KVL and KCL — Solved Problems

Set 3 — Kirchhoff's Laws

Twenty worked problems on the two statements that make circuit analysis possible: charge does not accumulate at a node, and energy per unit charge returns to zero around a loop. Series–parallel reduction works only on circuits with convenient structure; Kirchhoff's laws work on every circuit ever drawn. The price is simultaneous equations, and the discipline is signs.

Textbook Chapter 2 · 20 solved · 12 practice · 3 challenge · 12 MCQs

i Method Recap

KCL: \(\sum i = 0\) at every node. It follows from conservation of charge — charge cannot pile up at a point. Choose one convention (currents leaving are positive, say) and never mix it within a single equation.
KVL: \(\sum v = 0\) around every closed loop. It follows from conservation of energy — a unit charge carried round a loop returns to its starting potential. Walk the loop in one direction and record the sign of the terminal you enter first.
Assign reference directions before you know the answers. A negative result is not a mistake; it means the true direction is opposite to your arrow. Never redraw the arrow to make an answer positive — the sign is information.
How many equations? A circuit with \(n\) nodes and \(b\) branches gives \(n-1\) independent KCL equations and \(b-n+1\) independent KVL equations. Together that is exactly \(b\) — enough to find every branch current and no more.
A dependent source obeys KVL and KCL like any other element. Write it into the equations by its symbol, then add the constraint that defines its controlling variable. The extra equation is what makes the system solvable.
Check with power. Once every current is known, verify \(\sum p = 0\). It costs one line and catches sign errors that the equations themselves will happily absorb.

VideoWalkthrough

Problem 1Exam levelKCL & Dependent Source

A 9 A source feeds a single node. Three branches leave it: a 2 Ω resistor carrying \(i_0\), a current-controlled current source of value \(i_0/4\), and an 8 Ω resistor. Find \(i_0\) and the node voltage \(v_0\).

9 A into a node with a current-controlled current source

Solution

There is only one unknown node, so one KCL equation will do. Taking currents leaving node \(A\) as positive, the 9 A source is an inflow and the three branches are outflows:

\[ 9 = i_0 + \frac{i_0}{4} + \frac{v_0}{8} \]

Note that the dependent source was written into the equation exactly like an independent one. Its value happens to be unknown, but that is a matter for the next step.

Now supply the constraint that defines the controlling variable. The node voltage sits across the 2 Ω resistor, and \(i_0\) is the current through it, so Ohm's law links them:

\[ v_0 = 2 i_0 \]

Substituting to eliminate \(v_0\):

\[ 9 = i_0 + \frac{i_0}{4} + \frac{2i_0}{8} = i_0\left(1 + \frac{1}{4} + \frac{1}{4}\right) = \frac{3}{2}i_0 \]

Hence

\[ i_0 = 6\ \text{A},\qquad v_0 = 2(6) = 12\ \text{V} \]

KCL check with the numbers restored:

\[ 6 + \frac{6}{4} + \frac{12}{8} = 6 + 1.5 + 1.5 = 9\ \text{A}\;\checkmark \]

The pattern to internalise: one unknown node voltage needs one KCL equation; each dependent source adds one unknown and one defining constraint, so the count never goes wrong. Write the KCL first with the dependent source treated as a black box, and only then ask what its controlling variable equals.

Answer\(i_0 = 6\ \text{A},\quad v_0 = 12\ \text{V}\)

Problem 2Exam levelKVL & Dependent Source

In the single loop shown, a voltage-controlled voltage source of value \(3V_R\) drives a 5 Ω resistor, a 4 V source and a 2 Ω resistor, where \(V_R\) is the voltage across the 2 Ω. Find the loop current \(I\) and verify conservation of power.

Single loop containing a VCVS

Solution

Walk the loop clockwise starting at the bottom-left corner. The VCVS is a rise of \(3V_R\); the 5 Ω and the 4 V source are drops; and the 2 Ω is a drop of \(V_R\):

\[ 3V_R - 5I - 4 - V_R = 0 \;\Longrightarrow\; 2V_R = 5I + 4 \]

The controlling variable is the drop across the 2 Ω, taken in the direction of \(I\):

\[ V_R = 2I \]

Substituting:

\[ 2(2I) = 5I + 4 \;\Longrightarrow\; 4I = 5I + 4 \;\Longrightarrow\; I = -4\ \text{A} \]

The negative sign says the current really circulates anticlockwise at 4 A. The assumed direction was wrong; the algebra was not.

Back-substituting, \(V_R = 2(-4) = -8\ \text{V}\) and the dependent source is \(3V_R = -24\ \text{V}\). Verify KVL with the numbers:

\[ -24 - 5(-4) - 4 - (-8) = -24 + 20 - 4 + 8 = 0\;\checkmark \]

Now the power balance. The resistors absorb:

\[ P_{5\Omega} = I^2(5) = 80\ \text{W},\qquad P_{2\Omega} = I^2(2) = 32\ \text{W} \]

Both sources appear as rises against the true current direction, so both deliver:

\[ P_{\text{VCVS}} = (3V_R)(I) = (-24)(-4) = 96\ \text{W supplied} \]

\[ P_{4\text{V}} = (4)(-I) = 16\ \text{W supplied},\qquad 96 + 16 = 112 = 80 + 32\;\checkmark \]

A dependent source can dominate a circuit. Here the 4 V independent source contributes only 16 W of the 112 W dissipated; the VCVS supplies the rest. That is exactly what an amplifier does — a small input controls a large energy flow drawn from elsewhere. Circuit theory models the flow but not its ultimate source, which is why a dependent source is never the whole story of a real device.

Answer\(I = -4\ \text{A},\quad V_R = -8\ \text{V}\); 112 W supplied and absorbed

Problem 3Exam levelTwo Loops, Controlled Source

Two loops share only a common reference node. The left loop contains a 5 V source driving two 1 Ω resistors in series, with node \(a\) at their junction — that is, at the top of the lower 1 Ω. The right loop contains a voltage-controlled voltage source of value \(4V_{ab}\) driving a 3 Ω resistor in series with a 1 Ω resistor, with node \(b\) at the top of that 1 Ω. Find the current \(i\) in the right loop.

