⚡ Kirchhoff's Junction Rule (KCL)

Part 1 of 7 — Conservation of Charge at Nodes

When wires meet at a junction (node), charge cannot pile up or vanish. Every coulomb that flows in must flow out. This simple conservation law — Kirchhoff's Current Law (KCL) — lets us analyze circuits that series/parallel rules alone cannot handle.

The Junction Rule

At any junction (node) in a circuit:

$\sum I_{\text{in}} = \sum I_{\text{out}}$

Equivalently, if we assign signs (positive for currents entering, negative for currents leaving):

$\sum_{k} I_k = 0$

Why It Works

The junction rule is a direct consequence of conservation of charge. In steady-state (DC) circuits, charge does not accumulate at any point. If 5 A flows into a node, exactly 5 A must flow out — otherwise charge would build up at the junction, which doesn't happen in steady state.

Key Vocabulary

• Node / Junction: A point where three or more wires meet
• Branch: A path between two nodes containing one or more circuit elements
• KCL: Kirchhoff's Current Law (the junction rule)

Example: Three-Branch Junction

Consider a node where three wires meet:

• Branch 1 carries $I_1 = 3$ A into the node
• Branch 2 carries $I_2 = 5$ A into the node
• Branch 3 carries $I_3$ out of the node

Applying KCL:

$I_1 + I_2 = I_3$ $3 + 5 = I_3$ $I_3 = 8 \text{ A}$

Four-Branch Example

Now add a fourth branch:

• $I_1 = 6$ A in
• $I_2 = 2$ A out
• $I_3 = 1$ A out
• $I_4 = ?$

$I_{\text{in}} = I_{\text{out}}$ $6 = 2 + 1 + I_4$ $I_4 = 3 \text{ A (out)}$

💡 Tip: If you guess a current direction wrong when solving a problem, you'll get a negative answer — that just means the current flows opposite to your assumed direction!

Junction Rule Concept Check

Multiple Junctions in a Circuit

Real circuits have many junctions. KCL applies at each one independently.

Example: Two-Node Circuit

Consider a circuit with two nodes (A and B) and three branches:

Node A: $I_1 = I_2 + I_3$

Node B: $I_2 + I_3 = I_1$

Notice that the equation at node B gives the same information as node A — this is always the case. For a circuit with $N$ nodes, KCL gives only $N - 1$ independent equations.

Practical Rule

📝 In a circuit with $N$ nodes, you get $N - 1$ independent KCL equations.

This is important when setting up systems of equations for complex circuits — you'll need additional equations from the loop rule (Part 2) to solve for all unknowns.

Junction Rule Drill

A circuit node has five branches connected to it. The currents are:

• Branch 1: 8 A into the node
• Branch 2: 3 A out of the node
• Branch 3: $I_3$ into the node
• Branch 4: 6 A out of the node
• Branch 5: 4 A out of the node

1. Total current flowing out of the node (in A, not counting $I_3$):
2. Value of $I_3$ (in A):
3. A different node has currents 10 A in, 4 A in, $I_a$ out, and $I_b$ out. If $I_a = 3I_b$, find $I_b$ (in A):

Round all answers to 3 significant figures.

Exit Quiz

🔄 Kirchhoff's Loop Rule (KVL)

Part 2 of 7 — Conservation of Energy Around Loops

The second of Kirchhoff's laws says that the total voltage change around any closed loop is zero. This is conservation of energy applied to circuits — a charge that travels around a complete loop returns to its starting potential.

The Loop Rule

Around any closed loop in a circuit:

$\sum \Delta V = 0$

Or equivalently: the sum of all EMFs equals the sum of all voltage drops.

$\sum \varepsilon = \sum IR$

Why It Works

The loop rule is conservation of energy. Electric potential is like height — if you walk around a closed path and return to your starting point, your net change in height is zero. Similarly, a charge traversing a closed loop gains energy through batteries and loses it through resistors, but the net energy change is zero.

Sign Conventions

When traversing a loop in a chosen direction:

🔑 Memory Aid: Going through a battery from − to + is like climbing a hill (gain energy, $+\varepsilon$). Going through a resistor in the current direction is like sliding down (lose energy, $-IR$).

Example: Single-Battery Loop

A 12 V battery is connected to a 3 Ω and a 5 Ω resistor in series. Find the current.

