# Norton’s Theorem with Solved Examples

Table of Contents

## Norton’s Theorem

**Norton’s Theorem with Solved Examples –** Similar to Thevenin’s theorem, Norton’s theorem is also a method for converting complicated circuits into simple circuits thereby getting an easy solution. The basic difference between these two is that Norton’s theorem provides an equivalent current source, having an equivalent resistance being fixed across it. Whereas, Thevenin theorem provides an equivalent current source, having a resistance in its series.

Figure 2.85; Form of Norton’s equivalent circuit

An equivalent Norton’s circuit has been illustrated in fig. 2.85. No matter how complicated a given circuit, this theorem always reduces this circuit to the form shown in the figure. The equivalent current source is represented by I_{N} and equivalent resistance by R_{N}. In order to solve questions through the Norton’s theorem, we should know the method of determining I_{N} and R_{N}.

When I_{N} and R_{N} values of any given circuits have been determined, then a complex Norton’s circuit can be obtained through connecting them in parallel according to the figure. It ought to be remembered that it is possible to determine Norton’s equivalent through a Thevenin equivalent and a Thevenin’s equivalent from a Norton’s equivalent. That’s this theorem is a substitute for Thevenin theorem.

Norton’s theorem states that “the entire network connected to terminals A and B can be reduced by a single constant current source I_{N} in parallel with a single resistance R_{N}”

## Norton’s Equivalent Current

As has been discussed above, I_{N} is one part of a whole Norton’s equivalent circuit, whereas R_{N} is other part of this circuit. I_{N} can be defined as a short circuit current found between two points of any circuit. Whenever, a component is mounted between these two points, it seems just like a current source, having a value I_{N}, and R_{N} exist parallel to it.

Fig. 2.86

For further elaboration, assume a resistive circuit as shown in figure 2.86 (a). All its resistors have been installed between two points on the circuit. We have to convert this circuit to a Norton circuit. To determine I_{N}, short circuit the resistance existing between two points A and B and then calculate the value of I_{N}. This has been illustrated in figure 2.86 (b). The following example will prove helpful in this context;

**Example;** Determine I_{N} for the circuit within the dotted line shown in the figure 2.87

Figure 2.87

**Solution;**

Short terminals A and B are shown in fig. 2.87 (b). I_{N} is the current through the short and is calculated as follows;

First, the total resistance seen by the voltage source is;

R_{T} = R_{1} + R_{2} R_{3}/ R_{2} + R_{3}

= 47 + 47 x 100/ 47 + 100 = 78.97 ohms

The total current from the source is;

I_{T} = V_{S} / R_{T}

= 83.33 / 78.97 = 1.06 A

Now, apply the current divider formula to find I_{N} (the current through the short)

IN = [R2 / R2 + R3] IT

= [47 / 147] 1.06 = 0.339 A

This is the value for the equivalent Norton current source.

## Norton’s Equivalent Resistance

The value of R_{N} can also be determined just similar to the value of R_{th} (R_{th} and R_{N} both have same meaning, and both have same value), i.e., if internal resistances are installed instead of all sources, then R_{N} turns out to be the total resistance found between the two terminals of any circuit. It has been explained with the help of the following equation.

**Example;** Find R_{N} for the circuit within the dotted area of fig. 2.87.

**Solution;**

First reduce VS to zero by shorting it, as shown in fig. 2.88

Looking in at terminals A and B, we see that the parallel combination of R_{1} and R_{2} is in series with R_{3}. Thus,

R_{N} = R_{3} + R_{1} / R_{2} = 100 + 47 / 2 = 123.5 ohm

Fig. 2.88

Method for finding Norton’s Equivalent

The following method is usually adopted for finding out Norton’s equivalent.

1). First, disconnect the resistance fixed parallel to the two given circuits, and also short these two terminals mutually.

2). Determine current I_{N} passing through these shorted circuits. This current is called Norton’s current.

3). Remove all voltage sources and replace them with their internal resistances. Similarly, remove all current sources and replace them with open circuits.

4). Then find out an equivalent resistance (R_{N}) by viewing into the circuit from both the terminals (which are open). The method of finding out R_{N} is exactly similar to that of finding R_{th}, because R_{N} = R_{th.}

5). Connect the current source between the two terminals given parallel to the resistance R_{N}. Thus, an equivalent Norton circuit is abstracted. In other words, replace the original circuit given by joining I_{N} and R_{N} in parallel, with a whole Norton equivalent circuit. Remember that I_{N} can also be represented by I_{SC} and R_{N} by R_{i}.

