Basic Electrical Circuits & Machines (EE-107)

Course TeacherShaheena Noor

Assistant ProfessorComputer Engineering Department

Sir Syed University of Engineering & Technology.

VOLTAGE AND CURRENT LAWS

In this chapter, we discuss the behavior of electric circuits. Two simple laws, Kirchhoff’s

Current law and Kirchhoff’s voltage law form the foundation for circuit analysis

procedure.

Voltage and Current Laws

• Circuits– Series Circuit– Parallel Circuit

“Two components are connected in series if they have exactly one common terminal and if no other component has a terminal that shares that common connection.”

Figure (a) Figure (b)

Series Circuits

R1 R2

Common terminal

V

Commonterminal

R

• A series path is one in which every component in the path is in series with another component.

Analysis of Series Circuit:• Important property is that the current is the same in every

series-connected component.• Another fact is its total resistance. • Total resistance is the sum of all the series-connected

resistances.

RT or Req = R1 + R2 + R3 + . . .• When a voltage source is connected in series circuit, the

total current produced by that source is from Ohm’s Law.

Series Circuits

• Example # 01: Let R1 = 2Ω; R2 = 1 Ω; V = 5Volts; I = ?

• Example # 02: Find I and voltage across each resistor.

Series Circuits

V

R1

R2

R1 = 12 Ohm

24V

R3 = 10 Ohm

R2 = 6 Ohm

I

It states that “ The algebraic sum of the voltages around any closed path is zero.”

V1 + V2 + V3 + . . . . . . . + VN = 0 OR• “ The sum of the voltage drops around any

closed loop equals the sum of the voltage rises around the loop.”

Kirchhoff’s Voltage Law (KVL):

• Example 3.2Find vx and i .

Kirchhoff’s Voltage Law (KVL):

5V

+ -

7V

100 OhmV xi

Determine i and vx for the figure given below.

Drill Problem 3.2 ( page 34)

1V

3Vi

10 OhmV x

+ -

• Other Examples:

Kirchhoff’s Voltage Law (KVL):

• In the circuit, vs1 = 120V, vs2 = 30V, R1 = 30Ω and R2 = 15Ω. Compute the power absorbed by each element.

Drill Problem 3.4 (page 37)

V s1

V s2R1

R2

Drill Problem 3.5 (page 38) • In the Circuit , find the Power absorbed by each of

the five elements in the circuit.

For Dependent Sources:

7 Ohm

12 V

8 Ohm

V x30 Ohm

+ -

4Vx+-

• Determine i in the given circuit.

Drill Problem 3.9 ( page 45)

5V

5V

5 Ohm

5V

i15 Ohm 25 Ohm

• Break in a circuit path.• No current can flow through an open.• Since no current can flow through it, an open circuit

has an infinite resistance (R = ∞)

I = V/R = ?• Important: It is a common error that since the

current in an open circuit is zero, the voltage across the open must also be zero.

Open Circuit

For Example:

What is the voltage ‘V’ across the switch terminal when the switch is open.

+V-

20 Ohm

60 V

40 Ohm

Voltage Divider Rule (VDR)

R1

+ -

+ -

R2

EV1

V2

I = ?V1 = ?V2 = ?

• Use VDR to find V200Ω and V150 Ω.

• Verify this using KVL

For Example:

36V

50 Ohm

150 Ohm

100 Ohm 200 Ohm

• Two components are connected in parallel when they have 2 common terminal.

• For Example:

Parallel Circuits:

R3

R2R1

R2V R1R3

R1

R2

V

R2

R1

R3

R4

Analysis of Parallel Circuits:• Important property of parallel circuit is that

every parallel-connected component has the same voltage across it.

Parallel Circuits:

R2V

R1

• Find the current in each resistor.

For Example:

R2

4 Ohm

I3

R1

2 Ohm48V

I1

R3

6 Ohm

I2

• Resistance in Parallel:

• For 2 resistors (only)

Parallel Circuits:

R2V

R1

It states that: • “ The algebraic sum of the current entering

any node is zero”OR

• “The sum of all currents entering a junction or any portion of a circuit equals the sum of all currents leaving the same.”

Kirchhoff’s Current Law (KCL):

iA iB

iD ic

• Find the current in the 150Ω resistor

Example

330 Ohm

I3 = 0.1A

100 Ohm

270 Ohm

I4 = ?

I2 = 0.2A

150 Ohm

I1 = 0.8A

• Find ix in each of the circuits.

Q-5 (a) (page 55)

Vix

4A 1A

• Find ix; if iY = 2A and iZ = 0A

• Find iY; if iX = 2A and iZ = 2iYA

Q6 (page 55)

5A

3A

iZiY

iX

• Consider 2 parallel resistor

• Note: Parallel resistors must be branches between the same pairs of nodes.

Current Division Rule (CDR):

R1

I1

R2

I2

I

• Find I1 and I2 using the current divider rule.• Verify the result using KCL

Example:

R1 = 470 Ohm

I1

R2 = 330 Ohm

160mA

I2

• Find current across 3Ω resistor using CDR.

Example 3.13 (Page 52)

4 Ohm

3 Ohm12V 6 Ohm

• A short circuit is a path of zero resistance.• A component is said to be short-circuited

when there is a short circuit connected in parallel with it.

Short Circuit

R

IR = ?

I

Iss = ?