Solution

The left loop contains no dependent element, so solve it outright. Its total resistance is \(1 + 1 = 2\ \Omega\), so KVL gives

\[ 5 = 2I_1 \;\Longrightarrow\; I_1 = 2.5\ \text{A} \]

Node \(a\) sits above the lower 1 Ω, so its potential is that resistor's drop:

\[ V_a = (1)(I_1) = 2.5\ \text{V} \]

In the right loop, KVL round the dependent source and the 3 Ω plus 1 Ω gives

\[ 4V_{ab} = (3 + 1)\,i = 4i \;\Longrightarrow\; i = V_{ab} \]

The same current \(i\) flows through the 1 Ω resistor terminating at node \(b\), so

\[ V_b = (1)(i) = i \]

Now close the loop on the definition \(V_{ab} = V_a - V_b\):

\[ V_{ab} = 2.5 - i,\qquad\text{and}\qquad i = V_{ab} = 2.5 - i \]

\[ 2i = 2.5 \;\Longrightarrow\; i = 1.25\ \text{A} \]

Check: \(V_b = 1.25\ \text{V}\), so \(V_{ab} = 2.5 - 1.25 = 1.25\ \text{V} = i\;\checkmark\)

Why the two loops are coupled at all. No current passes between them — they share only the reference node. The coupling is entirely through the controlling variable: the right loop's source depends on a potential difference that spans both. This is the essential difference between a dependent source and a wire, and it is how a transistor amplifier isolates its input from its output while still letting one govern the other.

Answer\(i = 1.25\ \text{A},\quad V_{ab} = 1.25\ \text{V}\)

Problem 4CoreKCL at a Node

Four branches meet at a single node. A voltage source drives 1 A into the node, a 3 A source drives 3 A into it, a further branch carries 7 A out of it, and the fourth branch carries the unknown \(i_2\) out. Find \(i_2\).

Four branches at one node

Solution

Fix a convention and hold to it: take currents entering the node as positive. Then the two inflows are positive and the two outflows are negative:

\[ 1 + 3 - 7 - i_2 = 0 \]

Solving:

\[ i_2 = 1 + 3 - 7 = -3\ \text{A} \]

Equivalently, in the form "sum in = sum out":

\[ \underbrace{1 + 3}_{\text{in}} = \underbrace{7 + i_2}_{\text{out}} \;\Longrightarrow\; i_2 = 4 - 7 = -3\ \text{A} \]

Reading the sign. The negative answer means 3 A actually flows into the node along that branch, opposite to the drawn arrow. Notice that no resistance was needed anywhere: KCL is a statement about charge alone and knows nothing about what the branches contain. That is why it applies equally to a node in a resistive network, a node inside a transistor model, and a node in a power system.

Answer\(i_2 = -3\ \text{A}\) (3 A into the node)

Problem 5CoreKCL for a Node Voltage

A node of unknown voltage \(v\) is fed by a 1 A source and a 2 A source, and drains through two 5 Ω resistors to ground and through a 5 A source. Determine \(v\).

One node, five branches

Solution

Both resistors run from the same node to ground, so each carries \(v/5\) downward. Writing "in equals out":

\[ 1 + 2 = \frac{v}{5} + 5 + \frac{v}{5} \]

Collecting terms:

\[ 3 - 5 = \frac{2v}{5} \;\Longrightarrow\; -2 = 0.4\,v \;\Longrightarrow\; v = -5\ \text{V} \]

Check by substitution. Each resistor carries \(-5/5 = -1\ \text{A}\), meaning 1 A flows up into the node through each:

\[ 1 + 2 + 1 + 1 = 5\ \text{A}\;\checkmark \]

Why the node went negative. The 5 A source demands more current than the two supplies deliver, so the deficit has to come up through the resistors from ground — which is only possible if the node sits below ground potential. An ideal current source will do whatever it takes to its terminal voltage to enforce its current, including driving a node negative. That is precisely the modelling assumption that makes it "ideal".

Answer\(v = -5\ \text{V}\)

Problem 6ChallengeCCCS with a Sensing Branch

In the circuit shown, node \(A\) sits at potential \(V\). A 4 mA source draws current out of \(A\), a 5 kΩ resistor connects \(A\) to ground, and both a 20 kΩ resistor and a current-controlled current source of value \(3i_1\) connect \(A\) to node \(B\). Node \(B\) is tied to ground by a short-circuit sensing link carrying \(i_1\). Find \(i_x = i_1\).

CCCS controlled by a short-circuit sensing current

Solution

Node \(B\) is joined to ground by a plain wire, so \(V_B = 0\). That is the fact that makes the problem tractable: the 20 kΩ resistor therefore has the full \(V\) across it and carries \(V/20\text{k}\).

KCL at node \(A\), taking every current as leaving:

\[ 4\,\text{mA} + 3i_1 + \frac{V}{5\,\text{k}\Omega} + \frac{V}{20\,\text{k}\Omega} = 0 \tag{1} \]

KCL at node \(B\). Current arrives from the 20 kΩ and from the dependent source, and departs down the sensing link:

\[ i_1 = 3i_1 + \frac{V}{20\,\text{k}\Omega} \tag{2} \]

Rearranging (2) gives \(i_1\) in terms of \(V\):

\[ -2i_1 = \frac{V}{20\,\text{k}\Omega} \;\Longrightarrow\; i_1 = -\frac{V}{40\,\text{k}\Omega} \]

Substituting into (1), and putting everything over a common denominator of 40 kΩ:

\[ \begin{aligned} 4\,\text{mA} - \frac{3V}{40\,\text{k}} + \frac{8V}{40\,\text{k}} + \frac{2V}{40\,\text{k}} &= 0\\[2pt] 4\,\text{mA} + \frac{7V}{40\,\text{k}} &= 0 \end{aligned} \]

Hence

\[ V = -\frac{4 \times 10^{-3} \times 40 \times 10^{3}}{7} = -\frac{160}{7} = -22.86\ \text{V} \]

And finally

\[ i_x = i_1 = -\frac{V}{40\,\text{k}\Omega} = \frac{22.86}{40 \times 10^{3}} = 0.571\ \text{mA} \]

Unit discipline saves this problem. Every source is in milliamps and every resistance in kilohms, so \(\text{mA} \times \text{k}\Omega = \text{V}\) exactly, and the kilo and milli cancel throughout. Work in those units from the first line and no power of ten ever appears. The answer is 0.571 mA, not 0.571 A — a factor of a thousand that costs full marks.

Answer\(V = -22.86\ \text{V},\quad i_x = i_1 = 0.571\ \text{mA}\)

Problem 7CoreNode Voltages and V_ab

A 24 V source drives a series chain of 4 Ω, 8 Ω and 12 Ω to ground. Label the top of the chain \(d\), the junctions \(a\) and \(b\), and the ground node \(c\). Find every node potential, then evaluate \(V_{ac}\), \(V_{bd}\) and \(V_{ab} + V_{bc} + V_{cd} + V_{da}\).

