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Superposition and norton Theorem

This document introduces a presentation on the superposition theorem and Norton's theorem given by six students: Mahmudul Hassan, Mahmudul Alam, Sabbir Ahmed, Asikur Rahman, Omma Habiba, and Israt Jahan. The superposition theorem allows analysts to determine voltages and currents in circuits with multiple sources by considering each source independently and then summing their effects. Norton's theorem represents a linear two-terminal circuit as an equivalent circuit with a current source in parallel with a resistor. The document provides examples of applying both theorems to solve circuit problems.

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Welcome To Our Presentation PRESENTED BY: 1.MAHMUDUL HASSAN - 152-15-5809 2.MAHMUDUL ALAM - 152-15-5663 3.SABBIR AHMED – 152-15-5564 4.ASIKUR RAHMAN – 152-15-5948 5.OMMA HABIBA – 151-15- 5975 6.ISRAT JAHAN – 152 -15 - 5956
Presentation On Superposition theorem and Norton theorem
Superposition Theorem :  The superposition theorem extends the use of Ohm’s Law to circuits with multiple sources.  Definition :- The current through, or voltage across, an element in a linear bilateral network equal to the algebraic sum of the currents or voltages produced independently by each source.  The Superposition theorem is very helpful in determining the voltage across an element or current through a branch when the circuit contains multiple number of voltage or current sources.
Condition:  In order to apply the superposition theorem to a network, certain conditions must be met : 1. All the components must be linear, for e.g.- the current is proportional to the applied voltage (for resistors), flux linkage is proportional to current (in inductors), etc. 2. All the components must be bilateral, meaning that the current is the same amount for opposite polarities of the source voltage. 3. Passive components may be used. These are components such as resistors, capacitors, and inductors, that do not amplify or rectify. 4. Active components may not be used. Active components include transistors, semiconductor diodes, and electron tubes. Such components are never bilateral and seldom linear.
Procedure for applying Superposition Theorem:  Circuits Containing Only Independent Sources  Consider only one source to be active at a time.  Remove all other IDEAL VOLTAGE SOURCES by SHORT CIRCUIT & all other IDEAL CURRENT SOURCES by OPEN CIRCUIT. Voltage source is replaced by a Short Circuit Current source is replaced by a Open Circuit
Steps: 1) Select any one source and short all other voltage sources  and open all current sources if internal impedance is not known. If known replace them by their impedance. 2) Find out the current or voltage across the required element, due to the source under consideration. 3) Repeat the above steps for all other sources. 4) Add all the individual effects produced by individual sources to obtain the total current in or across the voltage element.
Example: Using the superposition theorem, determine the voltage drop and current across the resistor 3.3K as shown in figure below. Step 1: Remove the 8V power supply from the original circuit, such that the new circuit becomes as the following and then measure voltage across resistor.
Here 3.3K and 2K are in parallel, therefore resultant resistance will be 1.245K. Using voltage divider rule voltage across 1.245K will be V1= [1.245/(1.245+4.7)]*5 = 1.047V Step 2: Remove the 5V power supply from the original circuit such that the new circuit becomes as the following and then measure voltage across resistor.
Here 3.3K and 4.7K are in parallel, therefore resultant resistance will be 1.938K. Using voltage divider rule voltage across 1.938K will be V2= [1.938/(1.938+2)]*8 = 3.9377V Therefore voltage drop across 3.3K resistor is V1+V2 = 1.047+3.9377=4.9847 v
Circuit A of the sample circuit shown here has an independent voltage source and an independent current source. How do you find the output voltage vo as the voltage across resistor R2? Example : 2 Circuit A (with its two independent sources) breaks up into two simpler circuits, B and C, which have just one source each.
Circuit B has one voltage source because the current source was replaced with an open circuit. Circuit C has one current source because the voltage source was replaced with a short circuit. For Circuit B, you can use the voltage divider technique because its resistors, R1 and R2, are connected in series with a voltage source. So here’s the voltage vo1 across resistor R2:
For Circuit C, you can use a current divider technique because the resistors are connected in parallel with a current source. The current source provides the following current i22 flowing through resistor R2: You can use Ohm’s law to find the voltage output vo2 across resistor R2: Now find the total output voltage across R2 for the two independent sources in Circuit C by adding vo1 (due to the source voltage vs) and vo2 (due to the source current is). You wind up with the following output voltage:
General Idea: Nortons states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source IN in parallel with a resistor RN, where IN is the short-circuit current through the terminals and RN is the input or equivalent resistance at the terminals when the independent sources are turned off. Norton’s Theorem:
1. Find the norton current I(no) calculate the output current ,I(ab),with a short circuit as the load. This is I(no). 2. Find the norton resistance R(no) when there are no dependent sources there are two methods of determining the norton impedence R(no) (i) calculate the out put voltage , V(ab) when in open circuit condition .R(no) equals this V(ab) divided by I(no). (ii) Replace independent voltage with short circuit and independent current sources with open circuits. This total resistance across the output port is the norton impedence R(no) Steps of Norton Theorem:
Conversion to a Thevenin’s equivalent:
Mathematical problem: 1 Find the Norton equivalent circuit of the circuit at terminals a-b.
Solution: Solution: We find in the same way we find in the Thevenin equivalent circuit. Set the independent sources equal to zero. This leads to the circuit from which we find RN Thus, RN = 5 || (8 + 4 +8) =5 || 20 =20 || 5 =20 *5/25 = 4
To find IN we short-circuit terminals a and b, as shown in We ignore the 5 ohm resistor because it has been short-circuited. Applying mesh analysis, we obtain i1 =2 A, 20-i2 -4i1 – 12=0 From these equations, we obtain i2 =1 A=isc=IN
Mathematical problem: 2 To find the Nortons equivalent of the above circuit we firstly have to remove the centre 40Ω load resistor and short out the terminals A and B to give us the following circuit.
When the terminals A and B are shorted together the two resistors are connected in parallel across their two respective voltage sources and the currents flowing through each resistor as well as the total short circuit current can now be calculated as: If we short-out the two voltage sources and open circuit terminals A and B, the two resistors are now effectively connected together in parallel. The value of the internal resistor Rs is found by calculating the total resistance at the terminals A and B giving us the following circuit.
Having found both the short circuit current, Is and equivalent internal resistance, Rs this then gives us the following Nortons equivalent circuit.
Nortons equivalent circuit. Ok, so far so good, but we now have to solve with the original 40Ω load resistor connected across terminals A and B as shown below.
The voltage across the terminals A and B with the load resistor connected is given as: The voltage across the terminals A and B with the load resistor connected is given as: Then the current flowing in the 40Ω load resistor can be found as:
Superposition and norton Theorem