πŸ” Topic 5 β€” Lecture 8 Β· Tutorials 7–8

Kirchhoff's Laws

KCL Β· KVL Β· Voltage divider Β· Current divider Β· Dependent sources Β· All tutorial problems solved

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KCL & KVL

KCL β€” Kirchhoff's Current Law

Statement
Ξ£I_in = Ξ£I_out   (at any node)
Equivalent: Ξ£I = 0 at any node (with sign convention)
πŸ“Œ Physical basis: Conservation of charge

Charge cannot accumulate at a node. All current flowing in must flow out. Applies at every instant in time.

KVL β€” Kirchhoff's Voltage Law

Statement
ΣΔV = 0   (around any closed loop)
Traverse loop in any direction. Voltage rise = +, voltage drop = βˆ’.
πŸ“Œ Physical basis: Conservation of energy

Work done around any closed path = 0. Voltage supplied by sources = voltage dropped across resistors.

⚠️ Sign Convention for KVL
  • Traversing through a resistor in the direction of current β†’ voltage DROP (βˆ’)
  • Traversing through a voltage source from βˆ’ to + β†’ voltage RISE (+)
  • Negative result for current means it flows opposite to your assumed direction β€” this is OK!

Voltage & Current Dividers

Voltage Divider (Series)

Voltage across R₁
V₁ = Vs Γ— R₁/(R₁+Rβ‚‚)
The resistor you WANT is on top
Voltage across Rβ‚‚
Vβ‚‚ = Vs Γ— Rβ‚‚/(R₁+Rβ‚‚)
βœ” Memory rule

Voltage divider: same resistor on top. V₁ = Vs Γ— R₁/(R₁+Rβ‚‚). Higher R β†’ more voltage.

Current Divider (Parallel)

Current through R₁
I₁ = Is Γ— Rβ‚‚/(R₁+Rβ‚‚)
The OTHER resistor is on top!
Current through Rβ‚‚
Iβ‚‚ = Is Γ— R₁/(R₁+Rβ‚‚)
⚠️ Classic exam trap!

Current divider puts the opposite resistor on top: I₁ = Is Γ— Rβ‚‚/(R₁+Rβ‚‚). Higher R β†’ less current (current takes the easier path!).

Circuit Analysis Procedure

Lecture & Tutorial Problems β€” All Solved

Lecture Example 1 β€” Two-loop circuit: R₁=8Ξ©, Rβ‚‚=3Ξ©, R₃=6Ξ©, VS=30V
Find: i₁, iβ‚‚, i₃, and v₁, vβ‚‚, v₃.
KCL at top node: i₁ = iβ‚‚ + i₃  ...(1)
KVL loop 1 (left): βˆ’30 + v₁ + vβ‚‚ = 0 β†’ βˆ’30 + 8i₁ + 3iβ‚‚ = 0  ...(2)
KVL loop 2 (right): βˆ’vβ‚‚ + v₃ = 0 β†’ βˆ’3iβ‚‚ + 6i₃ = 0 β†’ i₃ = iβ‚‚/2  ...(3)

From (3) into (1): i₁ = iβ‚‚ + iβ‚‚/2 = 3iβ‚‚/2
Into (2): 8(3iβ‚‚/2) + 3iβ‚‚ = 30 β†’ 12iβ‚‚ + 3iβ‚‚ = 30 β†’ 15iβ‚‚ = 30 β†’ iβ‚‚ = 2A
i₃ = 1A, i₁ = 3A
v₁ = 8Γ—3 = 24V, vβ‚‚ = 3Γ—2 = 6V, v₃ = 6Γ—1 = 6V

Verify KVL loop 1: βˆ’30 + 24 + 6 = 0 βœ“
Verify KVL loop 2: βˆ’6 + 6 = 0 βœ“
Lecture Example 2 β€” Two batteries: Ρ₁=24V, Ξ΅β‚‚=12V, R₁=1Ξ©, Rβ‚‚=4Ξ©, R₃=2Ξ©
Find: I₁, Iβ‚‚, I₃ using KVL on multiple loops.
Assign: I₁ (loop1 current), Iβ‚‚ (through Rβ‚‚), I₃ (through R₃), with KCL: I₁ = Iβ‚‚ + I₃
KVL loop 1: βˆ’24 + 1Γ—I₁ + 4Γ—Iβ‚‚ = 0 β†’ I₁ + 4Iβ‚‚ = 24  ...(1)
KVL loop 2: βˆ’24 + 1Γ—I₁ + 2Γ—I₃ + 12 = 0 β†’ I₁ + 2I₃ = 12  ...(2)
KCL: I₁ = Iβ‚‚ + I₃  ...(3)

