ECE 211 · Lecture 2: KCL, Branch Currents & Equivalence

§1 · Kirchhoff's Current Law

Charge does not accumulate: KCL

At any node, and around any closed surface, the current flowing in equals the current flowing out. Three equivalent statements of the law are:

Σ I_in = 0 Σ I_out = 0 Σ I_in = Σ I_out

A four-terminal network containing internal elements. The branch currents were labeled using KCL during assignment, which explains their composite expressions. Select where KCL is applied:

\tm{k.i1}{-I_1} + \tm{k.ix}{I_x} + \tm{k.i1x}{(I_1 - I_x)} = 0 \;\;\Longrightarrow\;\; 0 = 0

Currents are summed out of node A, and all terms cancel: 0 = 0. This outcome is not an error. The branch labels were chosen using KCL at A, so the equation is satisfied automatically; labeling with the fewest unknowns consumes the KCL equations up front.

\tm{k.i2}{+I_2} - \tm{k.ix}{I_x} - \tm{k.ibot}{[I_1 - I_x + I_3 - I_4]} = 0 \;\;\Longrightarrow\;\; I_2 - I_1 - I_3 + I_4 = 0

Currents are summed out of node C: $I_2$ leaves the node, while $I_x$ and the bottom-branch current arrive. The $I_x$ terms cancel, and a clean relation among the four terminal currents is obtained.

\tm{k.i1}{-I_1} \tm{k.i3}{\;-\; I_3} \tm{k.i4}{\;+\; I_4} \tm{k.i2}{\;+\; I_2} = 0

A closed surface may be drawn around the entire network and KCL applied to the surface itself. Only currents that cross the dashed boundary appear in the equation; internal currents such as $I_x$ are excluded by construction. One surface equation therefore replaces a chain of node equations, and the result matches the node-C equation exactly.

§2 · The recipe

The branch-current method

Assign a current to each branch, using the fewest unknowns possible. Current-source branches are already known, and KCL should be applied at the nodes during labeling so that no additional symbols are introduced. $\#\text{UNK} = \#\text{BY} - \#\text{CS}$ : the number of unknowns equals the number of backyards minus the number of current sources.
Assign voltage polarities across the resistors. The positive mark is placed where the labeled current enters (the passive sign convention).
Write as many KVL equations as there are unknowns. Independent loops should be selected, and loops through current sources should be avoided because their voltages are unknown.
Solve for the branch currents.
Answer the question that was asked.

Vocabulary

A backyard (BY) is this course's nickname for a window pane of the circuit: an enclosed region of the drawing, analogous to a backyard on a plot map. The number of unknown currents equals the number of panes minus the number of current sources.

Hovering over the window panes and the sources demonstrates the count.

3 backyards − 1 current source = 2 unknown branch currents

§3 · Worked example A: dependent source

Solve for $V_x$

The diamond symbol denotes a current-controlled current source, which forces $2I_x$ through the right branch, where $I_x$ is the current in the 2 Ω resistor. The objective is $V_x$ , the voltage across this source. The recipe is applied step by step:

The problem

Find $V_x$ . Given: a 10 V source; 1 Ω, 2 Ω, and 4 Ω resistors; and the dependent current source $2I_x$ .

Step 1: Assign branch currents with the fewest unknowns

The 2 Ω current is labeled $I_x$ . The dependent source forces $2I_x$ through the right branch. KCL at the top node then determines the left branch: $I_x + 2I_x = 3I_x$ .

2 backyards − 1 current source = 1 unknown: Iₓ

Step 2: Mark polarities on the resistors

Under the passive sign convention, the positive mark is placed where the labeled current enters each resistor. Completing this step before writing any KVL equation largely eliminates sign errors.

Step 3: Apply KVL around the left backyard

\tm{x1.vs}{-10} + \tm{x1.r1 x1.i3x}{3I_x \cdot (1\,\Omega)} + \tm{x1.r2 x1.ix}{I_x \cdot (2\,\Omega)} = 0

Because there is one unknown, one KVL equation suffices. The left loop is selected because it avoids the current source, whose voltage is not yet known.