## Calculations About Norton’s Theorem

**Example 1;** Using Norton’s theorem, find the constant – current equivalent of the circuit shown in fig 2.89 (a).

**Solution;**

When terminals A and B are short – circuited as shown in fig. 2.89 (b), total resistance of the circuit, as seen by the battery, consists of a 10 Ω resistance in series with a parallel combination of 10 Ω and 15 Ω resistances.

∴ Total resistance = 10 + 15 x 10 / 15 + 10 = 16 Ω

∴ Battery current I = 100 / 16 = 6.25 A

Fig. 2.89 (a, b, c)

This current is divided into two parts at point C of fig. 2.89 (b).

Current through AB is I_{SC} = 6.25 x 10 / 25 = 2.5 A

Since the battery has no internal resistance, the input resistance of the network when viewed from A and B consists of a 15 Ω resistance in series with the parallel combination of 10 Ω and 10 Ω.

Hence, R_{i }= 15 + (10/2) = 20 Ω

Hence, the equivalent – current source is as shown in fig. ©

**Example 2;** Apply Norton’s theorem to calculate current flowing through 5 Ω resistor of fig. 2.90 (a).

**Solution;**

i). Remove 5 Ω resistor and put a short across terminals A and B as shown in fig. 2.90 (b). as seen. 10 Ω resistor also becomes short – circuited.

ii). Let us now find I_{SC.} The battery sees a parallel combination of 4 Ω and 8 Ω in series with a 4 Ω resistance. Total resistance seen by the battery = 4 + (4 x 8/4 + 80 = 20 / 3 Ω. Hence, I = 20 ÷ 20 /3 = 3 A. This current divides at point C of fig. 2.90 (b). Current going along path CAB gives I_{SC}. Its value = 3 x 4/12 = 1 A

Fig. 2.90 (a, b, c, d, e)

iii). In fig. ©, battery has been removed leaving behind its internal resistance which, in this case, is zero.

Resistance of the network looking into the terminals A and B is

R_{i} = 10 x 10 / 10 + 10 = 5 Ω

iv). Hence, fig.(e) gives the Norton’s equivalent circuit.

v). Now, join the 5 Ω resistance back across terminals A and B. The current flowing through it, obviously is

I_{AB} = 1 x 5 / 10 = 0.5 A

**Example 3;** Using Norton’s theorem, calculate the current flowing through the 15 Ω load resistor in the circuit of fig. 2.91 (a). All resistance values are in ohm.

**Solution;**

A). Short – Circuit Current I_{SC}

As shown in fig. 2.91 (b), terminals A and B have been shorted after removing 15 Ω resistor. We will use superposition theorem to find I_{SC. }

i). When only Current Source is Present

in this case, 30 V battery is replaced by a short – circuit. The 4 A current divides at point D between parallel combination of 4 Ω and 6 Ω. Current through 6 Ωresistor is;

I_{SC} = 4 x 4 / 4 + 6= 1.6 A … from B to A

ii). When only Battery is Present

in this case, current source is replaced by an open – circuit, so that no current flows in the branch CD. The current supplied by the battery constitutes the short – circuit current.

Fig. 2.91 (a, b, c, d)

∴ I_{SC} “= 30 / (4 + 6) = 3 A … from A to B

∴ I_{SC} = I_{SC}” – I_{SC} = 3 – 1.6 = 1.4 A … from A to B

B). Norton’s Parallel Resistance

As seen from fig. 2.91 ©, R_{i }= 4 + 6 = 10 Ω. The 8 Ω resistance does not come into the picture because of an “open” in the branch CD. Fig 2.9 (d) shows the Norton’s equivalent circuit along with the load resistor.

I_{L} = 1.4 x 10 / (10 + 15) = 0.56 A

**Example 4:** Using Norton’s current – source equivalent circuit of the network shown in fig 2.92 (a), find the current that would flow through the resistor R_{2} when it takes the values of 12, 24 and 36 Ω respectively.

