🎯⭐ INTERACTIVE LESSON

Kirchhoff's Laws

Learn step-by-step with interactive practice!

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Kirchhoff's Laws - Complete Interactive Lesson

Part 1: Junction Rule (KCL)

⚡ 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 the node

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: $I_1 = I_2 + I_3$

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

Total current flowing out of the node (in A, not counting $I_3$ ):
Value of (in A):

Exit Quiz

Part 2: Loop Rule (KVL)

🔄 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.

$\sum \varepsilon = \sum IR$

Part 3: Single-Loop Circuits

🔋 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 )

Part 4: Multi-Loop Circuits

🔗 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

Identify nodes and branches: Count junctions and separate paths
Assign current variables: One variable per branch, with an assumed direction (arrow)
Write KCL equations: At $N - 1$ nodes (where $N$ is the number of nodes)
Write KVL equations: For $B - N + 1$ independent loops

Part 5: Complex Circuits

🧮 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: $I_1, I_2, I_3, ...$

Part 6: RC Circuits

🔌 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

Part 7: Synthesis & AP Review

🏆 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

Identify the circuit type
- Single loop → KVL alone is sufficient
- Multi-loop → Need both KCL and KVL
- Contains capacitors → Consider time-dependent (RC) behavior
Label everything
- Assign current variables with directions to each branch
- Label nodes (junctions) with letters
- Identify independent loops
Write equations
- KCL at $N - 1$ nodes
- KVL around $B - N + 1$ loops

Branch 3 carries

I_3

out of the node

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

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!

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.

📝 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.

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.

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.

When traversing a loop in a chosen direction:

Element	Traversal Direction	Voltage Change
Battery	− to + (low to high)	$+\varepsilon$
Battery	+ to − (high to low)	$-\varepsilon$
Resistor	In direction of current $I$	$-IR$
Resistor	Against direction of current $I$	$+IR$

🔑 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: 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

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

Label currents: Assign a direction to each unknown current (guess if needed — a negative answer just means it flows the other way)
Choose loops: Identify independent loops that cover every branch
Pick a traversal direction: Usually clockwise for each loop
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

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 $\varepsilon_1$ )
$R_1 = 3\;\Omega$ , $R_2 = 5\;\Omega$ ,

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
$V_{_{}}$ V

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: $R_3 = 6\;\Omega$

Net EMF around the loop (in V):
Total resistance (in Ω):
Current in the loop (in A):
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)
$R_3 = 5\;\Omega$

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

Round all answers to 3 significant figures.

Solve the system: Substitution or elimination

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)

Step 2: KCL at top node

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

Step 3: KVL — Left Loop (clockwise)

$+\varepsilon_1 - I_1 R_1 - I_2 R_2 = 0$

Step 4: KVL — Right Loop (clockwise)

$+I_2 R_2 - I_3 R_3 - \varepsilon_2 = 0$

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)
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:

$I_2$ (in A):
$I_3$ (in A):
$I_1$ (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): $6I_2 - 2I_3 - 4 = 0$

Solve for $I_1$ (in A):
Solve for $I_2$ (in A):
Solve for $I_3$ (in A):

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
$R_1 = 2\;\Omega$ , $R_2 = 4\;\Omega$ ,
$R_4 = 6\;\Omega$ , $R_5 = 1\;\Omega$

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: $\varepsilon_1 - i_1 R_1 - (i_1 - i_2)R_2 = 0$

Loop 2: $-(i_2 - i_1)R_2 - i_2 R_3 - (i_2 - i_3)R_4 + \varepsilon_2 = 0$

Loop 3: $-(i_3 - i_2)R_4 - i_3 R_5 - \varepsilon_3 = 0$

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

$\det(\mathbf{A}) = 6[(-13)(-7) - (6)(6)] - (-4)[(4)(-7) - (6)(0)] + 0$

💡 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.

Mesh Current Drill

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

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

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

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

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

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

Time	$V_C$	$I$
$t = 0$	$0$	$\varepsilon / R$ (maximum)
$t = \tau$	$0.632\,\varepsilon$	$0.368\,\varepsilon/R$
$t \to \infty$	$\varepsilon$ (fully charged)	$0$ (no current)

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

Time	$V_C$	$I$
$t = 0$	$_{}$

Charging vs. Discharging Summary

	Charging	Discharging
$V_C$	$\varepsilon(1 - e^{-t/RC})$

💡 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 ( $R = 50\;\text{k}\Omega$ ).

