Kirchhoff's Laws are fundamental rules used to analyze electrical circuits. They are based on the conservation of charge and energy. These two laws help us solve complex circuits that cannot be solved using simple series-parallel combinations.
Kirchhoff's Current Law (KCL)
Statement: The algebraic sum of all currents entering and leaving a node (junction) is zero.
In simple words: The total current flowing INTO a junction equals the total current flowing OUT of that junction.
This law is based on the conservation of electric charge - charge cannot be created or destroyed at a junction.
KCL Formula
ΣI = 0
Or equivalently:
ΣIin = ΣIout
KCL at a Node
Current entering node = Current leaving node
Example: Applying KCL
Problem
At a node, three currents are entering: I₁ = 5A, I₂ = 3A, I₃ = 2A. One current I₄ is leaving the node. Find I₄.
Solution
Apply KCL: Sum of currents entering = Sum of currents leaving
I₁ + I₂ + I₃ = I₄
5 + 3 + 2 = I₄
I₄ = 10 A
Kirchhoff's Voltage Law (KVL)
Statement: The algebraic sum of all voltages around any closed loop (mesh) in a circuit is zero.
In simple words: If you travel around a closed loop and add up all the voltage rises and drops, you get back to where you started - meaning the total is zero.
This law is based on the conservation of energy - energy supplied equals energy consumed in a closed path.
KVL Formula
ΣV = 0
Or equivalently:
ΣVrise = ΣVdrop
Sign Convention for KVL
Moving from − to + terminal of a source: Voltage Rise (+)
Moving from + to − terminal of a source: Voltage Drop (−)
Moving through a resistor in direction of current: Voltage Drop (−IR)
Moving through a resistor opposite to current: Voltage Rise (+IR)
KVL in a Simple Loop
Sum of voltage drops across resistors = Source voltage
Example: Applying KVL
Problem
A circuit has a 24V battery connected in series with three resistors: R₁ = 2Ω, R₂ = 4Ω, R₃ = 6Ω. Find the current and voltage drop across each resistor.
2. Ideal and Practical Voltage and Current Sources
Voltage Sources
Ideal Voltage Source
An ideal voltage source maintains a constant terminal voltage regardless of the current drawn from it.
Internal resistance = 0Ω
Terminal voltage = EMF (always)
Can supply infinite current (theoretically)
Does not exist in practice
Practical Voltage Source
A practical voltage source has an EMF (E) with an internal resistance (r) in series.
Internal resistance r > 0
Terminal voltage V = E − Ir
Voltage drops as current increases
Examples: Batteries, generators
Practical Voltage Source Formula
Vterminal = E − I × r
Where: E = EMF, I = Current, r = Internal resistance
Ideal vs Practical Voltage Source
Current Sources
Ideal Current Source
An ideal current source maintains a constant current regardless of the voltage across it.
Internal resistance = ∞ (infinite)
Output current is constant
Can develop any voltage (theoretically)
Does not exist in practice
Practical Current Source
A practical current source has a source current (Is) with a high internal resistance (R) in parallel.
Internal resistance R is very high but finite
Some current leaks through internal resistance
Output current I = Is − V/R
Ideal vs Practical Current Source Symbols
Example: Practical Voltage Source
Problem
A battery has an EMF of 12V and internal resistance of 0.5Ω. Find the terminal voltage when it supplies a current of 4A.
Solution
Use the formula: Vterminal = E − Ir
Substitute values: Vterminal = 12 − (4 × 0.5)
Vterminal = 12 − 2
Vterminal = 10V
The terminal voltage is less than EMF because some voltage drops across the internal resistance.
3. Source Transformation
What is Source Transformation?
Source transformation is a technique to convert a voltage source in series with a resistance into an equivalent current source in parallel with the same resistance, and vice versa.
This technique simplifies circuit analysis by converting sources to a more convenient form.
Source Transformation Formulas
Voltage to Current Source:
Is = Vs / R
Current source Is in parallel with R
Current to Voltage Source:
Vs = Is × R
Voltage source Vs in series with R
Source Transformation Equivalence
Is = Vs/R and the resistance value remains the same
Important Rules for Source Transformation
The resistance value remains unchanged - only its position changes (series ↔ parallel)
Polarity matters: The current source arrow points from − to + of the equivalent voltage source
Ideal sources CANNOT be transformed (ideal voltage source has R=0, ideal current source has R=∞)
Source transformation preserves the terminal voltage and current
Example: Source Transformation
Problem
Convert a 24V voltage source in series with a 6Ω resistor into an equivalent current source.