From (1): I₁ = 24 βˆ’ 4Iβ‚‚   |   From (2): I₁ = 12 βˆ’ 2I₃
Substitute (3) into (1): Iβ‚‚+I₃ + 4Iβ‚‚ = 24 β†’ 5Iβ‚‚ + I₃ = 24  ...(4)
Into (2): Iβ‚‚+I₃ + 2I₃ = 12 β†’ Iβ‚‚ + 3I₃ = 12  ...(5)
From (4)βˆ’5Γ—(5): I₃ βˆ’ 15I₃ = 24βˆ’60 β†’ βˆ’14I₃ = βˆ’36 β†’ I₃ = 36/14 = 2.57A
Iβ‚‚ = (12 βˆ’ 3Γ—2.57)/1 = (12βˆ’7.71) = 4.29A
I₁ = 4.29+2.57 = 6.86A
Lecture Example 6 β€” Dependent source circuit: βˆ’24V + 10Ξ© + 12Ξ© + 4Γ—(Ioβˆ’I₁) + 4Ξ©
Given: Ξ΅=24V, R₁=10Ξ©, Rβ‚‚=12Ξ©, R₃=4Ξ©, dependent source = 4(Ioβˆ’I₁). Find Io and I₁.
KVL outer loop: βˆ’24 + 10Io + 12I₁ = 0  ...(1)
KVL inner loop: βˆ’12I₁ + 4(Ioβˆ’I₁) + 4Io = 0
β†’ βˆ’12I₁ + 4Io βˆ’ 4I₁ + 4Io = 0 β†’ 8Io βˆ’ 16I₁ = 0 β†’ Io = 2I₁  ...(2)

Substitute (2) into (1): βˆ’24 + 10(2I₁) + 12I₁ = 0
βˆ’24 + 20I₁ + 12I₁ = 0 β†’ 32I₁ = 24 β†’ I₁ = 3/4 A
Tutorial 7–8 Β· Problem 1 β€” Find Ix and I₁
Given: Circuit with multiple branches. Answer known from tutorial.
Apply KCL at each node and KVL around each loop.
Set up equations with unknowns Ix and I₁.
The system yields two equations in two unknowns β†’ solve simultaneously.
Tutorial 7–8 Β· Problem 2 β€” Find Req
Find R_eq for the given circuit.
Simplify series and parallel combinations step by step from the innermost branch outward.
Tutorial 7–8 Β· Problem 3 β€” Find unknown R given RAB=5Ξ©
Given: Find R such that RAB = 5Ξ©.
Express RAB in terms of R (series/parallel combinations).
Set equal to 5Ξ© and solve for R.
Tutorial 7–8 Β· Problem 4 β€” Find Vo
Given: Circuit with resistors and sources. Find Vo.
Apply KVL and voltage divider as appropriate.
Tutorial 7–8 Β· Problem 5 β€” Find i₁ and R given v=6V
Given: v=6V. Find i₁ and R. Also determine which supply absorbs power.
KVL: apply to circuit with known v=6V.
From KVL: resolve i₁ using Ohm's law.
Power check: P = VI for each source β€” positive = absorbing, negative = supplying.
Lec 8 Β· Voltage divider example
Given: VS=12V, R₁=4kΞ©, Rβ‚‚=8kΞ© in series. Find V₁ and Vβ‚‚.
V₁ = VS Γ— R₁/(R₁+Rβ‚‚) = 12 Γ— 4/(4+8) = 12 Γ— 4/12 = 4V
Vβ‚‚ = VS Γ— Rβ‚‚/(R₁+Rβ‚‚) = 12 Γ— 8/12 = 8V
Check: V₁ + Vβ‚‚ = 4 + 8 = 12V = VS βœ“
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EE-I Exam Prep Β· GIU Cairo Β· Created by Omar Nader