Series chain with four labelled nodes

Solution

One loop, one current. KVL round the chain:

\[ 24 = (4 + 8 + 12)\,I = 24\,I \;\Longrightarrow\; I = 1\ \text{A} \]

Take \(c\) as the reference at 0 V and walk down the chain, subtracting each drop in turn:

Node	Potential	Reasoning
\(d\)	24 V	directly at the source terminal
\(a\)	20 V	\(24 - (1)(4)\)
\(b\)	12 V	\(20 - (1)(8)\)
\(c\)	0 V	\(12 - (1)(12)\), the reference

Every double-subscript voltage is now a subtraction, using \(V_{xy} = V_x - V_y\):

\[ V_{ac} = 20 - 0 = 20\ \text{V},\qquad V_{bd} = 12 - 24 = -12\ \text{V} \]

For the closed chain of four terms, the potentials cancel in pairs:

\[ \begin{aligned} V_{ab} + V_{bc} + V_{cd} + V_{da} &= (V_a - V_b) + (V_b - V_c) + (V_c - V_d) + (V_d - V_a)\\[2pt] &= 8 + 12 + (-24) + 4 = 0 \end{aligned} \]

This is KVL. The cancellation in the last step needs no circuit at all — it is an algebraic identity about differences of any four numbers. That is the whole content of Kirchhoff's voltage law: because every node has a single well-defined potential, the drops around any closed path must sum to zero. Once you see KVL this way, "which sign do I use?" becomes "which node am I subtracting from which?"

Answer\(V_d = 24,\ V_a = 20,\ V_b = 12,\ V_c = 0\ \text{V}\); \(V_{ac} = 20\), \(V_{bd} = -12\), sum = 0

Problem 8Exam levelKCL Chain

A 1 A source injects current into node 1, which is connected to ground by a 4 Ω resistor and to node 2 by a 2 Ω resistor. At node 2 a 3 A source draws current to ground. Find the node voltages \(V_1\) and \(v_x = V_2\).

Two nodes connected by a single resistor

Solution

Start at node 2, where the answer is forced. Only two branches meet there: the 2 Ω resistor and the 3 A source. KCL therefore fixes the resistor current outright, with no algebra:

\[ I_{2\Omega} = 3\ \text{A} \quad \text{(flowing from node 1 to node 2)} \]

A node with only two branches is always worth finding first — it hands you a current for free.

Now KCL at node 1, with 1 A entering and two branches leaving:

\[ 1 = I_{4\Omega} + I_{2\Omega} = I_{4\Omega} + 3 \;\Longrightarrow\; I_{4\Omega} = -2\ \text{A} \]

Ohm's law on the 4 Ω gives the first node voltage:

\[ V_1 = (4)(-2) = -8\ \text{V} \]

Walking along the 2 Ω in the direction of its 3 A current, the potential falls:

\[ v_x = V_2 = V_1 - (2)(3) = -8 - 6 = -14\ \text{V} \]

Power check. The 1 A source sits at \(-8\) V and the 3 A source at \(-14\) V:

\[ \underbrace{(-8)(1)}_{-8\ \text{W}} + \underbrace{(-14)(-3)}_{+42\ \text{W}} = 34\ \text{W supplied by the 3 A source net} \]

\[ P_{4\Omega} + P_{2\Omega} = (2)^2(4) + (3)^2(2) = 16 + 18 = 34\ \text{W}\;\checkmark \]

Both nodes came out negative, and that is correct. The 3 A sink is stronger than the 1 A source, so the network has to pull the extra 2 A up from ground through the 4 Ω, which is only possible below 0 V. The 1 A source, sitting at a negative potential while pushing current out, is actually absorbing 8 W — another reminder that "source" is a name for a symbol, not a guarantee about energy flow.

Answer\(V_1 = -8\ \text{V},\quad v_x = -14\ \text{V}\)

Problem 9CoreKCL with a Voltage Source

A 10 mA source feeds node \(V\). From that node, one 100 Ω resistor runs to the positive terminal of a 1 V source whose negative terminal is grounded, and a second 100 Ω resistor runs directly to ground. Find \(V\) and the current \(I_x\) in the grounded resistor.

A node draining through two resistors, one to a 1 V source

Solution

The far end of the right-hand resistor is held at 1 V by the source, so the current in it is \((V-1)/100\). The left-hand resistor sees ground, so its current is \(V/100\). KCL at the node, with 10 mA entering:

\[ \frac{V - 1}{100} + \frac{V}{100} = 10\ \text{mA} = 0.01\ \text{A} \]

Multiplying through by 100:

\[ (V - 1) + V = 1 \;\Longrightarrow\; 2V = 2 \;\Longrightarrow\; V = 1\ \text{V} \]

The current in the grounded resistor is therefore

\[ I_x = \frac{V}{100} = \frac{1}{100} = 10\ \text{mA} \]

Check the other branch: \((1-1)/100 = 0\). All 10 mA goes down the grounded resistor and none flows into the 1 V source, which is consistent because both ends of that resistor sit at 1 V.

An accidental balance worth noticing. The node settles at exactly the source voltage, so the right-hand branch carries nothing at all. Change the 10 mA to 20 mA and the node rises to 1.5 V, pushing 5 mA into the source — charging it. Reduce it to 5 mA and the node falls to 0.5 V, so the source now supplies 5 mA. A single circuit, three qualitatively different behaviours, decided by one number.

Answer\(V = 1\ \text{V},\quad I_x = 10\ \text{mA}\)

Problem 10CoreSingle-Loop KVL

A 20 V source drives a clockwise current \(I\) through a 4 Ω and a 6 Ω resistor, opposed by a 5 V source. Find (a) the loop current, (b) the power delivered by the 20 V source, and (c) the voltage across the 6 Ω.

Single loop with opposing sources

Solution

aWalk clockwise from the bottom-left. The 20 V source is a rise; the two resistors are drops; the 5 V source is entered at its positive terminal and so is also a drop:

\[ 20 - 4I - 5 - 6I = 0 \;\Longrightarrow\; 15 = 10I \;\Longrightarrow\; I = 1.5\ \text{A} \]

bCurrent leaves the positive terminal of the 20 V source, so it delivers:

\[ P = VI = 20 \times 1.5 = 30\ \text{W} \]

cOhm's law on the 6 Ω:

\[ V_{6\Omega} = 6I = 9\ \text{V} \]

Full power audit. The 5 V source has current entering its positive terminal, so it absorbs:

\[ \underbrace{30}_{\text{20 V supplies}} = \underbrace{9}_{4\,\Omega} + \underbrace{13.5}_{6\,\Omega} + \underbrace{7.5}_{\text{5 V absorbs}}\;\checkmark \]

This is a battery charger. The 5 V source is being driven backwards by the stronger 20 V supply, and it takes 7.5 W out of the circuit rather than putting energy in. Every phone charger, every regenerative brake and every grid-tied inverter is the same arrangement — and the 4 Ω and 6 Ω are what stop the charging current from being catastrophic.

