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Text lesson

Lesson 3: Kirchhoffs First and Second Laws for Complex Circuits

Physics

Lesson 3: Kirchhoffs First and Second Laws for Complex Circuits

Overview

Overview

🌳 Concept tree

  • Lesson 3 Kirchhoffs First and Second Laws for Complex Circuits
  • ├── Kirchhoff's laws
  • ├── First law — current (junction) law
  • ├── Second law — voltage (loop) law
  • ├── Procedure for solving a network
  • ├── Worked examples (new values)
  • ├── Common mistakes to avoid
  • └── Summary of formulas
Lesson 3: Kirchhoffs First and Second Laws for Complex Circuits · 1 / 9
Kirchhoff’s laws

Kirchhoff’s laws

  • Some circuits (several sources, several loops, resistors that are neither clearly in series nor in parallel) cannot be reduced by the simple rules.
  • Kirchhoff’s two laws solve any network.
  • They are not new laws of nature they are the conservation of charge and the conservation of energy applied to circuits.
Lesson 3: Kirchhoffs First and Second Laws for Complex Circuits · 2 / 9
First law — current (junction) law

First law — current (junction) law

  • Based on conservation of electric charge (the free electrons flow steadily through the conductor and do not accumulate at any point).
  • At any junction (node), the sum of the currents entering equals the sum of the currents leaving.
  • Equivalently, the algebraic sum of the currents at a node is zero: ΣI = 0.
  • If you count currents entering as positive and leaving as negative, then I₁ + I₂ − I₃ − I₄ = 0, which is the same statement.
  • Example 5. Currents of 7 A and 2 A enter a junction and a current of 5 A leaves along one wire. The unknown current I in the third wire: 7 + 2 = 5 + I, so I = 4 A, leaving the junction.
  • Example 5a. At a node, 6 A and 3 A enter, and 4 A and an unknown current I leave. 6 + 3 = 4 + I so I = 5 A, leaving.
Lesson 3: Kirchhoffs First and Second Laws for Complex Circuits · 3 / 9
Second law — voltage (loop) law

Second law — voltage (loop) law

  • Based on conservation of energy.
  • The emf expresses the energy given to each coulomb in going once round the loop; the potential differences I R express the energy used by each coulomb in each resistor.
  • They must balance.
  • In any closed loop, the algebraic sum of the emfs equals the algebraic sum of the potential drops (I R terms): ΣVB = Σ I R.
Lesson 3: Kirchhoffs First and Second Laws for Complex Circuits · 4 / 9
Procedure for solving a network

Procedure for solving a network

  1. Assume a direction for the current in every branch (a wrong guess is harmless). It is wise to label them I₁, I₂, I₃.
  2. Apply the first law at as many junctions as needed (one fewer than the number of junctions is independent).
  3. Choose independent closed loops and a direction of travel for each (clockwise, say). Every branch should appear in at least one loop.
  4. an emf is positive if the direction of travel goes through the source from its negative to its positive terminal (the direction it would push current); otherwise negative. A term I R is positive if the assumed current is in the direction of travel, otherwise negative. Add the internal resistance r of a source to the resistors in the same branch.
  5. Solve the simultaneous equations (substitution or elimination: multiply one equation so that a coefficient is the same in both, then subtract). A positive answer means the assumed direction was right; a negative answer means the actual current flows the other way.
  6. A battery through which the current flows backwards (into its positive terminal) is being charged; its terminal voltage is then V = VB + I r. A battery delivering current has V = VB − I r.
  7. Check your answer by substituting into a loop you did not use, or by confirming that the junction law holds.
  • The number of equations must equal the number of unknown currents.
  • For a network of three branches with two nodes there are three unknowns: one junction equation and two loop equations.
Lesson 3: Kirchhoffs First and Second Laws for Complex Circuits · 5 / 9
Worked examples (new values)

Worked examples (new values)