Step 1: Choose a loop direction (clockwise).

Step 2: Assign current direction (clockwise, same as loop).

Step 3: Apply KVL starting from the battery's negative terminal:

$+\varepsilon - IR_1 - IR_2 = 0$ $+12 - I(3) - I(5) = 0$ $12 = 8I$ $I = 1.5 \text{ A}$

Step 4: Verify — Voltage drops: $V_1 = 1.5 \times 3 = 4.5$ V, $V_2 = 1.5 \times 5 = 7.5$ V. Total: $4.5 + 7.5 = 12$ V ✓

Checking Voltage at Each Point

Starting at the battery's negative terminal (call it 0 V):

• After battery: $0 + 12 = 12$ V
• After $R_1$: $12 - 4.5 = 7.5$ V
• After $R_2$: $7.5 - 7.5 = 0$ V ✓

We return to 0 V — the loop rule is satisfied!

Sign Convention Quiz

Systematic Approach

Follow these steps for every KVL problem:

The 4-Step Method

1. Label currents: Assign a direction to each unknown current (guess if needed — a negative answer just means it flows the other way)

2. Choose loops: Identify independent loops that cover every branch

3. Pick a traversal direction: Usually clockwise for each loop

4. Write KVL: Walk around each loop, summing voltage changes using the sign conventions:

   • Battery − to +: $+\varepsilon$
   • Battery + to −: $-\varepsilon$
   • Resistor with current: $-IR$
   • Resistor against current: $+IR$

How Many Loops?

For a circuit with $B$ branches and $N$ nodes:

$\text{Independent loops} = B - N + 1$

This is the number of KVL equations you can write. Combined with $N - 1$ KCL equations, you get $B$ equations total — exactly enough to solve for all $B$ unknown currents!

Loop Rule Practice

Exit Quiz

🔋 Single-Loop Circuits with Multiple Batteries

Part 3 of 7 — Opposing EMFs and Finding Current Direction

What happens when a circuit has two or more batteries that may oppose each other? The loop rule handles this elegantly — just be careful with signs!

Opposing EMFs

When two batteries face each other in a single loop, their EMFs partially cancel. The net EMF drives the current.

Example: Two Batteries in Opposition

A loop contains:

• Battery 1: $\varepsilon_1 = 12$ V
• Battery 2: $\varepsilon_2 = 5$ V (opposing $\varepsilon_1$)
• Resistor: $R = 7\;\Omega$

Step 1: Assume current flows clockwise (driven by the larger battery).

Step 2: Traverse clockwise, starting from the negative terminal of $\varepsilon_1$:

$+\varepsilon_1 - \varepsilon_2 - IR = 0$ $12 - 5 - 7I = 0$ $I = 1 \text{ A (clockwise)}$

The net EMF is $12 - 5 = 7$ V, and the current flows in the direction the larger battery "wants."

What If You Guess Wrong?

Suppose you assumed current flows counterclockwise:

$+\varepsilon_2 - \varepsilon_1 - IR = 0$ $5 - 12 - 7I = 0$ $I = -1 \text{ A}$

The negative sign tells you: the current is actually clockwise (opposite to your assumption). The magnitude is the same!

Three Batteries in a Loop

Consider a single loop with three batteries and three resistors:

• $\varepsilon_1 = 20$ V, $\varepsilon_2 = 8$ V (opposing), $\varepsilon_3 = 4$ V (aiding $\varepsilon_1$)
• $R_1 = 3\;\Omega$, $R_2 = 5\;\Omega$, $R_3 = 2\;\Omega$

Assume current clockwise. Traverse clockwise:

$+\varepsilon_1 - \varepsilon_2 + \varepsilon_3 - IR_1 - IR_2 - IR_3 = 0$

$20 - 8 + 4 = I(3 + 5 + 2)$

$16 = 10I$

$I = 1.6 \text{ A (clockwise)}$

Voltage Drops

• $V_{R_1} = 1.6 \times 3 = 4.8$ V
• $V_{R_2} = 1.6 \times 5 = 8.0$ V
• $V_{R_3} = 1.6 \times 2 = 3.2$ V
• Total drops: $4.8 + 8.0 + 3.2 = 16.0$ V = Net EMF ✓