Answer\(I = 1.5\ \text{A},\ P_{20\text{V}} = 30\ \text{W},\ V_{6\Omega} = 9\ \text{V}\)

Problem 11CoreKCL & Current Division

A 6 A source feeds node \(A\), which connects to ground through a 3 Ω and a 6 Ω resistor in parallel. Find \(V_A\) and the two branch currents, and verify KCL.

Current source feeding two parallel resistors

Solution

All 6 A must return to ground through the parallel pair, whose combined resistance is

\[ R_p = \frac{3 \times 6}{3+6} = 2\ \Omega \]

The node voltage follows immediately:

\[ V_A = I R_p = 6 \times 2 = 12\ \text{V} \]

Branch currents by Ohm's law:

\[ I_{3\Omega} = \frac{12}{3} = 4\ \text{A},\qquad I_{6\Omega} = \frac{12}{6} = 2\ \text{A} \]

KCL check at node \(A\):

\[ 4 + 2 = 6\ \text{A}\;\checkmark \]

Two routes to the same numbers. The current-divider formula gives them directly: \(I_{3\Omega} = 6 \times 6/(3+6) = 4\ \text{A}\), with the other resistance in the numerator. Using \(V_A\) as the intermediate step is slower but far harder to get wrong, and it generalises to any number of branches while the two-branch formula does not.

Answer\(V_A = 12\ \text{V},\ I_{3\Omega} = 4\ \text{A},\ I_{6\Omega} = 2\ \text{A}\)

Problem 12Exam levelBranch-Current Method

A 12 V source with a 3 Ω resistor and a 10 V source with a 4 Ω resistor share a common 2 Ω branch. Using branch currents \(I_1\), \(I_2\) and \(I_3\), find all three.

Two sources sharing a common branch

Solution

Three unknown branch currents need three equations. KCL at node \(P\) supplies the first — both source branches feed the middle branch:

\[ I_3 = I_1 + I_2 \]

KVL round the left loop, and round the right loop, supplies the other two:

\[ 12 = 3I_1 + 2I_3,\qquad 10 = 4I_2 + 2I_3 \]

Eliminate \(I_3\) at once by substitution, leaving two equations in two unknowns:

\[ \begin{aligned} 12 &= 3I_1 + 2(I_1 + I_2) = 5I_1 + 2I_2\\[2pt] 10 &= 4I_2 + 2(I_1 + I_2) = 2I_1 + 6I_2 \end{aligned} \]

From the second, \(I_1 = 5 - 3I_2\). Substituting into the first:

\[ 5(5 - 3I_2) + 2I_2 = 12 \;\Longrightarrow\; 25 - 13I_2 = 12 \;\Longrightarrow\; I_2 = 1\ \text{A} \]

Back-substituting:

\[ I_1 = 5 - 3(1) = 2\ \text{A},\qquad I_3 = 2 + 1 = 3\ \text{A} \]

Check both loops independently:

\[ 3(2) + 2(3) = 12\ \text{V}\;\checkmark \qquad 4(1) + 2(3) = 10\ \text{V}\;\checkmark \]

Why this method fell out of favour. The branch-current method is the most direct application of Kirchhoff's laws, and it works on anything. Its cost is that it carries \(b\) unknowns — here three — when the topology only requires two. Mesh analysis (Set 4) uses the loop equations alone and never writes KCL down at all; nodal analysis (Set 6) does the reverse. Both are bookkeeping improvements on exactly this calculation.

Answer\(I_1 = 2\ \text{A},\quad I_2 = 1\ \text{A},\quad I_3 = 3\ \text{A}\)

Problem 13CoreKVL Round Three Loops

A 24 V source drives a 4 Ω resistor into node \(A\), from which a 6 Ω and a 12 Ω resistor return to ground in parallel. Find all currents, then verify KVL around three different closed paths — including one that contains no source.

Solution

Reduce and solve:

\[ 6\,\|\,12 = 4\ \Omega,\qquad R_T = 4 + 4 = 8\ \Omega,\qquad I = \frac{24}{8} = 3\ \text{A} \]

\[ V_A = 24 - (3)(4) = 12\ \text{V},\qquad I_{6\Omega} = 2\ \text{A},\qquad I_{12\Omega} = 1\ \text{A} \]

Loop 1 — source, 4 Ω, 6 Ω:

\[ 24 - (3)(4) - (2)(6) = 24 - 12 - 12 = 0\;\checkmark \]

Loop 2 — source, 4 Ω, 12 Ω:

\[ 24 - (3)(4) - (1)(12) = 24 - 12 - 12 = 0\;\checkmark \]

Loop 3 — up through the 6 Ω and down through the 12 Ω, touching no source at all:

\[ (2)(6) - (1)(12) = 12 - 12 = 0\;\checkmark \]

The three equations are not independent. Loop 3 is the difference of Loops 1 and 2, so only two of them carry information — exactly the \(l = b - n + 1 = 3 - 2 + 1 = 2\) predicted by the topology.

Loop 3 is parallel resistance in disguise. Saying that two elements are in parallel and saying that KVL round the loop they form gives zero are the same statement. That is why the parallel rule needed no separate justification: it was Kirchhoff's voltage law all along, applied to the simplest possible loop.

Answer\(I = 3\ \text{A},\ V_A = 12\ \text{V},\ I_{6\Omega} = 2\ \text{A},\ I_{12\Omega} = 1\ \text{A}\)

Problem 14Exam levelGeneralised KCL

A sealed sub-network has four leads crossing its boundary. Measurement shows 12 A entering on lead 1, 5 A leaving on lead 2 and 3 A leaving on lead 3. Find the current in lead 4, stating its direction. What can you say about the contents of the box?

KCL applied to a closed surface, not a point

Solution

Kirchhoff's current law is usually stated for a node, but it holds for any closed surface. The reason is the same: charge cannot accumulate inside the surface, so whatever flows in must flow out. Shrink the surface onto a point and you recover the node version; expand it to enclose an entire sub-network and nothing changes.

Taking currents leaving the surface as positive:

\[ -12 + 5 + 3 + I_4 = 0 \]

Hence

\[ I_4 = 12 - 5 - 3 = 4\ \text{A}\ \text{leaving the box} \]

As for the contents: nothing whatsoever can be deduced. The box may hold two resistors or two hundred, sources or none, linear or nonlinear elements. KCL constrains only the boundary.

