OHM's Law, Voltage Devider Law - Series Circuit - Parallel Circuit- Basic lecture -3- Basic Electrical Engineering- EEE

Subject: Basic Electrical Engineering
Topic: OHM's Law, Voltage Devider Law - Series Circuit - Parallel Circuit- Basic lecture -3-
Subject Code: EEE-4201
Teacher: Md. Rezaul Karim 

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BASIC ELECTRICAL ENGINEERING

Contents

•Problems on Equivalent Resistance
•Voltage Divider Rule
•Color coding of Resistor
•Laws of Resistance and Problems
•Network Analysis
•Kirchhoff’s Laws
•Problems on Kirchhoff’s Laws

Problem: Equivalent Resistance 

Example: Find the equivalent resistance of the circuit given in fig 1.51(a) between following points (i) A and B(ii) and D(iii) E and F. 

Solution: 

(i) Resistance Between A and B

In this case, The entire circuit to the right side of AB is in parallel with 1 ohm resistance connected directly across points A and B.

As  seen, There are two parallel paths across points C and D, One having a resistance of 6 ohm's and the other of (4+2)=6 ohm's. As shown in fig: 1.51(c), the combined resistance between C and D is = 6॥6 = 3 ohm. Further simplification are shown in fig: 1.51 (d) and (e) as seen Rad= 5/6

Voltage Divider Rule

Since in series ckt, same current flows through each of the given resisitors, voltage drop varies directly wth its resistance. In Fig. 1.14 is shown a 24V battery connected across a series combination of three resistors.

Total resistance R= r1+r2+r3= 12 ohms.

 According to voltage divider rule, various voltage drops are

V1= (V*R1)/R = 4 V

V2=  (V*R2)/R = 8 V 

V3= (V*R3)/R = 12V

Overview: Series and Parallel Circuit

Series Circuit

I1=I2=I3+ ..........

VT= V1 + V2 + V3

RT = R1 + R2 + R3

I = V1/R1 = v2/R3 = V3/R3

Voltage Divider Rule  = V1= (Vt*R1)/Rt ,  V2 = (Vt*R2)/Rt

 Parallel Circuit

V1 = V2 = V3= ........

I = I1+ I2 + I3 = v/R1 + v/R2 + v/R3

1/R = 1/R1 + 1/R2 + 1/R3

Resistor: Color Code

Calculation of Color coding

Laws of Resistance

 The resistance R offered by a conductor depends on the following factors:

1. it varies directrly as its length, l
2. It varies inversely as the cross-Section A of the conductor
3. It depends on the nature of material. 
4. It also depends on the temperature of the conductor

Neglecting the last factor for the time being, we can say that

R∝l/A or R= (p*l)/A

Where p is a constant depending on the nature of the material of the conductor and is known as its 

specific resistance or resistivity

if in Eq (i) we put

L= 1 m and A=1 square m then R = p fig 1.4. 

Hence, Specific resistance of material may be defined as the resistance between the opposite face of metre cube of that material.

Problem based on Laws of Resistance

Example: The resistance of the wire used for telephone is 35 ohm per kilometre when the weight of the wire is 5 kg per kilometer. If the specific resistance of the matarial is 1.95 * 10^-8 ohm-m. 

what is the cross sectional area of the wire? What will be the resistance of a loop to a subscriber 8 km from the exchange if wire of the same martial but weighting 20 kg per kilometer is used? 

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