**Solution;**

In fig. 2.92 9a), terminals A and B have been short – circuited. Current in the shorted path due to E_{1} is 120 / 40 = 3 A from A to B. Current due to E2 is 180 / 60 = 3A from A to B. Hence, ISC = 6 A. With batteries removed, the resistance of the network when viewed from open – circuited terminals are;

= 40 x 60 / 40 + 60 = 24 Ω

(i). When R_{L} = 12 Ω I_{L }= 6 x 24 (24 + 12) = 4A

(ii). When R_{L} = 24 Ω I_{L} = 6/2 = 3A

(iii). When R_{L} = 36 Ω I_{L} = 6 x 24 / (24 + 36) = 2.4 A

Fig. 2.92 (a, b, c, d)

**Example 5;** Using Norton’s theorem, calculate the current in the 6 Ω resistor in the network of fig. 2.93 (a). All resistances are in ohms.

**Solution;**

Figure 2.93 (a, b, c, d, e)

When the branch containing 6 Ω resistance is short – circuited, the given circuit is reduced to that shown in fig 2.93 9b0 and finally to fig. 2.93 (c). As seen, the 12 A current divides into two unequal parts at point A. The current passing through 4 Ω resistor forms the short – circuit current I_{SC}.

I_{SC} = 12 x 8 / 8 + 4 = 8 A

Resistance R_{i }between points C and D when they are open – circuited, is;

R_{i} = (4 + 8) x (10 + 2) / (4 +8) + (10 + 2) = 6 Ω

It is so because the constant – current source has infinite resistance i.e.; it behaves like an open circuit as shown in fig. 2.93 (d). Hence, Norton’s equivalent circuit is as shown in fig. 2.93 (e). As seen, the current of 8 A is divided equally between the two equal resistances of 6 Ω each. Hence, current through the required 6 Ω resistor is 4 A.

## Delta/ Star Transformation

Complex networks (consisting of many branches) when solved through Kirchhoff’s laws, causes lot of difficulties due to formation of so many simultaneous equations. However, a simple and easy method for solving such complex networks is that delta meshes be replaced with equivalent star systems and star meshes with equivalent delta systems.

Let us suppose that we have been given three resistances R_{12}, R_{23}, and R_{31}, which have been joined to the delta between terminals 1, 2, and 3. This has been illustrated in fig. 2.94 (a). These three resistances joined to delta can be replaced with three resistances R_{1}, R_{2}, and R_{3} connected to star with respect to the corresponding terminals. This has been illustrated in fig. 2.94 (b). If the resistance found between any two terminals is constant or same, then both these arrangements tend to be electrically same. Its detail is as follows;

Figure 2.94

First take delta connection. Two paths exist here between terminals 1 and 2. Resistance of one of the paths is R_{12}, and that of the other is (R_{23} + R_{31}).

Thus, resistance between terminals 1 and 2 tends to be = R_{12} x (R_{23} + R_{31}) / R_{12} + (R_{23} + R_{31})

Now take star connection. Now the mean resistance between terminals 1 and 2 to is (R_{1} + R_{2}). If the resistances found between these two terminals are same both in delta as well as star situation, then

∴ R_{1} + R_{2} = R_{12} x (R_{23} + R_{31}) / R_{12} + (R_{23} +R_{31}) … (1)

Similarly, resistances found between terminals 2 and 3 and terminals 1 and 3 are as follows;

R_{2 }+ R_{3} = R_{23} (R_{31} + R_{12}) / R_{12} + R_{23 }+ R_{31} … (2)

R_{3 }+ R_{1} = R_{31} (R_{12 }+ R_{23}) / R_{12} + R_{23} + R_{31} … (3)

After subtracting equation 2 from equation 1, adding the result thus achieved to equation 3, following values for R_{1}, R_{2}, and R_{3 }are obtained;

R_{1} = R_{12} R_{31} / R_{12 }+ R_{23} + R_{31}; R_{2 }= R_{23} R_{12} / R_{12} + R_{23} + R_{31} and R_{3} = R_{31} R_{23} / R_{12} + R_{23} + R_{31}

## Star / Delta Transformation

If a given star network is desired to be transformed into delta network, this objective can be achieved by means of application of equations 1. 2, and 3 mentioned above. For this purpose, mutually multiply 1 by 2, 2 by 3 and 3 by 1 and then sum them up together. Then, following equations are obtained after abridgement.