This is the idea behind the supernode. When a voltage source sits between two nodes and blocks the usual node equation, you draw a surface enclosing both nodes and the source, and apply KCL to that surface instead — Set 7 does exactly this. It is also how a whole substation is treated as one node in power-system analysis, and how a transistor is handled as a three-terminal object with \(i_E = i_B + i_C\), a statement that is nothing but generalised KCL.

Answer\(I_4 = 4\ \text{A}\) leaving the box; contents undetermined

Problem 15Exam levelSign Discipline

A single loop contains a 10 V source, a 3 Ω resistor, a 20 V source connected in opposition, and a 7 Ω resistor. Taking \(I\) clockwise and starting at the 10 V source, find \(I\) and account for the power in every element.

Solution

KVL clockwise, with the 10 V source a rise and the 20 V source a drop:

\[ 10 - 3I - 20 - 7I = 0 \;\Longrightarrow\; -10 = 10I \;\Longrightarrow\; I = -1\ \text{A} \]

The true current is 1 A anticlockwise, driven by the stronger 20 V source against the 10 V one.

Resistors absorb regardless of direction:

\[ P_{3\Omega} = (1)^2(3) = 3\ \text{W},\qquad P_{7\Omega} = (1)^2(7) = 7\ \text{W} \]

The 20 V source has current leaving its positive terminal, so it supplies:

\[ P_{20\text{V}} = 20 \times 1 = 20\ \text{W supplied} \]

The 10 V source has current entering its positive terminal, so it absorbs:

\[ P_{10\text{V}} = 10 \times 1 = 10\ \text{W absorbed} \]

Balance:

\[ 20 = 3 + 7 + 10\;\checkmark \]

The three-step discipline that never fails. First assign a reference direction — arbitrarily, and before solving. Second, write KVL by recording the sign of the terminal you enter first, mechanically, without thinking about what the answer "should" be. Third, interpret the sign at the end. Students who redraw the arrow halfway through to keep the current positive lose track of which power is supplied and which absorbed, and then cannot make the balance close.

Answer\(I = -1\ \text{A}\); 20 V supplies 20 W, 10 V absorbs 10 W, resistors 10 W

Problem 16Exam levelKVL with a VCVS

A single loop contains an 18 V source, a 3 Ω resistor across which \(v_x\) is defined, a 6 Ω resistor, and a voltage-controlled voltage source of value \(2v_x\) connected in opposition. Find \(I\), \(v_x\), and the power in the dependent source.

Solution

KVL round the loop, with the dependent source written as a symbol:

\[ 18 - 3I - 6I - 2v_x = 0 \]

The constraint that defines the controlling variable:

\[ v_x = 3I \]

Substituting turns one equation in two unknowns into one in one:

\[ 18 - 3I - 6I - 6I = 0 \;\Longrightarrow\; 18 = 15I \;\Longrightarrow\; I = 1.2\ \text{A} \]

Hence

\[ v_x = 3(1.2) = 3.6\ \text{V},\qquad 2v_x = 7.2\ \text{V} \]

Verify KVL numerically — the drops must exhaust the source:

\[ 3.6 + 7.2 + 7.2 = 18\ \text{V}\;\checkmark \]

The dependent source has current entering its positive terminal, so it absorbs:

\[ P_{\text{dep}} = (7.2)(1.2) = 8.64\ \text{W absorbed} \]

Full balance: the 18 V source supplies \(18 \times 1.2 = 21.6\ \text{W}\); the resistors take \(1.44(3) + 1.44(6) = 12.96\ \text{W}\); and \(12.96 + 8.64 = 21.6\ \text{W}\;\checkmark\)

Compare with Problem 2, where the dependent source supplied 96 W. The only difference is polarity: there the VCVS reinforced the loop current, here it opposes it. A dependent source has no intrinsic tendency to give or take energy — it does whatever its controlling variable and its orientation dictate. This particular arrangement, where the source opposes in proportion to the current, is negative feedback, and it is why the effective resistance seen by the 18 V source is 15 Ω rather than the 9 Ω the resistors alone would give.

Answer\(I = 1.2\ \text{A},\ v_x = 3.6\ \text{V}\); dependent source absorbs 8.64 W

Problem 17Exam levelNegative Branch Current

An 18 V source with a 2 Ω resistor and an 8 V source with a 4 Ω resistor share a common 6 Ω branch. Find the three branch currents and identify which source is being charged.

Solution

Assume both source currents flow towards the common node, so that \(I_3 = I_1 + I_2\). KVL round the two loops:

\[ 18 = 2I_1 + 6I_3,\qquad 8 = 4I_2 + 6I_3 \]

Substituting \(I_3 = I_1 + I_2\):

\[ \begin{aligned} 18 &= 8I_1 + 6I_2\\[2pt] 8 &= 6I_1 + 10I_2 \end{aligned} \]

Multiply the first by 5 and the second by 3, then subtract to eliminate \(I_2\):

\[ 90 = 40I_1 + 30I_2,\quad 24 = 18I_1 + 30I_2 \;\Longrightarrow\; 66 = 22I_1 \;\Longrightarrow\; I_1 = 3\ \text{A} \]

Back-substituting:

\[ I_2 = \frac{18 - 8(3)}{6} = \frac{-6}{6} = -1\ \text{A},\qquad I_3 = 3 + (-1) = 2\ \text{A} \]

Check both loop equations:

\[ 2(3) + 6(2) = 18\;\checkmark \qquad 4(-1) + 6(2) = -4 + 12 = 8\;\checkmark \]

Full power audit, with the negative \(I_2\) interpreted correctly:

\[ \begin{aligned} \text{18 V supplies} &= 18 \times 3 = 54\ \text{W}\\[2pt] \text{8 V absorbs} &= 8 \times 1 = 8\ \text{W}\\[2pt] \text{Resistors} &= 3^2(2) + 1^2(4) + 2^2(6) = 18 + 4 + 24 = 46\ \text{W}\\[2pt] 54 &= 8 + 46\;\checkmark \end{aligned} \]

The negative sign is the entire answer to the last part. \(I_2 = -1\ \text{A}\) means current flows into the positive terminal of the 8 V source, so it is being charged at 8 W by the 18 V source. Had you "fixed" the sign by redrawing the arrow, the numbers would still be right but the physical conclusion would be lost. This is a real configuration: two batteries of different voltage connected in parallel across a load, with the stronger one charging the weaker.

Answer\(I_1 = 3\ \text{A},\ I_2 = -1\ \text{A},\ I_3 = 2\ \text{A}\); the 8 V source is charging

Problem 18Exam levelOpen Branch

A 20 V source feeds a 5 Ω resistor into node \(A\). From \(A\), a 15 Ω resistor returns to ground, and a second path consisting of an open switch in series with a 30 Ω resistor also runs to ground. Find the source current, \(V_A\), the current in the 30 Ω, and the voltage across the open switch.