Solution
Use the formula: Is = Vs / R
Is = 24V / 6Ω = 4A
The equivalent is a 4A current source in parallel with 6Ω resistor
The 6Ω resistor that was in series with the voltage source is now in parallel with the current source.
4. Mesh Analysis
What is Mesh Analysis?
Mesh analysis (also called Loop analysis or Maxwell's Mesh Current Method) is a technique for analyzing circuits by defining mesh currents that circulate around each mesh (loop).
A mesh is a loop that does not contain any other loop inside it.
Steps for Mesh Analysis
Step 1: Identify all meshes in the circuit
Step 2: Assign mesh currents (usually clockwise) to each mesh
Step 3: Apply KVL to each mesh, writing equations in terms of mesh currents
Step 4: Solve the simultaneous equations to find mesh currents
Step 5: Calculate branch currents and voltages from mesh currents
Mesh Current Method
Two meshes with mesh currents I₁ and I₂ (both assumed clockwise)
Writing Mesh Equations
For a resistor in only one mesh: Voltage = (Mesh current) × R
For a resistor shared by two meshes: Voltage = (I₁ − I₂) × R
(Take the difference if currents flow in opposite directions through the shared element)
Example: Mesh Analysis
Problem
In the circuit shown above, V = 20V, R₁ = 5Ω, R₂ = 10Ω, R₃ = 4Ω, R₄ = 6Ω. Find the mesh currents I₁ and I₂.
Nodal analysis (also called Node Voltage Method) is a technique for analyzing circuits by defining node voltages at each node with respect to a reference node (ground).
A node is any point where two or more circuit elements meet.
Steps for Nodal Analysis
Step 1: Identify all nodes and select one as the reference (ground) node (usually the one with most connections)
Step 2: Label the node voltages (V₁, V₂, ...) with respect to ground
Step 3: Apply KCL at each non-reference node
Step 4: Express branch currents in terms of node voltages using Ohm's law: I = V/R
Step 5: Solve the simultaneous equations to find node voltages
Nodal Analysis Setup
Two nodes V₁ and V₂ with respect to ground reference
Writing Nodal Equations
At any node, apply KCL: ΣIin = ΣIout
Current through a resistor between nodes: I = (Vhigher − Vlower) / R
I = (V₁ − V₂) / R
Example: Nodal Analysis
Problem
In the circuit above, Is = 5A, R₁ = 4Ω, R₂ = 2Ω, R₃ = 8Ω. Find the node voltages V₁ and V₂.
Solution
At Node 1 (apply KCL): Current in = Current out
Is = V₁/R₁ + (V₁ − V₂)/R₂
5 = V₁/4 + (V₁ − V₂)/2
5 = 0.25V₁ + 0.5V₁ − 0.5V₂
0.75V₁ − 0.5V₂ = 5 ... (1)
At Node 2 (apply KCL): Current in = Current out
(V₁ − V₂)/R₂ = V₂/R₃
(V₁ − V₂)/2 = V₂/8
4(V₁ − V₂) = V₂
4V₁ − 4V₂ = V₂
4V₁ = 5V₂ → V₁ = 1.25V₂ ... (2)
The equivalent Star network has R₁ = 30Ω, R₂ = 15Ω, R₃ = 10Ω
7. Superposition Theorem
Statement of Superposition Theorem
In a linear circuit with multiple independent sources, the current through (or voltage across) any element is the algebraic sum of the currents (or voltages) produced by each independent source acting alone, while all other independent sources are replaced by their internal resistances.
Steps for Applying Superposition Theorem
Step 1: Select one source and deactivate all other sources
• Replace voltage sources with short circuits (0Ω)
• Replace current sources with open circuits (∞Ω)
Step 2: Calculate the current/voltage due to the selected source
Step 3: Repeat for each source
Step 4: Add all individual contributions algebraically (consider directions!)
Deactivating Sources
Superposition theorem applies only to linear circuits. It CANNOT be used to calculate power directly (power is not a linear function).
Example: Superposition Theorem
Problem
A circuit has a 12V voltage source and a 3A current source connected to a 4Ω resistor. Using superposition, find the current through the 4Ω resistor.
