Apply Kirchhoff's Current Law (KCL): Exercises — Hardware deep dive

Level 1 — Recognition

Exercise 1.1

Three wires meet at node A. Two carry current into A: $i_{1} = 4 A$ and $i_{2} = 5 A$ . One wire carries $i_{3}$ out. What is $i_{3}$ ?

Recall Solution 1.1

WHAT we do: write "in equals out." WHY: no charge accumulates at A, so every ampere that enters must leave. $i_{1} + i_{2} = i_{3}$ $4 + 5 = i_{3} \Rightarrow i_{3} = 9 A (out) .$

Exercise 1.2

Look at the figure below. Four labelled branches touch node N. State which of them are on the same node as N and which are truly separate nodes.

Figure — Apply Kirchhoff's Current Law (KCL)

Recall Solution 1.2

WHAT we do: collapse every point joined by pure wire into one node. WHY: an ideal wire has zero resistance, so there is no voltage drop across it — every point on it is electrically the same point.

The red highlighted stretch (top wire, both bends, up to the junction dot) is all one node N. Branches $a$ and $b$ attach to it.
The resistor $R$ sits between two different regions, so the wire on its far side is a separate node M.
Branches $c$ and $d$ join at M, not N. So node N has branches ${a, b, and the top of R}$ ; node M has ${c, d, and the bottom of R}$ .

Level 2 — Application

Exercise 2.1

At node B: $i_{1} = 7 A$ in, $i_{2} = 3 A$ in, $i_{3} = 4 A$ out, and unknown $i_{4}$ . Find $i_{4}$ and state its direction.

Recall Solution 2.1

WHAT: sum of ins = sum of outs. $i_{1} + i_{2} = i_{3} + i_{4}$ $7 + 3 = 4 + i_{4} \Rightarrow i_{4} = 6 A .$ WHY the sign is positive-out: we assumed $i_{4}$ leaves. $10$ A entered, only $4$ A has left via $i_{3}$ , so the remaining $6$ A must leave through branch 4. Direction confirmed: out.

Exercise 2.2

A $12 A$ current source pushes current into node D. It splits into three resistor branches carrying $i_{1} = 5 A$ , $i_{2} = 4 A$ , and $i_{3}$ . Find $i_{3}$ .

Recall Solution 2.2

WHAT: the source current is fully distributed among the exits — none is lost. $12 = i_{1} + i_{2} + i_{3}$ $12 = 5 + 4 + i_{3} \Rightarrow i_{3} = 3 A (out) .$ WHY: a node has no storage; the $12$ A that enters must be exactly accounted for by what leaves.

Exercise 2.3

At node E you guess that $i_{x}$ flows into the node. Known: $6 A$ enters, $10 A$ leaves. Using "leaving $= +$ ", write $\sum i = 0$ and solve for $i_{x}$ . Interpret the sign.

Recall Solution 2.3

WHAT: write every branch with its sign. Entering currents are negative, leaving are positive; our guessed $i_{x}$ enters, so it enters the sum as $- i_{x}$ . $- 6 - i_{x} + 10 = 0$ Solve: $i_{x} = 4 A .$ WHY positive is fine: $i_{x} = + 4$ with our "in" arrow means it really does flow into E at $4$ A. If it had come out negative, we would keep the magnitude and flip the arrow — the algebra self-corrects a wrong guess.

Level 3 — Analysis

Exercise 3.1

Currents at node F are given in terms of an unknown $I$ :

$i_{1} = 2 I$ enters,
$i_{2} = 8 A$ enters,
$i_{3} = 3 I$ leaves,
$i_{4} = 5 A$ leaves.

Find $I$ , then compute each branch current.