Solution

An open branch carries no current, so the 30 Ω branch is simply absent from the current calculation:

\[ I_{30\Omega} = 0 \]

The remaining circuit is a single loop:

\[ I = \frac{20}{5 + 15} = 1\ \text{A},\qquad V_A = (1)(15) = 15\ \text{V} \]

Now the switch voltage. Because no current flows in the 30 Ω, there is no drop across it, so its far terminal sits at the same potential as the switch's lower contact — that is, at ground:

\[ V_{\text{lower contact}} = 0 - (0)(30) = 0\ \text{V} \]

The upper contact is node \(A\), so KVL across the open branch gives

\[ V_{\text{switch}} = V_A - 0 = 15\ \text{V} \]

Closing the switch would change everything: the load becomes \(15\,\|\,30 = 10\ \Omega\), so \(I = 20/15 = 1.33\ \text{A}\) and \(V_A\) falls to 13.3 V.

Zero current does not mean zero voltage. This is the single most common conceptual error with open branches. The full 15 V stands across the switch contacts precisely because nothing flows — and it is why the switch must be rated for that voltage even though it carries no current, and why a broken wire in a mains circuit is still lethal to touch.

Answer\(I = 1\ \text{A},\ V_A = 15\ \text{V},\ I_{30\Omega} = 0,\ V_{\text{switch}} = 15\ \text{V}\)

Problem 19ChallengeCurrent Source in a Shared Branch

A 30 V source with a 5 Ω resistor forms the left branch; a 10 V source with a 5 Ω resistor forms the right branch; and the shared middle branch contains a 2 A current source directed downwards. Find the two outer branch currents and the voltage across the current source.

Solution

The current source removes one unknown before you start: the middle branch current is given as 2 A. What it does not give you is the voltage across itself, which becomes the new unknown. Call it \(V_{cs}\).

KCL at the top node, with both outer branches feeding the middle one:

\[ I_1 = 2 + I_2 \]

KVL round the left loop, taking the drop across the source as \(V_{cs}\):

\[ 30 = 5I_1 + V_{cs} \tag{1} \]

KVL round the right loop:

\[ V_{cs} = 5I_2 + 10 \tag{2} \]

Substitute (2) into (1) and use the KCL relation to eliminate \(I_1\):

\[ 30 = 5(2 + I_2) + 5I_2 + 10 = 10 + 5I_2 + 5I_2 + 10 = 20 + 10I_2 \]

Hence

\[ I_2 = 1\ \text{A},\qquad I_1 = 3\ \text{A},\qquad V_{cs} = 5(1) + 10 = 15\ \text{V} \]

Check the left loop: \(5(3) + 15 = 30\ \text{V}\;\checkmark\). Power audit:

\[ \underbrace{30(3)}_{90\ \text{W}} = \underbrace{5(9)}_{45} + \underbrace{5(1)}_{5} + \underbrace{15(2)}_{30} + \underbrace{10(1)}_{10} = 90\ \text{W}\;\checkmark \]

The current source absorbs 30 W and the 10 V source absorbs 10 W; the 30 V source supplies all 90 W.

The lesson that carries into mesh analysis. A current source in a shared branch is what makes plain mesh analysis fail, because the source's voltage is unknown and cannot be written in terms of mesh currents. The remedy in Set 5 is the supermesh: write KVL round a path that avoids the source entirely, and recover the missing equation from the source's own value. What you did here by introducing \(V_{cs}\) is the same thing, done the long way.

Answer\(I_1 = 3\ \text{A},\ I_2 = 1\ \text{A},\ V_{cs} = 15\ \text{V}\)

Problem 20ChallengePower from Terminal Quantities

A three-terminal box has terminal currents (all defined as entering the box) \(I_1 = 4\ \text{A}\) and \(I_2 = -6\ \text{A}\), and terminal potentials measured against an external reference of \(V_1 = 10\ \text{V}\), \(V_2 = 4\ \text{V}\) and \(V_3 = 0\ \text{V}\). Find \(I_3\) and the power absorbed by the box, and prove that the answer does not depend on which reference was chosen.

Solution

Generalised KCL over a surface enclosing the box gives the third current at once:

\[ I_1 + I_2 + I_3 = 0 \;\Longrightarrow\; I_3 = -4 + 6 = 2\ \text{A} \]

Each terminal current enters at the potential of that terminal, so the total power absorbed is the sum of the products:

\[ P = \sum_k V_k I_k = (10)(4) + (4)(-6) + (0)(2) = 40 - 24 + 0 = 16\ \text{W} \]

Reference independence. Suppose every potential is shifted by an arbitrary constant \(V_0\), as happens whenever the reference node is moved. The new power is

\[ P' = \sum_k (V_k + V_0)I_k = \sum_k V_kI_k + V_0\sum_k I_k \]

But the second sum is zero, by the KCL statement already used. Therefore

\[ P' = P \]

Verify numerically with \(V_0 = 5\ \text{V}\): \((15)(4) + (9)(-6) + (5)(2) = 60 - 54 + 10 = 16\ \text{W}\;\checkmark\)

Two laws, one conclusion. The power a box absorbs is a physical fact and cannot depend on where an engineer chose to put the ground symbol — but that invariance is not obvious from the formula \(P = \sum V_kI_k\), which is full of reference-dependent potentials. It survives only because KCL forces the terminal currents to sum to zero. This is the same argument, in miniature, that proves Tellegen's theorem for an entire network, which you will meet in Set 14.

Answer\(I_3 = 2\ \text{A}\); box absorbs 16 W, independent of reference

Practice

Practice Problems

Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not.

P1. Five branches meet at a node. Four carry 3 A in, 7 A in, 2 A out and 6 A out. Find the fifth current and its direction.

Show answer
In = 10 A, out = 8 A, so the fifth carries 2 A out of the node.
P2. A loop contains a 15 V source, a 2 Ω, a 3 Ω and a 9 V source in opposition. Find the current and state which source absorbs.

Show answer
\(15 - 9 = 5I\), so \(I = 1.2\) A. The 9 V source absorbs 10.8 W; the 15 V supplies 18 W; resistors take 7.2 W.
P3. A node at potential \(V\) is fed by a 4 A source and drains through a 2 Ω and a 3 Ω resistor to ground. Find \(V\) and both branch currents.

Show answer
\(R_p = 1.2\ \Omega\), \(V = 4.8\) V, currents 2.4 A and 1.6 A.
P4. A single loop contains a 24 V source, a 4 Ω resistor, and a current-controlled voltage source of value \(4I\) in opposition. Find \(I\).

