Electrical Circuit Analysis Lab
JIITU
EXPERIMENT No.4
AIM:
To get experienced with Thevenin’s and Norton’s Theorems for the Resistive load.
APPARATUS REQUIRED:
Sr. No
1.
2.
3.
Apparatus
DC power supply
Multimeter
BreadBoard
Specification
Quantity
1
1
1
Specification
1KΩ,1.2KΩ,2.2KΩ,
3.3KΩ,
Quantity
1 each
COMPONENTS REQUIRED:
Sr. No.
1.
Component
Resistance
THEORY:
Thevenin and Norton equivalents are circuit simplification techniques that focus on
terminal behavior. They are 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. They are extremely
valuable aids in analysis. Their equivalent circuits may be used to represent any circuit
made up of linear elements. These two theorems are two equally valid methods of
reducing a complex network down to something simpler to analyze. A Thevenin
equivalent circuit can be readily converted to a Norton equivalent circuit and, vice versa.
Figure 1 pictures Thevenin and Norton equivalents of a circuit.
Figure 1
Thévenin's Theorem states that we can replace entire network, exclusive of the load, by
an equivalent circuit that contains only an independent voltage source in series with an
impedance (resistance) such that the current-voltage relationship at the load is unchanged
(see Figure 2)
07B11EC701
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Deptt of ECE
Electrical Circuit Analysis Lab
JIITU
Figure 2: Thevenin Theorem
Norton's Thereom is identical to Thévenin's Theorem except that the equivalent circuit is
an independent current source in parallel with an impedance (resistor). Hence, the Norton
equivalent circuit is a source transformation of the Thévenin equivalent circuit (see
Figure 3)
Figure 3: Norton Theorem
For a given complex electrical circuit, following steps are done in order to find Thevenin
and Norton equivalents of that circuit.
1. Pick a good breaking point in the circuit (cannot split a dependent source and its
control variable).
2. Thevenin: Compute the open circuit voltage, VOC.
Norton: Compute the short circuit current, ISC.
3. Compute the Thevenin equivalent resistance, RTh (or impedance, ZTh).
a. If there are only independent sources, then short circuit all the voltage
sources and open circuit the current sources (just like superposition).
b. If there are only dependent sources, then must use a test voltage or current
source in order to calculate RTh= vTest/iTest (or ZTh=VTest/ITest).
c. If there are both independent and dependent sources, then compute RTh (or
ZTh) from RTh= vOC/iSC (or ZTh=VOC/ISC).
4. Replace circuit with Thevenin/Norton equivalent.
Thevenin: VOC in series with RTh (or ZTh).
Norton: ISC in parallel with RTh (or ZTh).
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Deptt of ECE
Electrical Circuit Analysis Lab
JIITU
Note: for figure 3 the equivalent network is merely RTh (or ZTh), that is, no current
or voltage sources.
Let’s take a look in the following circuit and try to find Thevenin and Norton equivalents
for the shaded part of it.
First, we remove the capacitor from the circuit; we can re-introduce it once we have
derived the equivalent circuit.
The open-circuit voltage Voc is found from the voltage divider rule:
The short-circuit current Isc is found by determining the current flowing between the
terminals when they are shorted:
The equivalent resistance can be found by de-activing the independent voltage source by
replacing it by a short circuit, as shown:
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Deptt of ECE
Electrical Circuit Analysis Lab
JIITU
The resistance seen at the two external terminals is the parallel combination:
We can verify the standard relationship Voc = I sc ⋅ Rt is satisfied: hence, we could have
derived only two of these three parameters.
Consequently, the Thevenin and Norton equivalent circuits are as follows:
CIRCUIT DIAGRAM:
1. Find Thevenin’s and Norton Equivalent for the circuit as shown in Figure 4.
Figure 4
The Resistance RL = 1KΩ
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Deptt of ECE
Electrical Circuit Analysis Lab
JIITU
2.The student can practice the circuit as shown in Figure 5.
.
Figure 5
OBSERVATION:
Value
Quantity, and Determination-Method
= Voc By DMM Measurment
= Isc By DMM Measurment
ƒ Hint: The DMM itself acts as the Short Circuit
= RTH,VI = Voc/Isc
ƒ Use DMM Measured Voc and Isc
= RTH,SD by Source Deactivation
ƒ Deactivate the Voltage Sources by REMOVING them from the
Ckt, and REPLACING them with a wire
ƒ Measure using the DMM the resulting Resistance as seen from
the Voc terminals
= RTH,avg = [RTH,VI + RTH,SD]/2
07B11EC701
5
Deptt of ECE
Electrical Circuit Analysis Lab
JIITU
CALCULATION:
Student will do calculation as outlined above in theory and compare with experimental
values.
RESULT:
To be written by student
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Deptt of ECE