Step 4: Solve

-10 + 5 I_x = 0 \;\Longrightarrow\; \boxed{I_x = 2\ \text{A}}

Every branch current follows: $3I_x = 6\ \text{A}$ and $2I_x = 4\ \text{A}$ .

Step 5: Answer the question with KVL around the right backyard

\tm{x1.r2 x1.ix}{-I_x \cdot (2\,\Omega)} + \tm{x1.vx}{V_x} + \tm{x1.r4 x1.i2x}{2I_x \cdot (4\,\Omega)} = 0 \;\Longrightarrow\; -4 + V_x + 16 = 0 \;\Longrightarrow\; \boxed{V_x = -12\ \text{V}}

This loop does pass through the current source, which is acceptable here because its voltage $V_x$ is precisely the quantity requested.

Check: The power balance

The powers of Lecture 1 provide the check: $\sum P = 0$ over the elements, each evaluated under the passive sign convention. Hovering over a row highlights the corresponding element:

Element	p = v · i	Power
Dependent source	(−12 V)(4 A)	−48 W
10 V source	−(10 V)(6 A)	−60 W
1 Ω	(6 A)² · 1 Ω	+36 W
2 Ω	(2 A)² · 2 Ω	+8 W
4 Ω	(4 A)² · 4 Ω	+64 W
Total		0 W ✓

Both sources supply power in this circuit (negative values under the passive convention), and the resistors absorb the entire 108 W.

§4 · Worked example B: two current sources

Solve for $I_x$

The same recipe is applied to a busier circuit that contains a dependent current source $\tfrac{1}{2} I_x$ together with an independent 2 A source. It is worth noting how much labeling work the current sources perform for free.

5 branches 6 loops 3 backyards − 2 current sources = 1 unknown

The problem

Find $I_x$ , the current in the 20 Ω resistor. The diamond source pushes $\tfrac{1}{2} I_x$ upward, and the 2 A source pulls current downward.

Step 1: Label with the fewest unknowns

Both current-source branches become known the moment $I_x$ is named. KCL at node a requires the left branch to carry $I_x - \tfrac{1}{2} I_x = \tfrac{1}{2} I_x$ , and KCL at node b assigns $I_x - 2$ to the 5 Ω resistor. One symbol therefore covers all five branches.

Step 2: Mark polarities

The positive mark is placed where the labeled current enters each resistor, as before.

Step 3: Apply KVL around the outside

\tm{x2.vs}{-10} + \tm{x2.r10 x2.ihx}{\tfrac{1}{2} I_x \cdot (10\,\Omega)} + \tm{x2.r20 x2.iix}{I_x \cdot (20\,\Omega)} + \tm{x2.r5 x2.ix2}{[I_x - 2] \cdot (5\,\Omega)} = 0

With one unknown, one KVL equation suffices. The outer loop is the only loop that avoids both current sources, which is precisely why it is selected.

Step 4: Solve

-10 + 5 I_x + 20 I_x + 5 I_x - 10 = 0 \;\Longrightarrow\; 30 I_x = 20 \;\Longrightarrow\; \boxed{I_x = \tfrac{2}{3}\ \text{A}}

The left branch therefore carries $\tfrac{1}{3}$ A, and the 5 Ω resistor carries $\tfrac{2}{3} - 2 = -1\tfrac{1}{3}$ A; the negative sign indicates that this current actually flows upward. A negative answer is information, not an error.

§5 · Equivalence

Same v–i characteristics, simpler circuit

Two networks are equivalent when they present identical voltage-current characteristics at their terminals; the external circuit cannot distinguish between them. Two cases are used constantly:

Series: same current · one node shared exclusively Parallel: same voltage · two nodes shared

One loop implies everything is in series

A single loop carries a single current through all four elements. Therefore the 10 V, 10 Ω, 5 V, and 20 Ω elements are all in series, and series elements may be reordered freely.

Regroup the likes

The sources are grouped together and the resistors are grouped together. The polarities deserve attention: around the loop, the 5 V source opposes the 10 V source.