Solution
Case 1: Only 12V source active (current source → open) With current source open, the resistor is directly across the voltage source
I₁ = V/R = 12/4 = 3A (flowing from + to −, let's say left to right)
Case 2: Only 3A source active (voltage source → short) With voltage source shorted, all 3A flows through the short, so
I₂ = 0A through the resistor (current takes the path of zero resistance)
Wait - let's reconsider: if the 4Ω is in parallel with the short circuit... Actually, the resistor gets shorted, so I₂ = 0A through resistor
Total current by superposition: Itotal = I₁ + I₂ = 3 + 0 = 3A
The exact result depends on the circuit topology. In this simple case, when the voltage source is shorted, the current source's current bypasses the resistor.
8. Thevenin's Theorem
Statement of Thevenin's Theorem
Any linear circuit containing voltage sources, current sources, and resistances can be replaced by an equivalent circuit consisting of:
A single voltage source called Thevenin Voltage (VTH)
In series with a single resistance called Thevenin Resistance (RTH)
This equivalent circuit, when connected to the same load, will produce the same current and voltage as the original circuit.
Thevenin Equivalent Circuit
Steps to Find Thevenin Equivalent
Step 1: Remove the load resistor (RL) from the circuit
Step 2: Find VTH = Open circuit voltage across terminals A-B
Step 3: Find RTH by deactivating all sources and calculating equivalent resistance across A-B
• Replace voltage sources with short circuits
• Replace current sources with open circuits
Step 4: Draw the Thevenin equivalent circuit with VTH and RTH
Load Current using Thevenin Equivalent
IL = VTH / (RTH + RL)
Example: Finding Thevenin Equivalent
Problem
Find the Thevenin equivalent circuit across terminals A-B for the following circuit: A 20V source is in series with a 5Ω resistor, and another 10Ω resistor is connected in parallel to this combination. Terminals A and B are across the 10Ω resistor.
Solution
Find VTH (open circuit voltage): With terminals open, no current flows through the 10Ω resistor (it's in parallel with open terminals)
Current through 5Ω = 0 (no complete circuit)
Wait - let me reconsider the topology...
If 20V is in series with 5Ω, and 10Ω is across terminals A-B:
VTH = Voltage across 10Ω with terminals open
Since no current flows through 5Ω (open circuit), VTH = 20V
Find RTH (deactivate sources): Replace 20V source with short circuit
Looking from terminals A-B: 5Ω is in parallel with the short (= 0Ω)
So RTH = 10Ω || (5Ω in series with 0Ω) = 10Ω || 5Ω
RTH = (10 × 5)/(10 + 5) = 50/15 = 3.33Ω
Thevenin Equivalent: VTH = 20V RTH = 3.33Ω
9. Norton's Theorem
Statement of Norton's Theorem
Any linear circuit containing voltage sources, current sources, and resistances can be replaced by an equivalent circuit consisting of:
A single current source called Norton Current (IN)
In parallel with a single resistance called Norton Resistance (RN)
Norton's theorem is the dual of Thevenin's theorem.
Norton Equivalent Circuit
Steps to Find Norton Equivalent
Step 1: Remove the load resistor from the circuit
Step 2: Find IN = Short circuit current through terminals A-B
Step 3: Find RN (same as RTH) by deactivating all sources
Step 4: Draw the Norton equivalent circuit
Relationship between Thevenin and Norton
IN = VTH / RTH
RN = RTH
This means you can easily convert between Thevenin and Norton equivalents!
Example: Finding Norton Equivalent
Problem
For the Thevenin example above (VTH = 20V, RTH = 3.33Ω), find the Norton equivalent.
Solution
Use the relationship: IN = VTH / RTH
IN = 20 / 3.33 = 6A
RN = RTH = 3.33Ω
Norton Equivalent: 6A current source in parallel with 3.33Ω resistor
10. Maximum Power Transfer Theorem
Statement of Maximum Power Transfer Theorem
Maximum power is transferred from a source to a load when the load resistance equals the source's internal resistance (or Thevenin resistance).
In other words, for maximum power transfer:
RL = RTH
Maximum Power Transfer
Maximum power is transferred when RL = RTH
Maximum Power Transfer Formulas
Condition for Maximum Power:
RL = RTH
Maximum Power Delivered to Load:
Pmax = VTH² / (4 × RTH)
Current at Maximum Power:
I = VTH / (2 × RTH)
Efficiency at Maximum Power Transfer = 50%
(Half the power is dissipated in RTH, half in RL)
Maximum power transfer is important for communication systems where signal strength matters more than efficiency. For power systems, we aim for high efficiency (close to 100%), not maximum power transfer.
Example: Maximum Power Transfer
Problem
A circuit has Thevenin equivalent VTH = 48V and RTH = 6Ω. Find:
(a) The load resistance for maximum power transfer
(b) The maximum power delivered to the load