Recall Solution 3.1

WHAT: balance ins against outs; this becomes one linear equation in $I$ . $in 2 I + 8 = out 3 I + 5$ WHY solve for $I$ : KCL gives exactly one constraint, and there is exactly one unknown, so it pins $I$ down. $8 - 5 = 3 I - 2 I \Rightarrow 3 = I \Rightarrow I = 3 A .$ Then $i_{1} = 2 I = 6 A$ (in), $i_{3} = 3 I = 9 A$ (out). Check: in $= 6 + 8 = 14$ , out $= 9 + 5 = 14$ . ✓

Exercise 3.2 — Two coupled nodes

Look at the figure. A $10 A$ source feeds node P. From P, a branch $i_{a}$ goes to node Q, and a branch $i_{b} = 4 A$ leaves P to ground. At node Q, the incoming $i_{a}$ splits into $i_{c} = 2 A$ and $i_{d}$ (both leave Q). Find $i_{a}$ and $i_{d}$ .

Recall Solution 3.2

WHAT: write KCL at each node separately, then chain them. WHY two equations: two nodes, two unknowns ( $i_{a}$ , $i_{d}$ ) — one KCL equation per node gives exactly enough.

Node P (in = out): $10 = i_{a} + i_{b} = i_{a} + 4 \Rightarrow i_{a} = 6 A .$ Node Q (in = out): the current arriving from P is $i_{a}$ , which splits: $i_{a} = i_{c} + i_{d} \Rightarrow 6 = 2 + i_{d} \Rightarrow i_{d} = 4 A .$ WHAT IT LOOKS LIKE: trace the red path in the figure — $6$ A leaves P toward Q along the accent branch; at Q it forks into $2$ A and $4$ A. Nothing is stored anywhere along the way. This is the seed of Nodal Analysis: one KCL per node, solved as a system.

Level 4 — Synthesis

Exercise 4.1 — Bridge junction with symbolic branches

At node G, four branches carry currents defined by directions:

$i_{1} = 9 A$ in,
$i_{2}$ out,
$i_{3} = 2 A$ in,
$i_{4}$ out,

with the extra physical constraint that branch 4 carries twice the current of branch 2: $i_{4} = 2 i_{2}$ . Find $i_{2}$ and $i_{4}$ .

Recall Solution 4.1

WHAT: KCL gives one equation; the constraint $i_{4} = 2 i_{2}$ gives a second. Two equations, two unknowns. WHY we need the constraint: KCL alone ( $9 + 2 = i_{2} + i_{4}$ ) is one equation in two unknowns — underdetermined. The extra physical relation closes the system.

KCL (in = out): $9 + 2 = i_{2} + i_{4} .$ Substitute $i_{4} = 2 i_{2}$ : $11 = i_{2} + 2 i_{2} = 3 i_{2} \Rightarrow i_{2} = \frac{11}{3} A \approx 3.67 A .$ Then $i_{4} = 2 i_{2} = \frac{22}{3} A \approx 7.33 A .$ Check: out $= \frac{11}{3} + \frac{22}{3} = \frac{33}{3} = 11 =$ in. ✓

Exercise 4.2 — Capacitor at a node (time-varying)

A node H has: a steady $5 A$ flowing in, a resistor branch taking $3 A$ out, and a capacitor branch. The capacitor has $C = 2 F$ and its voltage at this instant is changing at $\frac{d v}{d t} = 1.5 V/s$ . The capacitor current (leaving the node into the capacitor) is $i_{C} = C \frac{d v}{d t}$ . Verify KCL holds at H by finding the required source current if instead the source is unknown, given the other two branches. First compute $i_{C}$ , then find the source current $i_{S}$ entering H.

Recall Solution 4.2

WHAT: a capacitor does not break KCL. Its terminal current $i_{C} = C \frac{d v}{d t}$ enters one plate and an equal current leaves the other; at the node it is just another branch (see Capacitor i-v Relationship). WHY the formula $i_{C} = C d v / d t$ : current is charge-per-second, $i = d q / d t$ ; on a capacitor the stored charge is $q = C v$ , so $i_{C} = \frac{d ( C v )}{d t} = C \frac{d v}{d t}$ . This is where the derivative enters — we need the rate of voltage change, because a constant voltage stores constant charge and pushes no current.