Combine

10\ \text{V} - 5\ \text{V} = \tm{sq.vs5c}{5\ \text{V}} \qquad\quad 10\ \Omega + 20\ \Omega = \tm{sq.r30}{30\ \Omega}

Voltage sources add algebraically in series, and resistances add in series. Either circuit drives $i = 5/30 = 1/6$ A; the external behavior is unchanged.

§6 · Voltage division

Series resistors split the volts by resistance

V_x = \frac{R_x}{\sum R} \cdot V_s

Clicking a resistor in the schematic selects which voltage is tapped; the sliders then drive the division. The largest resistance always receives the largest share.

V_s (V) 10

R₁ (Ω) 10

R₂ (Ω) 20

R₃ (Ω) 30

V₃ = 3060 · 10 V = 5.00 V

V₁ = 1.67 V V₂ = 3.33 V V₃ = 5.00 V

Derivation of the formula (KVL)

One loop yields one current: $-V_s + R_1 i + R_2 i + R_3 i = 0$ , so $i = V_s/(R_1 + R_2 + R_3)$ and $V_x = R_x \cdot i$ . With the default values, $i = 10/60 = 1/6$ A and $V_3 = 30 \cdot (1/6) = 5$ V.

§7 · Parallel combinations & current division

Branches on shared rails

Everything here is in parallel

All four branches connect the same two rails, that is, the same two nodes. Consequently, every branch sees the same voltage, and the branches may slide along the rails freely without changing the circuit.

Regroup

Sources are grouped with sources and resistors with resistors. The 5 A (down) and the 2 A (up) sources oppose each other.

Combine

5 - 2 = \tm{pp.cs3}{3\ \text{A} \downarrow} \qquad\quad 40 \parallel 10 = \frac{40 \cdot 10}{40 + 10} = \tm{pp.r8}{8\ \Omega}

Current sources add in parallel; voltage sources add in series. For two resistors, the product-over-sum form is the fastest route to the parallel combination.

Invalid configurations

A source concedes nothing about the variable it fixes, so any second element that insists on a different value of that same variable leaves the circuit with no solution. Four configurations are ruled out on this ground, and the board notes strike them out for good reason:

Unequal current sources in series, a KCL contradiction: the current in the connecting wire cannot satisfy both sources.
Unequal voltage sources in parallel, the dual statement, which violates KVL.
A short circuit across a voltage source, which asserts $v = 0$ against the value the source asserts.
An open circuit in series with a current source, which asserts $i = 0$ against the value the source asserts.

The last two are prohibitions on the ideal model rather than on laboratory practice, where the internal resistance of a real source supplies the freedom the ideal model lacks.

Current division

I_x = \frac{R_{\text{opposite}}}{\sum R} \cdot I_s

Two parallel resistors share a source current. The contrast with voltage division deserves emphasis: the opposite resistance appears in the numerator, so the larger share of the current takes the easier path. The moving dots represent the current, and the split follows the controls.

I_s (A) 3

R₁ (Ω) 10

R₂ (Ω) 40

I_x = 1050 · 3 A = 0.60 A (and I₁ = 2.40 A through R₁)

§8 · Y-Δ transformation

When neither series nor parallel applies

Some resistor triples are neither in series nor in parallel; bridge networks are the classic case. A Y (three legs meeting at a center node) may be exchanged for a Δ (a triangle across the same three terminals) without affecting the rest of the circuit. For the balanced case:

Y (also called T): three legs to a center node

Δ (drawn as Π): a triangle over A, B, C

R_Δ (Ω) 45

R_Y = R_Δ3 = 15 Ω R_Δ = 3R_Y

The bridge that requires it

The equivalent resistance between the two ports is sought. No two resistors are in series or in parallel until the top Δ is transformed.

Identify the Δ

The three 15 Ω resistors form a balanced Δ over nodes A, B, and C (dashed). Nothing is yet in series or in parallel, which may be verified directly: no two elements share a lone node or the same node pair.

Transform Δ → Y

The Δ is replaced by a Y with $R_Y = 15/3 = 5\ \Omega$ legs meeting at a new internal node $n$ . The terminals A, B, and C, together with everything attached to them, are untouched.