Show answer
\(24 - 4I - 4I = 0\), so \(I = 3\) A. The CCVS behaves exactly like an extra 4 Ω of resistance.
P5. A circuit has 8 branches and 5 nodes. How many independent KCL and KVL equations does it provide, and does that suffice to find all branch currents?

Show answer
KCL: \(n - 1 = 4\). KVL: \(b - n + 1 = 4\). Total 8, exactly matching the 8 unknown branch currents — yes.
P6. Two 12 V batteries of internal resistance 0.3 Ω and 0.2 Ω are connected in parallel across a 5 Ω load. Find the load current.

Show answer
The two sources are identical in emf, so no circulating current flows. Combined internal resistance \(0.3\,\|\,0.2 = 0.12\ \Omega\), giving \(I = 12/5.12 = 2.34\) A.
P7. In a two-loop network the branch currents come out as \(I_1 = 5\) A, \(I_2 = -2\) A. What is the current in the shared branch, if \(I_3 = I_1 + I_2\)?

Show answer
\(I_3 = 3\) A. Note that \(I_2\) being negative reduces the shared-branch current rather than adding to it.
P8. A node has a 5 A source entering and a dependent source of value \(2i_1\) leaving, plus a 4 Ω resistor to ground carrying \(i_1\). Find the node voltage.

Show answer
KCL: \(5 = i_1 + 2i_1 = 3i_1\), so \(i_1 = 1.667\) A and \(V = 4i_1 = 6.67\) V.
P9. A 6 V source, a 2 Ω resistor and an open switch are in series. What current flows, and what voltage stands across the switch?

Show answer
Current is zero, so there is no drop across the resistor and the full 6 V appears across the switch.
P10. Terminal currents entering a four-terminal box are 3 A, −5 A and 6 A on three of the leads. Find the fourth.

Show answer
By generalised KCL the four must sum to zero, so \(I_4 = -4\) A — that is, 4 A leaving.
P11. Around a loop the element voltages are \(+12\), \(-5\), \(+x\) and \(-9\) V. Find \(x\).

Show answer
\(12 - 5 + x - 9 = 0\), so \(x = 2\) V.
P12. A 40 V source with a 4 Ω resistor and a 20 V source with a 2 Ω resistor share an 8 Ω branch. Find the three branch currents.

Show answer
\(40 = 4I_1 + 8I_3\), \(20 = 2I_2 + 8I_3\), \(I_3 = I_1+I_2\). Solving: \(I_1 = 3.33\) A, \(I_2 = -0.67\) A, \(I_3 = 2.67\) A — the 20 V source is being charged.

Challenge

Challenge Problems

Each of these needs an idea rather than a formula. Decide what the governing principle is before opening the answer.

C1. Explain, using Kirchhoff's laws alone, why (a) two ideal voltage sources of different value must never be connected in parallel, and (b) two ideal current sources of different value must never be connected in series. What actually happens if you try it with real components?

Show answer
(a) Two ideal voltage sources in parallel share both nodes, so KVL round the loop containing only those two sources demands \(V_1 - V_2 = 0\). If \(V_1 \ne V_2\) the equation has no solution: the circuit is inconsistent, not merely hard.
(b) Two ideal current sources in series form a node between them at which KCL demands \(I_1 = I_2\). If they differ, again no solution exists.

With real components the contradiction is resolved by the modelling assumption failing. Real voltage sources have internal resistance, so a circulating current \((V_1-V_2)/(r_1+r_2)\) flows — very large if the resistances are small, which is why paralleling mismatched batteries is dangerous. Real current sources have finite output resistance, and the node between them takes whatever voltage forces the currents to agree, often driving one source out of compliance. The lesson generalises: when a circuit model yields "no solution", the fault is in the model, and the physical outcome is found by restoring whatever non-ideality was discarded.
C2. A network contains \(b\) branches and \(n\) nodes. Show that writing KCL at every node gives one equation too many, and identify exactly what the redundant equation says.

Show answer
Write KCL at every node with currents leaving taken as positive, and add all \(n\) equations together. Every branch has two ends, at two different nodes; it appears in one equation with a \(+\) and in the other with a \(-\). Every term therefore cancels, and the sum is the identity \(0 = 0\).

So the \(n\) equations satisfy one linear relation — their sum — and only \(n-1\) are independent. The redundant equation carries no new information because it says only that charge is conserved for the network as a whole, which is already implied by conservation at each of the other nodes. This is why one node is always chosen as the reference and its equation discarded, and why nodal analysis has exactly \(n-1\) unknowns.
C3. A black box with two terminals is measured and found to satisfy \(v = 5i + 10\) volts for every value of \(i\) imposed on it. Using KVL, propose a two-element circuit inside the box that reproduces this exactly. Is the proposal unique? At what current does the box change from absorbing to supplying?

Show answer
KVL round a loop containing a 10 V source in series with a 5 Ω resistor gives exactly \(v = 5i + 10\), so a 10 V source in series with 5 Ω works.

Not unique. A 2 A current source in parallel with 5 Ω has the same terminal relation (this is the source transformation of Set 12), as does any network with the same Thévenin equivalent. Terminal measurements fix the equivalent, never the contents — the same conclusion reached in Set 2, C3.

Power crossover: the box absorbs when \(p = vi = 5i^2 + 10i > 0\). The roots are \(i = 0\) and \(i = -2\) A, so the box supplies power only for \(-2 < i < 0\) A and absorbs otherwise. The maximum delivered power, at \(i = -1\) A, is 5 W — which is \(V_{\text{oc}}^2/4R\), the maximum-power-transfer result of Set 13, arrived at here by nothing but KVL and a quadratic.

