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# Kirchoff’s Voltage Law And Kirchoff’s Current Law (Coursework Sample)

Instructions:

The two Ohms law, Kirchoff's voltage Law and Kirchoff's Current Law are important in the study and understanding of linear circuitry. These laws make use of resistance in a circuit where voltage and current sources and resistors are important component of study.

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Content:

Kirchoff’s Current Law and Kirchoff’s Voltage Law Applications
Name
Institution
Lab 5
Title: Kirchoff’s Current law and Kirchoff’s Voltage Law applications.
Introduction
The two Ohms law, Kirchoff’s voltage Law and Kirchoff’s Current Law are important in the study and understanding of linear circuitry. These laws make use of resistance in a circuit where voltage and current sources and resistors are important component of study.
Kirchoff’s voltage law state that the algebraic sum of voltage drop in a complete closed loop is zero.
Kirchoff’s current law states that the algebraic of all current in a junction is zero.
Apparatus
Dc supply 4volts
Resistors R1=R2=R3=R4=1 kilo ohm
Method
The circuit was connected in the breadboard as below:
Questions
Determining the current and voltages across all resistors
Solutions
Resistor R3 and R4 are in series. An equivalent resistance for both is given by:
R =R3 + R4
= 1 ohm + 1 ohm
=2 ohms.
The equivalent resistor of R4 and R3, R is in parallel with R2
Combining R equivalent of R3, R4 and R2 we have
=1/ R2 + 1/R
= 1 + ½
= 0.67 ohms.
The resultant equivalent resistor of R2, R3 and R4 is in series with R1. The total resistance of the circuit is given by:
R total = R1 + the equivalent resistor of R2, R3, R4
= 1 + 0.67
= 2.5 ohms.
Calculating current through R1
I = V/R
= 4/2.5
= 1.6A
Voltage through R1
V= I * R1
= 1.6 * 1
= 1.6 V.
Current through R2= 2/3 * 1.6
= 1.067 A.
Voltage across R2 = IR
= 1 * 1.067
= 1.067 V.
Current through R4 and R3 is given by
1/3 * 1.6
= 0.533 A.
Voltage through R4 and R3 is equal to 0.533 V since they have the same resistance.
Applying the Kirchoff’s current and voltage laws to verify the answers
I3R3 – I4R4 – I2R2 = 0
0.53 + 0.53 – 1.06 = 0
0 = 0.
Hence verified.
Three currents and four voltages should be calculated.
Measurements using a multimeter.

IS (A)

I1 (A)

I2 (A)

V-AB

V-BD

V-BC

V-CD

Theoretical values

1.60

1.07

0.53

1.60

1.07

0.53

1.06

measured

1.60

1.00

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