Reduce by series and parallel combinations

R_{\text{eq}} = \tm{bd.y1}{5} + (\tm{bd.y2 bd.rad}{5 + 10}) \parallel (\tm{bd.y3 bd.rdc}{5 + 15}) = 5 + 15 \parallel 20 = 5 + \tfrac{60}{7} = \boxed{13.57\ \Omega}

Alternate route: Y → Δ yields the same answer

The three 15 Ω legs meeting at C form a Y over A, B, and D. Converting this Y to a Δ with $R_\Delta = 3 \cdot 15 = 45\ \Omega$ places 45 Ω in parallel with the original 15 Ω (A-B) and with the 10 Ω (A-D), plus 45 Ω directly across B-D:

R_{\text{eq}} = [15 \parallel 45 + 10 \parallel 45] \parallel 45 = [11.25 + 8.18] \parallel 45 = \boxed{13.57\ \Omega\ \checkmark}

A different route reaches the same destination; agreement between independent routes is a useful self-check.

§9 · Your turn

Practice: find $V_o$

This problem was left on the board unsolved. Two sources are tied to ground references, so two node voltages are known immediately: the labels −10 V and +5 V are provided. The problem should be attempted on paper before the hints are opened, in order.

Both ground symbols denote the same reference node at 0 V. Solve for $V_o$ .

Hint 1: What is already known

With ground at 0 V, the 10 V source pins its upper terminal at −10 V (note its polarity), and the 5 V source pins the right node at +5 V. Only two node voltages remain unknown: $V_M$ and $V_o$ .

Hint 2: The plan

KCL is written at node M and at node $V_o$ , with every resistor current expressed by Ohm's law as a voltage difference divided by resistance. Two equations in two unknowns result. (This procedure previews node-voltage analysis.)

Hint 3: The two equations

\text{@}M:\quad \tm{pr.r10}{\frac{V_M - (-10)}{10}} + \tm{pr.r20}{\frac{V_M - V_o}{20}} = \tm{pr.cs2}{2}

\text{@}V_o:\quad \tm{pr.r20}{\frac{V_o - V_M}{20}} + \tm{pr.r30}{\frac{V_o - 5}{30}} + \tm{pr.r40}{\frac{V_o}{40}} = 0

Solution

Clearing denominators (×20 for the first equation, ×120 for the second) gives $3 V_M - V_o = 20$ and $13 V_o - 6 V_M = 20$ . Substituting $V_M = (20 + V_o)/3$ :

13 V_o - 2(20 + V_o) = 20 \;\Longrightarrow\; 11 V_o = 60 \;\Longrightarrow\; \boxed{V_o = 60/11 \approx 5.45\ \text{V}} \qquad (V_M = 280/33 \approx 8.48\ \text{V})

A check at node M confirms the result: $(8.48 + 10)/10 = 1.85$ A leaves through the 10 Ω resistor and $(8.48 - 5.45)/20 = 0.15$ A leaves through the 20 Ω resistor, totaling 2.00 A, exactly the current the source injects. ✓

Charge does not accumulate: KCL

The branch-current method

Solve for VxV_xVx​

The problem

Step 1: Assign branch currents with the fewest unknowns

Step 2: Mark polarities on the resistors

Step 3: Apply KVL around the left backyard

Step 4: Solve

Step 5: Answer the question with KVL around the right backyard

Check: The power balance

Solve for IxI_xIx​

The problem

Step 1: Label with the fewest unknowns

Step 2: Mark polarities

Step 3: Apply KVL around the outside

Step 4: Solve

Same v–i characteristics, simpler circuit

One loop implies everything is in series

Regroup the likes

Combine

Series resistors split the volts by resistance

Branches on shared rails

Everything here is in parallel

Regroup

Combine

Current division

When neither series nor parallel applies

The bridge that requires it

Identify the Δ

Transform Δ → Y

Reduce by series and parallel combinations

Practice: find VoV_oVo​

Solve for $V_x$

Solve for $I_x$

Practice: find $V_o$