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

Verify: total equations = total unknowns

Solve the system

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

Check your answer

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

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

❌ Mistake 5: Unit Errors in RC

$R$ in ohms, $C$ in farads → $\tau$ in seconds
$\text{k}\Omega \times \mu\text{F}$ : $(10^{3}) (10^{- 6}) =$ s = ms

✅ 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): $6I_2 - 6I_3 - 6 = 0$

Find $I_1$ (in A):
Find $I_2$ (in A):
Find $I_3$ (in A):

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.

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

After

R_2

7.5 - 7.5 = 0

V ✓

V_{R_{2}} = 1.6 \times 5 = 8.0

V_{R_3} = 1.6 \times 2 = 3.2

Total drops:

4.8 + 8.0 + 3.2 = 16.0

V = Net EMF ✓

I_2

: through

R_2

(middle branch, downward)

I_3

: through

\varepsilon_2

and

R_3

(right branch, downward)

10 - 2I_1 - 3I_2 = 0 \quad \text{...(2)}

3I_2 - 4I_3 - 6 = 0 \quad \text{...(3)}

12 - 2i_1 - 4(i_1 - i_2) = 0

12 - 6i_1 + 4i_2 = 0

6i_1 - 4i_2 = 12 \quad \text{...(1)}

4i_1 - 13i_2 + 6i_3 = -6 \quad \text{...(2)}

6i_2 - 7i_3 = 9 \quad \text{...(3)}

= 6[91 - 36] + 4[-28]

Discharging:

V_C = V_0 e^{-t/RC}

— starts at

V_0

, falls to 0

Don't confuse the two!

(1 0^{3}) (1 0^{- 6}) = 1 0^{- 3}

Always convert to base SI units first!

Power dissipated by

R_2

(in W):

Kirchhoff's Laws

Kirchhoff's Laws - Complete Interactive Lesson

Part 1: Junction Rule (KCL)

⚡ Kirchhoff's Junction Rule (KCL)

The Junction Rule

Why It Works

Key Vocabulary

Example: Three-Branch Junction

Multiple Junctions in a Circuit

Example: Two-Node Circuit

Part 2: Loop Rule (KVL)

🔄 Kirchhoff's Loop Rule (KVL)

The Loop Rule

Part 3: Single-Loop Circuits

🔋 Single-Loop Circuits with Multiple Batteries

Opposing EMFs

Example: Two Batteries in Opposition

Part 4: Multi-Loop Circuits

🔗 Multi-Loop Circuits

Systematic Approach for Multi-Loop Circuits

Step-by-Step Method

Part 5: Complex Circuits

🧮 Complex Circuits & Matrix Methods

Systematic Labeling

Node Labeling

Part 6: RC Circuits

🔌 RC Circuits Basics

Charging a Capacitor

Applying KVL at Any Instant

Part 7: Synthesis & AP Review

🏆 Synthesis & AP Review

Complete Circuit Analysis Strategy

Master Checklist

Four-Branch Example

Practical Rule

Why It Works

Sign Conventions

Example: Single-Battery Loop

Checking Voltage at Each Point

Systematic Approach

The 4-Step Method

How Many Loops?

What If You Guess Wrong?

Three Batteries in a Loop

Voltage Drops

Counting Check

Classic Two-Loop Problem

Step 1: Label Currents

Step 2: KCL at top node

Step 3: KVL — Left Loop (clockwise)

Step 4: KVL — Right Loop (clockwise)

Step 5: Solve

Interpretation

Branch Counting

Example: Three-Loop Circuit

Three-Loop Example

Mesh Equations

Matrix Form

Cramer's Rule Preview

Determinant of the Coefficient Matrix

Charge on the Capacitor

Voltage Across the Capacitor

Current in the Circuit

Key Behaviors

The Time Constant τ=RC\tau = RCτ=RC

Physical Interpretation

Example

Discharging a Capacitor

KVL for Discharging

Solutions

Key Behaviors

Charging vs. Discharging Summary

Common Shortcuts

Common Mistakes on the AP Exam

❌ Mistake 1: Wrong Sign Convention

❌ Mistake 2: Forgetting a Branch

❌ Mistake 3: Inconsistent Current Directions at Nodes

❌ Mistake 4: RC Circuit Mix-ups

❌ Mistake 5: Unit Errors in RC

✅ Pro Tip for AP FRQs

The Time Constant $\tau = RC$