Step 1 — capacitor current: $i_{C} = C \frac{d v}{d t} = 2 \times 1.5 = 3 A (leaving H into the capacitor) .$ Step 2 — KCL at H (in = out): $i_{S} = resistor 3 + i_{C} 3 = 6 A .$ So the source must supply $i_{S} = 6 A$ into H. Charge stores on the plates inside the component, not at the node — the node still balances instant by instant.

Level 5 — Mastery

Exercise 5.1 — Three-node system, fully symbolic

A circuit has three nodes X, Y, Z (plus ground). A $14 A$ source enters X. Branches:

X → Y carries $i_{1}$ ,
X → ground carries $i_{2} = 3 A$ ,
Y → Z carries $i_{3}$ ,
Y → ground carries $i_{4} = 5 A$ ,
Z → ground carries $i_{5} = 4 A$ ,
Z also has a branch back to ground $i_{6}$ .

Find $i_{1}$ , $i_{3}$ , and $i_{6}$ by applying KCL node-by-node.

Recall Solution 5.1

WHAT: write KCL (in = out) at each of X, Y, Z in order. Each node has exactly one unknown once we solve the previous. WHY node-by-node works here: the branch currents chain forward — solving X gives $i_{1}$ , which feeds Y, and so on. This is exactly the Nodal Analysis philosophy of one balance per node.

Node X: in $= 14$ ; out $= i_{1} + i_{2}$ . $14 = i_{1} + 3 \Rightarrow i_{1} = 11 A .$ Node Y: in $= i_{1} = 11$ ; out $= i_{3} + i_{4}$ . $11 = i_{3} + 5 \Rightarrow i_{3} = 6 A .$ Node Z: in $= i_{3} = 6$ ; out $= i_{5} + i_{6}$ . $6 = 4 + i_{6} \Rightarrow i_{6} = 2 A .$ Global check (conservation over the whole circuit): all $14$ A from the source must exit to ground. Ground receives $i_{2} + i_{4} + i_{5} + i_{6} = 3 + 5 + 4 + 2 = 14 A$ . ✓ Everything that entered has left.

Exercise 5.2 — Degenerate / zero case

At node W: two branches enter with $i_{1}$ and $i_{2}$ , one branch leaves with $i_{3} = 0 A$ (an open, broken wire — no current can flow), and one more leaves with $i_{4}$ . If $i_{1} = 6 A$ enters and $i_{2} = 6 A$ leaves (its arrow points out even though named as "enter" — trust the direction: it leaves), what is $i_{4}$ ? Handle the zero branch carefully.

Recall Solution 5.2

WHAT: identify true directions first, then apply KCL. A broken (open) branch forces its current to exactly zero — an open circuit carries no current because charge has nowhere to go. WHY $i_{3} = 0$ matters: it does not mean "ignore the node." It means that branch contributes $0$ to the balance; the other branches must still sum correctly.

True picture: in = $i_{1} = 6$ A. out = $i_{2}$ (6 A leaves) $+ i_{3}$ (0) $+ i_{4}$ . $in 6 = i_{2} 6 + i_{3} 0 + i_{4} \Rightarrow i_{4} = 0 A .$ Interpretation: the $6$ A entering is entirely carried away by $i_{2}$ ; with the open branch dead ( $i_{3} = 0$ ), branch 4 also carries zero. A degenerate but perfectly valid KCL result — the law holds even when currents vanish.

Connections

Parent: Apply KCL — the theory these exercises drill.
Nodal Analysis — Exercises 3.2 and 5.1 are its foundation: one KCL per node.
Conservation of Charge — why every balance above must close.
Current and Current Density — the meaning of $i = d q / d t$ used throughout.
Capacitor i-v Relationship — the tool for Exercise 4.2.
Ideal Wires and Nodes — the node-identification skill of Exercise 1.2.
Kirchhoff's Voltage Law (KVL) — the loop-based partner law.