Concept Check

Single-Loop Drill

A single loop contains (going clockwise):

• Battery 1: $\varepsilon_1 = 24$ V (+ terminal at top, drives current clockwise)
• Resistor: $R_1 = 4\;\Omega$
• Battery 2: $\varepsilon_2 = 6$ V (opposes $\varepsilon_1$, drives current counterclockwise)
• Resistor: $R_2 = 8\;\Omega$
• Resistor: $R_3 = 6\;\Omega$

1. Net EMF around the loop (in V):
2. Total resistance (in Ω):
3. Current in the loop (in A):
4. Voltage across $R_2$ (in V):

Challenge: Three-Battery Loop

A single loop contains:

• $\varepsilon_1 = 16$ V (drives clockwise)
• $R_1 = 2\;\Omega$
• $\varepsilon_2 = 10$ V (drives counterclockwise)
• $R_2 = 3\;\Omega$
• $\varepsilon_3 = 6$ V (drives clockwise)
• $R_3 = 5\;\Omega$

1. Net EMF with clockwise positive (in V):
2. Current magnitude (in A):
3. Voltage across $R_3$ (in V):

Round all answers to 3 significant figures.

Exit Quiz

🔗 Multi-Loop Circuits

Part 4 of 7 — Simultaneous Equations for Two-Loop Problems

When a circuit has multiple loops that share branches, a single KVL equation isn't enough. You need to combine KCL (junction rule) and KVL (loop rule) to build a system of simultaneous equations.

Systematic Approach for Multi-Loop Circuits

Step-by-Step Method

1. Identify nodes and branches: Count junctions and separate paths

2. Assign current variables: One variable per branch, with an assumed direction (arrow)

3. Write KCL equations: At $N - 1$ nodes (where $N$ is the number of nodes)

4. Write KVL equations: For $B - N + 1$ independent loops

5. Solve the system: Substitution or elimination

Counting Check

Total unknowns = number of branches $B$

Total equations = $(N-1) + (B - N + 1) = B$ ✓

You always have exactly enough equations!

Classic Two-Loop Problem

Consider a circuit with:

• Left loop: $\varepsilon_1 = 10$ V, $R_1 = 2\;\Omega$
• Right loop: $\varepsilon_2 = 6$ V, $R_3 = 4\;\Omega$
• Shared middle branch: $R_2 = 3\;\Omega$

Step 1: Label Currents

• $I_1$: through $\varepsilon_1$ and $R_1$ (left branch, downward)
• $I_2$: through $R_2$ (middle branch, downward)
• $I_3$: through $\varepsilon_2$ and $R_3$ (right branch, downward)

Step 2: KCL at top node

$I_1 = I_2 + I_3 \quad \text{...(1)}$

Step 3: KVL — Left Loop (clockwise)

$+\varepsilon_1 - I_1 R_1 - I_2 R_2 = 0$ $10 - 2I_1 - 3I_2 = 0 \quad \text{...(2)}$

Step 4: KVL — Right Loop (clockwise)

$+I_2 R_2 - I_3 R_3 - \varepsilon_2 = 0$ $3I_2 - 4I_3 - 6 = 0 \quad \text{...(3)}$

Step 5: Solve

Substitute (1) into (2): $10 - 2(I_2 + I_3) - 3I_2 = 0 \Rightarrow 10 - 5I_2 - 2I_3 = 0$ ...(2')

From (3): $3I_2 - 4I_3 = 6$ ...(3)

From (2'): $5I_2 + 2I_3 = 10$ ...(2')

Multiply (2') by 2: $10I_2 + 4I_3 = 20$

Add to (3): $13I_2 = 26 \Rightarrow I_2 = 2$ A

From (3): $3(2) - 4I_3 = 6 \Rightarrow I_3 = 0$ A

From (1): $I_1 = 2 + 0 = 2$ A

Interpretation

• $I_1 = 2$ A (left branch)
• $I_2 = 2$ A (middle branch)
• $I_3 = 0$ A (right branch — no current through the right battery!)