Comparison aspectMain ideaKeywords
Example 6.
  • Two cells feed a shared resistor. Cell 1: emf 10 V
  • negligible internal resistance
  • in series with 2 Ω. Cell 2: emf 4 V
  • negligible internal resistance
  • in series with 1 Ω. A 4 Ω resistor is connected across the junctions a and b. Let I₁ and I₂ be the currents supplied by cells 1 and 2 towards the common resistor
  • the current in the common 4 Ω is I₃ = I₁ + I₂ (first law)
cellsfeedsharedresistor
Loop with cell 110 = 2 I₁ + 4 (I₁ + I₂) → 10 = 6 I₁ + 4 I₂
Loop with cell 24 = 1 × I₂ + 4 (I₁ + I₂) → 4 = 4 I₁ + 5 I₂
Multiply the first by 5 and the second by 4
  • 50 = 30 I₁ + 20 I₂ and 16 = 16 I₁ + 20 I₂. Subtract: 34 = 14 I₁
  • so I₁ = 2.43 A
Subtract
•Then 4 = 4(2.43) + 5 I₂ gives I₂ = −1.14 AThengives
•The negative sign means cell 2 is actually being charged by cell 1 (current flows into its positive terminal)negativesignmeanscell
•I₃ = 2.43 − 1.14 = 1.29 A
Check in loop 16(2.43) + 4(−1.14) = 14.58 − 4.56 ≈ 10.0 V ✔
Comparison aspectMain ideaKeywords
Example 7 — both currents positive.
  • Use the same layout
  • but now cell 1 has emf 12 V in series with 2 Ω
  • cell 2 has emf 9 V in series with 1 Ω (both internal resistances negligible)
  • and the shared resistor is 4 Ω
samelayoutcellseries
First lawI₃ = I₁ + I₂
Loop 112 = 2 I₁ + 4 (I₁ + I₂) = 6 I₁ + 4 I₂
Loop 29 = I₂ + 4 (I₁ + I₂) = 4 I₁ + 5 I₂
Multiply loop 1 by 5 and loop 2 by 4
  • 60 = 30 I₁ + 20 I₂ and 36 = 16 I₁ + 20 I₂. Subtract: 24 = 14 I₁
  • so I₁ = 12/7 = 1.714 A
Subtract
•Then I₂ = (9 − 4 × 1.714)/5 = 0.429 A (= 3/7 A)Then
•
  • Both are positive
  • so both assumed directions are right
  • and both cells are discharging. I₃ = 15/7 = 2.14 A
Bothpositiveassumeddirections
•Voltage across the shared resistor = 4 × 2.14 = 8.57 VVoltageacrosssharedresistor
Check in loop 16 × 1.714 + 4 × 0.429 = 10.29 + 1.71 = 12.0 V ✔
Comparison aspectMain ideaKeywords
Example 8 — with internal resistances, and a battery that is charged.
  • Two batteries are connected in parallel to a 4 Ω resistor. Battery 1 has emf 20 V and internal resistance 1 Ω
  • battery 2 has emf 8 V and internal resistance 1 Ω. Take I₁ and I₂ as the currents leaving the positive terminals of batteries 1 and 2
  • and I₃ = I₁ + I₂ the current in the resistor
batteriesconnectedparallelresistor
Loop through battery 1 and the resistor20 = 1 × I₁ + 4 (I₁ + I₂) = 5 I₁ + 4 I₂
Loop through battery 2 and the resistor8 = 1 × I₂ + 4 (I₁ + I₂) = 4 I₁ + 5 I₂
Add the two equations
  • 28 = 9 (I₁ + I₂)
  • so I₃ = I₁ + I₂ = 28/9 = 3.11 A
Subtract the second from the first12 = I₁ − I₂
•So I₁ = (3.11 + 12)/2 = 7.56 A and I₂ = (3.11 − 12)/2 = −4.44 A
•The negative sign for I₂ means battery 2 is being charged (the current enters its positive terminal)negativesignmeansbattery
Terminal voltages
  • battery 1: V = 20 − 7.56 × 1 = 12.44 V. Battery 2 (charging): V = 8 + 4.44 × 1 = 12.44 V. Resistor: 4 × 3.11 = 12.44 V. All three agree
  • as they must since the three branches are connected between the same two points
batterychargingResistorthree
Example 9 — finding the potential difference between two points.→In a single-loop circuit a 12 V battery of internal resistance 1 Ω drives a current through a 5 Ω and a 6 Ω resistor in series. The current is I = 12 / (5 + 6 + 1) = 1 A. The voltage across the 5 Ω resistor is 5 V; across the 6 Ω resistor 6 V; the terminal voltage of the battery is 12 − 1 = 11 V, which equals 5 + 6 ✔.
Lesson 3: Kirchhoffs First and Second Laws for Complex Circuits · 6 / 9
Common mistakes to avoid

