Thevenins and Norton's Theorem

  Norton's  Theorem

Norton's theorem states that a network consists of  several voltage sources, current sources and resistors with two terminals, is electrically equivalent to an ideal current source " I_NO" and a single parallel  resistor, R_NO. The theorem can be applied to both A.C and D.C cases. The Norton equivalent of a circuit consists of an ideal current source in parallel with an ideal impedance (or resistor for non-reactive circuits).

      

The Norton equivalent circuit is a current source with current "I_NO" in parallel with a resistance R_NO.To find its Norton equivalent circuit,

1. Find the Norton current "I_NO". Calculate the output current, "I_AB", when a short circuit is the load (meaning 0 resistances between A and B). This is INo.
2. Find the Norton resistance RNo. When there are no dependent sources (i.e., all current and voltage sources are independent), there are two methods of determining the Norton impedance RNo.

• Calculate the output voltage, VAB, when in open circuit condition (i.e., no load resistor — meaning infinite load resistance). RNo equals this VAB divided by INo.
  or
• Replace independent voltage sources with short circuits and independent current sources with open circuits. The total resistance across the output port is the Norton impedance RNo.
  However, when there are dependent sources the more general method must be used. This method is not shown below in the diagrams.
• Connect a constant current source at the output terminals of the circuit with a value of 1 Ampere and calculate the voltage at its terminals. The quotient of this voltage divided by the 1 A current is the Norton impedance RNo. This method must be used if the circuit contains dependent sources, but it can be used in all cases even when there are no dependent sources.

Example 1:-

Consider this circuit-

To find the Norton’s 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:

With A-B Shorted :

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.

Find the Equivalent Resistance (Rs):

10Ω Resistor in parallel with the 20Ω Resistor

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.

Again, the two resistors are connected in parallel across the terminals A and B which gives us a total resistance of:

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:

Steps to follow for Norton's Theorem

• (1) Find the Norton source current by removing the load resistor from the original circuit and calculating current through a short (wire) jumping across the open connection points where the load resistor used to be.
• (2) Find the Norton resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
• (3) Draw the Norton equivalent circuit, with the Norton current source in parallel with the Norton resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
• (4) Analyze voltage and current for the load resistor following the rules for parallel circuits.

Learnings:

1.Norton's theorem and Thevenin's theorem are equivalent,and the equivalence leads to source transformation in electrical circuits.

2.For an electric-circuit,the equivalence is given by  V_Th= Ith x Rth ie Thevenin's volage=Norton's current x Thevenin's resistance.

3.The applications of Norton theorem is similar to that of Thevenin's theorem.The main application is nothing but the simplification of electrical circuit by introducing source transoformation.

Thevenin Theorem

This theorem is very conceptual. If we think deeply about an electrical circuit, we can visualize the statements made in Thevenin theorem. Suppose we have to calculate theelectric current through any particular branch in a circuit. This branch is connected with rest of the circuits at its two terminal. Due to active sources in the circuit, there is one electric potential difference between the points where the said branch is connected. The currentthrough the said branch is caused by this electric potential difference that appears across the terminals. So rest of the circuit can be considered as a single voltage source, that's voltage is nothing but the open circuit voltage between the terminals where the said branch is connected and the internal resistance of the source is nothing but the equivalent resistanceof the circuit looking back into the terminals where, the branch is connected. So theThevenin theorem can be stated as follows,

1. When a particular branch is removed from a circuit, the open circuit voltage appears across the terminals of the circuit, is Thevenin equivalent voltage and,
2. The equivalent resistance of the circuit network looking back into the terminals, isThevenin equivalent resistance.
3. If we replace the rest of the circuit network by a single voltage source , then thevoltage of the source would be Thevenin equivalent voltage and internal resistanceof the voltage source would be Thevenin equivalent resistance which would be connected in series with the source as shown in the figure below.

To make Thevenin theorem easy to understand, we have shown the circuit below,

Here two resistors R₁ and R₂ are connected in series and this series combination is connected across one voltage source of emf E with internal resistance R_i as shown. One resistive branch of R_L is connected across the resistance R₂ as shown. Now we have to calculate the current through R_L.

First, we have to remove the resistor R_L from the terminals A and B.

Second, we have to calculate the open circuit voltage or Thevenin equivalent voltage V_Tacross the terminals A and B.

The electric current through resistance R₂,

Hence voltage appears across the terminals A and B i.e.

Third, for applying Thevenin theorem, we have to determine the Thevenin equivalentelectrical resistance of the circuit, and for that; first we have to replace the voltage sourcefrom the circuit, leaving behind only its internal resistance R_i. Now view the circuit inwards from the open terminals A and B. It is found the circuits now consist of two parallel paths - one consisting of resistance R₂ only and the other consisting of resistance R₁ and R_i in series.

Thus the Thevenin equivalent resistance R_T is viewed from the open terminals A and B is given as. As per Thevenin theorem, when resistance R_L is connected across terminals A and B, the network behaves as a source of voltage V_T and internal resistance R_T and this is called Thevenin equivalent circuit. The electric current through R_L is given as,

Thevenin Equivalent Circuit