Unit i electrical circuits

ELECTRICAL CIRCUITS
5/20/2020 1NSR/AP/ECE/PSNACET/BEEIE/R2017

Syllabus
Basic circuit components - Ohms Law - Kirchhoff’s
Law – Instantaneous Power – Inductors - Capacitors –
Independent and Dependent Sources - steady state
solution of DC circuits – Nodal analysis, Mesh
analysis- Thevinin’s Theorem, Norton’s Theorem,
Maximum Power transfer theorem-Linearity and
Superposition Theorem.

Basic circuit components

Ohms Law

Kirchhoff’s Law

Resistance connected in series

Parallel connected resistances

Problem

Independent and Dependent Sources

Dependent sources

Independent sources

Source transformation
Voltage source to current source I= V/R
Current source to voltage source V= IR5/20/2020 29NSR/AP/ECE/PSNACET/BEEIE/R2017

Voltage division in series resistive
circuit

Current division rule in parallel
circuit

Mesh analysis example 1

THE UNKNOWN CURRENTS ARE DETERMINED USING GRAMMER RULE

Nodal analysis
Thevenin’s Theorem

Delta to Star and Star to Delta transformation

Nodal analysis

Thevenin’s Theorem
Thevenin theorem is an analytical
method used to change a complex
circuit into a simple equivalent circuit
consisting of a single resistance in
series with a source voltage

Thevenin’s Theorem statement
Thevenin’s Theorem states that “Any
linear circuit containing several
voltages and resistances can be
replaced by just one single voltage in
series with a single resistance
connected across the load“

Thevenin’s equivalent circuit
v

Thevenin’s theorem example
To find current through A to B
Using thevenins theorem

Remove load resistance and short circuit
the voltage sources

Find the thevenins voltage (Vth)

MESH ANALYSIS
PROBLEMS

1.Find the current through 8 Ω
resistor using mesh analysis

solution
Mark the meshes and mesh currents

Writing the KVL equations for
meshs,we get
 Mesh ABGHA
 -5 I1 -15(I1 – I2)+4 = 0
 20I1-15I2 = 4 ------ (1)

 Mesh BCFGB
 - 10 I2 – 8 (I2+I3)+15 (I1-I2) = 0
 15 I1-33I2-8I3 = 0 ------------------- (2)

 Mesh CDEFC
 12I3-6+8(I2+I3)= 0
 8I2+20I3 = 6 -------- (3)

Solving (1),(2) and (3) we get
 I1 = 0.22A
 I2 = 0.03A
 I3 = 0.29A
 CURRENT IN 8 Ω RESISTOR IS
 I2 +I3 =0.03 +0.29 = 0.32 A

2.Find the current through 6Ω resistor
of the given circuit using mesh method

Mark the current direction
Mark the meshes and mesh currents5/20/2020 69NSR/AP/ECE/PSNACET/BEEIE/R2017

Writing the KVL equations for
meshs,we get
 Mesh ABGHA
 - 8 I 1 – 4 (I1-I2) + 100 = 0
 12I 1 + 4 I 2 = 100 ----------------- (1)

 Mesh BCFGB
 -6 I2 – 10 (I 2 + I 3) + 4 (I 1 – I 2 ) = 0
 - 6 I2 – 10 I 2 – 10 I 3 + 4 I1 – 4 I 2 = 0
 4 I1 – 20 I2 – 10 I3 = 0 ---------------- (2)

 Mesh CDEFC
 4 I3+10 (I2+I3)-40 = 0
 10I2+14I3 = 40 ---------- (3)

Solving (1),(2) and (3) we get
 I1 = 8.47 A
 I2 = 0.413 A
 I3 = 2.56 A
 CURRENT IN 6 Ω RESISTOR IS I2
 I2 = 0.413 A

MESH ANALYSIS
NORTONS THEOREM
MAXIMUM POWER TRANSFER THEOREM
SUPER POSITION THEOREM
PROBLEMS

3.Using kirchoff’s law ,Calculate the
current delivered by the battery

ANSWER : Mark the current direction and meshes

Apply KVL to the mesh ABEFA
 5(I2 – I3) +4 (I1-I3)- 4I3 = 0
 5I2-5I3+4I1-4I3-4I3 = 0
 4I1+5I2-13I3 = 0 ------------------- (1)

Apply KVL to the mesh BCDEB
 3I2 + 4I3 + 3 I1 = 10
 3I1+3I2+4I3 = 10 ----------------(2)

Apply KVL to the mesh ABCA
 5(I2-I3)+3I2 -6(I1-I2)= 0
 5I2-5I3+3I2-6I1+6I2 = 0
 -6I1 + 14 I2- 5I3 = 0 ----------------------- (3)

 4I1+5I2-13I3 = 0 ------------------- (1)
 3I1+3I2+4I3 = 10 ----------------------(2)
 -6I1 + 14 I2- 5I3 = 0 ----------------- (3)
Write the equations (1),(2) and (3)

Rearrange these equations (1),(2) and (3) in
matrix form

Find the current I1 using crammer rule
The current delivered by the battery
Similarly I2, and I3 currents also determined using crammer rule

4.Use mesh analysis to determine the current in the 4 Ω
resistor in circuit shown below.

NORTONS THEOREM

MAXIMUM POWER TRANSFER THEOREM

SUPER POSITION THEOREM

Unit i electrical circuits

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