Multi-Loop Concept Check

Two-Loop Problem Drill

A two-loop circuit has:

• Left loop: $\varepsilon_1 = 12$ V battery, $R_1 = 2\;\Omega$ resistor
• Shared middle branch: $R_2 = 4\;\Omega$ resistor
• Right loop: $\varepsilon_2 = 8$ V battery, $R_3 = 4\;\Omega$ resistor

Currents: $I_1$ (left branch, down), $I_2$ (middle, down), $I_3$ (right branch, down).

KCL at top node: $I_1 = I_2 + I_3$

KVL Left (clockwise): $12 - 2I_1 - 4I_2 = 0$

KVL Right (clockwise): $4I_2 - 4I_3 - 8 = 0$

Solve:

1. $I_2$ (in A):
2. $I_3$ (in A):
3. $I_1$ (in A):

Two-Loop Problem #2

Two-loop circuit:

• Left loop: $\varepsilon_1 = 14$ V, $R_1 = 4\;\Omega$
• Shared branch: $R_2 = 6\;\Omega$
• Right loop: $\varepsilon_2 = 4$ V, $R_3 = 2\;\Omega$

KCL: $I_1 = I_2 + I_3$

Left loop (clockwise): $14 - 4I_1 - 6I_2 = 0$

Right loop (clockwise): $6I_2 - 2I_3 - 4 = 0$

1. Solve for $I_1$ (in A):
2. Solve for $I_2$ (in A):
3. Solve for $I_3$ (in A):

Exit Quiz

🧮 Complex Circuits & Matrix Methods

Part 5 of 7 — Three-Loop Problems and Systematic Solutions

When circuits get complex, organization becomes essential. In this part we tackle three-loop circuits and preview how linear algebra (matrices) can streamline the solution process.

Systematic Labeling

For complex circuits, adopt a consistent labeling scheme:

Node Labeling

• Label every junction with a letter: A, B, C, D, ...
• Mark every branch with a current arrow: $I_1, I_2, I_3, ...$

Branch Counting

A circuit with $N$ nodes and $L$ independent loops has: $B = L + N - 1 \text{ branches}$

Example: Three-Loop Circuit

• 4 nodes (A, B, C, D)
• 6 branches ($I_1$ through $I_6$)
• Independent loops: $6 - 4 + 1 = 3$
• KCL equations: $4 - 1 = 3$
• Total equations: $3 + 3 = 6$ ✓

📝 Always verify: number of equations = number of unknown currents before solving!

Three-Loop Example

Consider a circuit with three loops sharing branches:

Given:

• $\varepsilon_1 = 12$ V, $\varepsilon_2 = 6$ V, $\varepsilon_3 = 9$ V
• $R_1 = 2\;\Omega$, $R_2 = 4\;\Omega$, $R_3 = 3\;\Omega$
• $R_4 = 6\;\Omega$, $R_5 = 1\;\Omega$

Using the mesh current method (each loop gets its own current variable):

• Loop 1 current: $i_1$ (clockwise)
• Loop 2 current: $i_2$ (clockwise)
• Loop 3 current: $i_3$ (clockwise)

Mesh Equations

Loop 1: $\varepsilon_1 - i_1 R_1 - (i_1 - i_2)R_2 = 0$ $12 - 2i_1 - 4(i_1 - i_2) = 0$ $12 - 6i_1 + 4i_2 = 0$ $6i_1 - 4i_2 = 12 \quad \text{...(1)}$

Loop 2: $-(i_2 - i_1)R_2 - i_2 R_3 - (i_2 - i_3)R_4 + \varepsilon_2 = 0$ $4i_1 - 13i_2 + 6i_3 = -6 \quad \text{...(2)}$

Loop 3: $-(i_3 - i_2)R_4 - i_3 R_5 - \varepsilon_3 = 0$ $6i_2 - 7i_3 = 9 \quad \text{...(3)}$

Matrix Form

The three mesh equations can be written as a matrix equation $\mathbf{A}\vec{i} = \vec{b}$:

$\begin{pmatrix} 6 & -4 & 0 \\ 4 & -13 & 6 \\ 0 & 6 & -7 \end{pmatrix} \begin{pmatrix} i_1 \\ i_2 \\ i_3 \end{pmatrix} = \begin{pmatrix} 12 \\ -6 \\ 9 \end{pmatrix}$

Cramer's Rule Preview

For a $3 \times 3$ system, each variable can be found using determinants:

$i_k = \frac{\det(\mathbf{A}_k)}{\det(\mathbf{A})}$

where $\mathbf{A}_k$ is the matrix $\mathbf{A}$ with column $k$ replaced by $\vec{b}$.

