \tfrac{24}{17}$ A $= i_1-i_2 \Rightarrow i_2 = \tfrac{98}{17}$ A ✓, and $i_3 = B/8 = \tfrac{30}{17}$ A ✓ — all three match, and $i_2 - i_3 = \tfrac{68}{17} = 4$ ✓.

The source voltage is $B = \tfrac{240}{17} \approx 14.12$ V, top positive. Its 4 A flows out of that $+$ terminal into the rest of the circuit, so it is delivering $\tfrac{240}{17}\times 4 = \tfrac{960}{17} \approx 56.5$ W.

i_1 \approx 7.18

i_2 \approx 5.76

i_3 \approx 1.76

A; the current source has

\approx 14.12

V across it (top +), delivering

\approx 56.5

Before the exam

If you can do four things without hesitating, this chapter is done: (1) write a KCL equation with every term as this node minus that node; (2) recognize a floating voltage source and draw the enclosure without being told; (3) walk a mesh clockwise writing $(i_{\text{mine}} - i_{\text{neighbour}})R$ ; (4) spot a shared current source and reach for the outer loop. Everything else in this chapter is arithmetic.