Self-Test

Multiple-Choice Questions

MCQ 1. Kirchhoff's current law is a consequence of the conservation of:
(a) energy (b) charge (c) momentum (d) power

Show answer
(b) charge. KVL is the one that follows from conservation of energy — a frequent swap in examinations.
MCQ 2. A circuit has 7 branches and 4 nodes. The number of independent KVL equations is:
(a) 3 (b) 4 (c) 6 (d) 7

Show answer
(b) 4, from \(l = b - n + 1 = 7 - 4 + 1\). Option (a) is the number of independent KCL equations, \(n-1\).
MCQ 3. A branch current is calculated as \(-3\) A. This means:
(a) the calculation is wrong (b) 3 A flows opposite to the assumed direction (c) the branch absorbs power (d) the circuit is inconsistent

Show answer
(b). The reference direction was assigned before the answer was known; a negative result simply reverses it. The sign carries information and should never be "corrected".
MCQ 4. Kirchhoff's current law applies to:
(a) nodes only (b) any closed surface (c) planar circuits only (d) resistive circuits only

Show answer
(b) any closed surface — the generalised form of Problem 14, which is what makes the supernode possible.
MCQ 5. Two ideal 12 V and 10 V sources are connected directly in parallel. The circuit:
(a) settles at 11 V (b) settles at 12 V (c) has no solution (d) settles at 22 V

Show answer
(c) has no solution. KVL round the loop containing only the two sources requires \(12 - 10 = 0\), which is false. The model is inconsistent; with real sources, internal resistance permits a large circulating current — see Challenge C1.
MCQ 6. Current enters the positive terminal of a 9 V source at 2 A. The source:
(a) supplies 18 W (b) absorbs 18 W (c) supplies 4.5 W (d) absorbs 4.5 W

Show answer
(b) absorbs 18 W. The passive sign convention is satisfied, so \(p = vi = +18\) W. The source is being charged.
MCQ 7. An open branch in a circuit:
(a) carries no current and has no voltage (b) carries no current but may have voltage (c) carries current but no voltage (d) must be removed before analysis

Show answer
(b). Zero current, but KVL still fixes the voltage across the gap — Problem 18, where 15 V stood across an open switch.
MCQ 8. Writing KCL at all \(n\) nodes of a circuit gives:
(a) \(n\) independent equations (b) \(n-1\) independent equations (c) \(n+1\) (d) it depends on the sources

Show answer
(b) \(n-1\). The \(n\) equations sum identically to zero, so one is redundant — Challenge C2.
MCQ 9. A dependent source in a circuit:
(a) always supplies power (b) always absorbs power (c) may do either (d) obeys KVL but not KCL

Show answer
(c) may do either — compare Problem 2, where the VCVS supplied 96 W, with Problem 16, where a VCVS absorbed 8.64 W. Both obey both laws.
MCQ 10. In the branch-current method applied to a circuit with \(b\) branches, the number of simultaneous equations to be solved is:
(a) \(b\) (b) \(n-1\) (c) \(b-n+1\) (d) 2

Show answer
(a) \(b\), one per unknown branch current. Mesh analysis reduces this to (c) and nodal analysis to (b), which is the entire motivation for those methods.
MCQ 11. Three currents entering a node are 5 A, −8 A and 2 A. A fourth branch carries:
(a) 1 A entering (b) 1 A leaving (c) 15 A entering (d) 5 A leaving

Show answer
(a) 1 A entering. The four must sum to zero: \(5 - 8 + 2 + I_4 = 0\), so \(I_4 = +1\) A, and positive means entering under the stated convention.
MCQ 12. The power absorbed by a multi-terminal box computed as \(\sum V_kI_k\) is independent of the reference node because:
(a) the voltages are all positive (b) the currents sum to zero (c) the box is linear (d) it is not independent

Show answer
(b) the currents sum to zero. Shifting every potential by \(V_0\) adds \(V_0\sum I_k = 0\) — Problem 20. Linearity is not required.

Reference

Key Formulas

Statement	Relation	Notes
KCL (node)	\(\sum_{k} i_k = 0\)	Conservation of charge; fix one sign convention
KCL (surface)	\(\sum_{\text{leads}} i_k = 0\)	Holds for any closed surface — basis of the supernode
KVL (loop)	\(\sum_{k} v_k = 0\)	Conservation of energy; walk the loop one way
Double subscript	\(V_{ab} = V_a - V_b = -V_{ba}\)	Every KVL sign question reduces to this
Independent KCL	\(n - 1\)	The \(n\)th equation is the sum of the others
Independent KVL	\(l = b - n + 1\)	Meshes, for a planar circuit
Total equations	\((n-1) + (b-n+1) = b\)	Exactly enough for \(b\) branch currents
Passive sign convention	\(p = vi\) if \(i\) enters \(+\)	Positive \(p\) means absorbing
Power balance	\(\sum p = 0\)	Use as a check on every solved circuit
Series branch	same \(i\), voltages add	KCL at the intermediate node forces it
Parallel branch	same \(v\), currents add	KVL round the two-element loop forces it
Dependent source	one symbol + one constraint	Write KVL/KCL first, then the defining equation
Ideal source rules	no unequal \(V\) in parallel; no unequal \(I\) in series	Violating either makes the equations inconsistent
Terminal power	\(P = \sum_k V_kI_k\)	Reference-independent because \(\sum I_k = 0\)

Diagnostics

Common Mistakes

Mixing sign conventions within one equation. Decide once whether currents leaving or entering are positive, and apply it to every term. Half the terms one way and half the other is the commonest single error in this topic.
Redrawing an arrow to make a negative answer positive. The sign is the answer to "which way does it actually flow" and, in Problem 17, to "which battery is being charged". Reversing the arrow mid-solution destroys that information.
Assuming an open branch has zero voltage. It has zero current. KVL still fixes the voltage across it — Problem 18.
Assuming a source always supplies. A source with current entering its positive terminal absorbs, as in Problems 10, 15 and 17. So can a dependent source, as in Problem 16.
Forgetting the constraint equation for a dependent source. Writing KVL or KCL with the source symbol in it leaves you one equation short. The controlling variable must be expressed in circuit quantities before you can solve.
Writing KCL at every node and expecting all of them to be independent. Only \(n-1\) are; the last is their sum. Choosing a reference node and discarding its equation is not an approximation.
Trying to write the voltage across a current source in terms of the resistors. It has no such expression — it is an independent unknown determined by the rest of the circuit, as Problem 19 shows.
Losing a factor of a thousand between mA and A, or kΩ and Ω. Work consistently in mA and kΩ so that their product is volts, as in Problem 6.
Treating two nodes joined by a plain wire as different nodes. An ideal wire is a single node however it is drawn, which is exactly the fact that unlocks Problem 6.
Skipping the power check. It takes one line and catches sign errors that the equations themselves will solve happily and wrongly.

Looking Ahead

Everything in this set has been the direct application of two laws, one equation at a time. The method is universal but wasteful: Problem 12 needed three unknowns where the topology only demanded two, and a circuit with a dozen branches would need a dozen simultaneous equations written by hand.

The next four sets are organised assaults on that waste. Mesh analysis chooses loop currents that satisfy KCL automatically, so only KVL need be written — \(b-n+1\) equations. Nodal analysis chooses node voltages that satisfy KVL automatically, so only KCL need be written — \(n-1\) equations. Neither adds any new physics; both are bookkeeping schemes built on exactly the two laws you have just used. Where a source sits awkwardly, as the current source did in Problem 19, the supermesh and supernode extensions handle it.