Determinant of the Coefficient Matrix

$\det(\mathbf{A}) = 6[(-13)(-7) - (6)(6)] - (-4)[(4)(-7) - (6)(0)] + 0$ $= 6[91 - 36] + 4[-28]$ $= 6(55) - 112$ $= 330 - 112 = 218$

💡 On the AP exam, you won't need to compute $3 \times 3$ determinants — but you WILL need to set up the equations correctly and solve $2 \times 2$ systems.

Concept Check

Mesh Current Drill

Two mesh currents $i_1$ and $i_2$ (both clockwise) satisfy:

$5i_1 - 3i_2 = 9 \quad \text{...(1)}$ $-3i_1 + 7i_2 = 5 \quad \text{...(2)}$

1. Multiply equation (1) by 7 and equation (2) by 3, then add. What is $i_1$ (in A)?
2. Substitute back to find $i_2$ (in A):
3. The current through the shared resistor is $i_1 - i_2$. Find this value (in A):

Exit Quiz

🔌 RC Circuits Basics

Part 6 of 7 — Charging, Discharging, and the Time Constant

So far we've analyzed circuits in steady state (DC). But what happens when you flip a switch and a capacitor begins charging or discharging? The currents and voltages change with time — and Kirchhoff's laws still apply at every instant!

Charging a Capacitor

A battery ($\varepsilon$), resistor ($R$), and initially uncharged capacitor ($C$) are connected in series. At $t = 0$ the switch closes.

Applying KVL at Any Instant

$\varepsilon - V_R - V_C = 0$ $\varepsilon - IR - \frac{q}{C} = 0$

Since $I = \frac{dq}{dt}$, this becomes a differential equation whose solution is:

Charge on the Capacitor

$q(t) = C\varepsilon\left(1 - e^{-t/RC}\right)$

Voltage Across the Capacitor

$V_C(t) = \varepsilon\left(1 - e^{-t/RC}\right)$

Current in the Circuit

$I(t) = \frac{\varepsilon}{R}\,e^{-t/RC}$

Key Behaviors

The Time Constant $\tau = RC$

The time constant sets the timescale for charging and discharging:

$\tau = RC$

• Units: $\Omega \cdot \text{F} = \text{seconds}$
• After $1\tau$: capacitor is 63.2% charged
• After $2\tau$: 86.5% charged
• After $3\tau$: 95.0% charged
• After $5\tau$: 99.3% charged (effectively "fully charged")

Physical Interpretation

• Large $R$: Current is small → charging is slow → large $\tau$
• Large $C$: More charge to store → takes longer → large $\tau$
• Small $RC$: Fast charging/discharging

Example

$R = 10\;\text{k}\Omega = 10{,}000\;\Omega$, $C = 50\;\mu\text{F} = 50 \times 10^{-6}\;\text{F}$

$\tau = RC = (10{,}000)(50 \times 10^{-6}) = 0.5 \text{ s}$

The capacitor is effectively fully charged after about $5\tau = 2.5$ s.

Discharging a Capacitor

A capacitor initially charged to voltage $V_0$ discharges through a resistor $R$ (no battery in the loop).

KVL for Discharging

$V_C - IR = 0 \quad \text{(capacitor acts as the source)}$

Solutions

$V_C(t) = V_0\,e^{-t/RC}$

$I(t) = \frac{V_0}{R}\,e^{-t/RC}$

$q(t) = CV_0\,e^{-t/RC}$

Key Behaviors

Charging vs. Discharging Summary

💡 Both processes have current that starts at a maximum and decays exponentially with the same time constant $\tau = RC$.

RC Circuit Concept Quiz

RC Circuit Calculations

A $20\;\text{V}$ battery charges a capacitor ($C = 100\;\mu\text{F}$) through a resistor ($R = 50\;\text{k}\Omega$).

