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UNIVERSITI TEKNOLOGI MARA

ELECTRICAL ENGINEERING LABORATORY 1

( EEE 230 )

EXPERIMENT 4

THEVENIN’S THEOREM AND WHEATESTONE BRIDGE

OBJECTIVES

1. To analyze dc resistive circuit using thevenin’s theorem. 2. To analyze an unbalanced wheatstone bridge using thevenin’s theorem

LIST OF REQUIREMENTS

Equipment

DC Power Supply

Galvanometer

Digital Multimeter

Ammeter

Components

Diode : IN4002

Resistor : 2.2kΩ, 1.2kΩ, 10kΩ, 3.3kΩ

Decade Resistance

1

THEORY

THEVENIN’S THEOREM

Thevenin's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load. The qualification of "linear" is identical to that found in the Superposition Theorem, where all the underlying equations must be linear (no exponents or roots). If we're dealing with passive components (such as resistors, and later, inductors and capacitors), this is true. However, there are some components (especially certain gas-discharge and semiconductor components) which are nonlinear: that is, their opposition to current changes with voltage and/or current. As such, we would call circuits containing these types of components, nonlinear circuits. Thevenin's Theorem is especially useful in analyzing power systems and other circuits where one particular resistor in the circuit (called the "load" resistor) is subject to change, and re-calculation of the circuit is necessary with each trial value of load resistance, to determine voltage across it and current through it.

According to thevenin’s theorem, in order to simplify a complex circuit to a two resistor single voltage, we need to find the thevenin voltage by removing the load resistor and replace with open circuit the voltage across the open circuit ia a thevenin voltage, VTH

As in the figure above, thevenin voltage can be calculated by applying Voltage divider rule, as the a-b terminaL is an open circuit terminal, there is no current flow on R2 and there is no voltage drop across it, thus the voltage across the open circuit terminal (VOC=VTH), can be calculated as followed, VTH=R3R3+R4V

2

Again with the load disconnected, and terminal a-b open circuited, we turn off all independent source. The input ressistance (equivalent resistance) of the dead circuit, at the terminal a-b must be equal to RTH because the original circuit and the simplified circuit are equivalent. Thus, RTH is the input resistance at the terminals when the independent source are turned off, RTH=Rin . to find the thevenin voltage, the load resistor RL is removed again and replace with open circuit(O/C), and the dependent voltage source is removed and replace with short circuit (S/C).

Rin=RTH

To find the thevenin resistance RTH in the above circuit, (R4//R3)+R2.

RTH=R3R4R3+R4+R2

Therefore, the thevenin equivalent circuit can be obtain by putting the VTH as a single voltage source, and RTH as a single resistor. The load resistance RL is simply connected across terminal a-b.

Thevenin’s Equivalent Circuit

3

The current through the load resistance, RL is ,

IR=VTHRTH+R

And the voltage across the load resistance, RL is

VR=IR=VTHRVTH+R

WHEATSTONE BRIDGE

Wheatstone bridge is the most accurate method for measuring resistance and is popular for laboratory used.It consists of four resistors, a dc voltage source and a galvanometer arranged as shown in Figure 4.4.The galvanometer is used to detect the current flow from point ‘c’ to ‘d’.

A bridge is said to be in balance condition or null condition if there is no current flows through the galvanometer.The current across galvanometer is said to be equal to zero when voltage from point ‘a’ to ‘d’ is equal to voltage at the point ‘a’ to ‘c’.The same goes to voltage from point ‘b’ to ‘c’ is equal to voltage at the point ‘b’...