R_{12} = R_{1} R_{2} + R_{2} R_{3} + R_{3 }R_{1} / R_{3} = R_{1 }+ R_{2} + R_{1} R_{2} / R_{3}

R_{23} = R_{1} R_{2} + R_{2} R_{3} + R_{3} R_{1} / R_{1} = R_{2} + R_{3} + R_{2} R_{3} / R_{1}

R_{31} = R_{1} R_{2} + R_{2} R_{3} + R_{3 }R_{1} / R_{2} = R_{1} + R_{3} + R_{1} R_{3 }/ R_{2}

## Calculations About Star–Delta and Delta–Star Transformation

**Example 1;** Find the input resistance of the circuit between the points A and B of fig. 2.95 (a) below;

Fig. 2.95 (a, b, c)

**Solution;**

For finding R_{AB}, we will convert the delta CDE of fig. 2.95 (a) into its equivalent star as shown in fig. 2.95 (b).

R_{CS} = 8 x 4 / 4 + 6 + 8 = 32 / 18 = 16 / 9 Ω; R_{ES} =8 x 6 / 18 = 24 / 9 Ω; R_{DS} = 6 x 4 / 18 = 12 / 9 Ω

The two parallel resistances between S and B can be reduced to a single resistance of 35 / 9 Ω.

As seen from fig. (c), R_{AB} = 4 + 16 / 9 + 35 / 9 = 87 / 9 Ω

**Example 2;** Calculate the equivalent resistance between the terminals A and B in the network shown in fig. 2.96 (a) below;

Fig. 2.96 (a, b, c)

**Solution;**

The given circuit can be redrawn as shown in fig. 2.96 (b). When the delta BCD is converted to its equivalent start, the circuit becomes as shown in fig. 2.96 (c).

Each arm of the delta has a resistance of 10 Ω. Hence, each arm of the equivalent star has a resistance = 10 x 10 / 30 = 10 /3 Ω. As seen, there are two parallel paths between points A and n, each having a resistance of [10 + 10/3] = 40 / 3 Ω. Their combined resistance is 20 / 3 Ω.

Hence, R_{AB} = (20 / 3) + 10 / 3 = 10 Ω

**Example 3;** Find the equivalent resistance of the complex network illustrated between points A and D by first using a delta – star transformation on mesh ABC.

Figure 2.97

**Solution;**

Using delta – star transformation on mesh ABC of the given circuit the simplified circuit 2.98 (a) results;

R_{A} = 30 x 10 / 30 + 10 + 50 = 3.33 Ω

Fig 2.98 (a, b)

R_{B} = 50 x 10 / 30 + 10 + 50 = 5.56 Ω

R_{C} = 50 x 30 / 50 + 30 + 10 = 16.67 Ω

The circuit (a) therefore reduces to that of (b) equivalent resistance between A and D is;

3.33 + [55.67 x 25.56/ 56.67 + 25.56] = 20 .94 Ω

**Example 4;** Reduce the circuit shown at 2.99 (a) to that of (b)

**Solution;**

A star – mesh transformation is first carried out on mesh JCDF giving the circuit shown at 2.99 ©.

R_{A} = 40 x 10 / 40 + 10 + 50 = 4 Ω

R_{B} = 50 x 40 / 100 = 20 Ω

R_{C} = 50 x 10 / 100 = 5 Ω

Fig. 299 (c)

Carrying out a further transformation on mesh ADJ, the circuit at 9d) results;

R_{a} = 2 x9 / 2 + 9 + 5 = 9 / 8 Ω

R_{b} = 2 x 5 / 16 = 8/ 5 Ω

R_{c} = 9 x 5 / 16 = 45 / 16 Ω

R_{a} + R_{c} + 20 = (9/ 8) + (45 / 16) + 20 = 23 .93 Ω, the result is equivalent to ©.

Fig. 2.99 (d, e)

**Example 5;** A network of conductors is arranged as shown. Determine the resistance between A and C by using delta – star and star – delta transformations, or symmetry argument.

Fig. 2.100

**Solution;**

If star – connected 1 Ω resistors are replaced by an equivalent delta connection circuit, below is obtained;

If the meshes AED, ECF, and DFB are replaced by equivalent star connections circuit (b) results and this reduces the circuit (c)

Mesh connection OPQ may be replaced by an equivalent star to give circuit (d)

Resistance between A and C = 2 [ 2 x 1 ^{2/3}] Ω = 3 ^{1/3} Ω

Fig. 2.101

**Alternative Solution**

Replace the star between B and C and the star point of the given circuit by a delta connection. This gives circuit 2.102 (a) below;

Fig. 2.102

Repeating this procedure on the star between A, B and the star point and A, C and the star point gives circuit (b) which simplifies the circuit.

Resistance between A and C = (5 x 10) / (5 + 10) Ω = 3 1/ 3 Ω

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