1. Time constant $\tau$ (in seconds):
2. Initial current $I_0$ at $t = 0$ (in mA):
3. Voltage across the capacitor after one time constant, $V_C(\tau)$ (in V, to one decimal place):
4. After a long time, the capacitor is disconnected and discharged through a $25\;\text{k}\Omega$ resistor. New time constant (in seconds):

Exit Quiz

🏆 Synthesis & AP Review

Part 7 of 7 — Complete Circuit Analysis Strategy

You now have all the tools: KCL, KVL, single-loop analysis, multi-loop systems, and RC circuits. This final part brings everything together with a complete analysis strategy, common mistakes to avoid, and AP-style problems.

Complete Circuit Analysis Strategy

Master Checklist

1. Identify the circuit type

   • Single loop → KVL alone is sufficient
   • Multi-loop → Need both KCL and KVL
   • Contains capacitors → Consider time-dependent (RC) behavior

2. Label everything

   • Assign current variables with directions to each branch
   • Label nodes (junctions) with letters
   • Identify independent loops

3. Write equations

   • KCL at $N - 1$ nodes
   • KVL around $B - N + 1$ loops
   • Verify: total equations = total unknowns

4. Solve the system

   • For 2 unknowns: substitution or elimination
   • For 3+ unknowns: systematic elimination or matrices
   • Negative answers → current flows opposite to assumed direction

5. Check your answer

   • Do the currents satisfy KCL at every node?
   • Does KVL hold for every loop?
   • Are the signs and magnitudes physically reasonable?

Common Shortcuts

• Resistors in series: $R_{\text{eq}} = R_1 + R_2 + \cdots$
• Resistors in parallel: $\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$
• Two in parallel: $R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}$

Simplify with series/parallel first, then use Kirchhoff's laws on what remains!

Common Mistakes on the AP Exam

❌ Mistake 1: Wrong Sign Convention

Going through a battery from + to − is $-\varepsilon$, not $+\varepsilon$. Going through a resistor with the current is $-IR$.

❌ Mistake 2: Forgetting a Branch

Every branch needs a current variable. If you miss one, your system is underdetermined and you'll get infinite solutions or contradictions.

❌ Mistake 3: Inconsistent Current Directions at Nodes

Make sure the same current ($I_k$) flowing into a node in your KCL equation is the same one that appears in the KVL equation for any loop containing that branch.

❌ Mistake 4: RC Circuit Mix-ups

• Charging: $V_C = \varepsilon(1 - e^{-t/RC})$ — starts at 0, rises to $\varepsilon$
• Discharging: $V_C = V_0 e^{-t/RC}$ — starts at $V_0$, falls to 0
• Don't confuse the two!

❌ Mistake 5: Unit Errors in RC

• $R$ in ohms, $C$ in farads → $\tau$ in seconds
• $\text{k}\Omega \times \mu\text{F}$: $(10^3)(10^{-6}) = 10^{-3}$ s = ms
• Always convert to base SI units first!

✅ Pro Tip for AP FRQs

Show your work clearly: state which rule you're applying (KCL or KVL), write the equation, then solve. Graders give partial credit for correct setups even if the algebra goes wrong.

AP-Style FRQ: Multi-Loop Circuit

A circuit has two loops sharing a middle branch:

• Left loop: $\varepsilon_1 = 24$ V, $R_1 = 4\;\Omega$ (left branch)
• Middle branch: $R_2 = 6\;\Omega$
• Right loop: $\varepsilon_2 = 6$ V, $R_3 = 6\;\Omega$ (right branch)

Assume $I_1$ flows down through the left branch, $I_2$ flows down through the middle, $I_3$ flows down through the right branch.

KCL: $I_1 = I_2 + I_3$

Left loop (CW): $24 - 4I_1 - 6I_2 = 0$

Right loop (CW): $6I_2 - 6I_3 - 6 = 0$

1. Find $I_1$ (in A):
2. Find $I_2$ (in A):
3. Find $I_3$ (in A):
4. Power dissipated by $R_2$ (in W):

RC Circuit Review

A $30\;\text{V}$ battery is connected in series with a $200\;\Omega$ resistor and a $500\;\mu\text{F}$ capacitor. The capacitor is initially uncharged.

1. Time constant $\tau$ (in s):
2. Maximum current $I_0$ (in A):
3. Current at $t = 0.2$ s (in A, to three decimal places):
4. Voltage across the capacitor at $t = 0.1$ s (in V, to one decimal place):

Mastery Quiz