Common mistakes to avoid

  • Forgetting the internal resistance of a cell that is part of the loop.
  • Mixing time units (minutes against seconds) in I = Q / t.
  • Using the parallel formula as if it were the series formula, or forgetting to take the reciprocal at the end of 1/R.
  • Giving the sign of the emf wrongly in a Kirchhoff loop. Always decide by the direction of travel.
  • Treating a negative current as an error it only tells you the real direction.
  • Saying "the voltmeter reads the emf" without adding "when the circuit is open".
Lesson 3: Kirchhoffs First and Second Laws for Complex Circuits · 7 / 9
Summary of formulas

Summary of formulas

Quantity Formula
Current I = Q/t
Potential difference V = W/Q
Ohm’s law V = I R
Resistance of a wire R = ρ L / A
Conductivity σ = 1/ρ
Series R = R₁ + R₂ + … ; N equal: R = N R₀
Parallel
  • 1/R = 1/R₁ + 1/R₂ + …
  • two: R₁R₂/(R₁+R₂)
  • N equal: R₀/N
Closed circuit I = VB/(R + r)
Terminal voltage V = VB − I r
Kirchhoff 1 ΣI = 0 (in = out)
Kirchhoff 2 ΣVB = Σ I R around a loop
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Lesson 3: Kirchhoffs First and Second Laws for Complex Circuits · 8 / 9
Part 9

Part 9

  • 📖

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Physics Thanaweya Amma (2026/2027)

  • Lesson 3 – Kirchhoffs First and Second Laws for Complex Circuits
  • When a circuit contains multiple loops with batteries in different branches, Ohm's law is insufficient.
  • Gustav Kirchhoff formulated two fundamental conservation laws: the Junction Rule (Conservation of Charge) and the Loop Rule (Conservation of Energy).

🌳 Hierarchy Classification & Concept Tree

Kirchhoff's Circuit Analysis

Comparison aspectMain ideaKeywords
•├── 1. Kirchhoff's First Law (Junction Rule / Node Law)KirchhoffFirstJunctionRule
│ ├── StatementThe algebraic sum of currents entering any node equals the sum leaving (Σ Iin = Σ Iout)algebraiccurrentsenteringnode
│ └── Physical PrincipleConservation of Electric Charge (No charge accumulates at a junction)ConservationElectricChargeaccumulates
•└── 2. Kirchhoff's Second Law (Loop Rule / Mesh Law)KirchhoffSecondLoopRule
├── StatementThe algebraic sum of EMFs around any closed loop equals the sum of potential drops (Σ VB = Σ I·R)algebraicEMFsaroundclosed
├── Physical PrincipleConservation of Energy (Work done moving around a closed path is zero)ConservationEnergyWorkdone
•└── Sign ConventionsSignConventions
├── Battery
  • If traversal emerges from positive plate → +VB
  • if from negative plate → -VB
traversalemergespositiveplate
└── Resistor
  • If traversal direction matches branch current → +I·R
  • if opposite → -I·R
traversaldirectionmatchesbranch

📋 Master Teacher's Blackboard & Exam Secrets

Comparison aspectMain ideaKeywords
RULE 1Kirchhoff's First Law represents the law of Conservation of Electric ChargeKirchhoffFirstrepresentsConservation
RULE 2Kirchhoff's Second Law represents the law of Conservation of EnergyKirchhoffSecondrepresentsConservation
RULE 3
  • If the calculated value of a branch current is negative
  • it simply means its actual physical direction is opposite to the initially assumed direction
calculatedvaluebranchcurrent
EXAM TRAP
  • Don't forget internal resistance! When summing I·R around a loop
  • include r in the branch: I·(R + r)
forgetinternalresistancesumming
📊 Comparative AnalysisKirchhoff Sign Convention SummaryKirchhoffSignConventionSummary
Circuit Element Direction of Traversal vs. Element Sign Assigned in Equation
Battery EMF Traversed from (-) plate to (+) plate + VB
Battery EMF Traversed from (+) plate to (-) plate – VB
Resistor (R + r) Traversed in same direction as assumed current + I · (R + r)
Resistor (R + r) Traversed in opposite direction to assumed current – I · (R + r)
Lesson 3: Kirchhoffs First and Second Laws for Complex Circuits · 9 / 9