Capacitor

Inductor

What is Norton’s Theorem?

Norton’s Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load. Just as with Thevenin’s Theorem, the qualification of “linear” is identical to that found in the Superposition Theorem: all underlying equations must be linear (no exponents or roots).

Simplifying Linear Circuits

Contrasting our original example circuit against the Norton equivalent: it looks something like this:

. . . after Norton conversion . . .

Remember that a current source is a component whose job is to provide a constant amount of current, outputting as much or as little voltage necessary to maintain that constant current.

Thevenin’s Theorem vs. Norton’s Theorem

As with Thevenin’s Theorem, everything in the original circuit except the load resistance has been reduced to an equivalent circuit that is simpler to analyze. Also similar to Thevenin’s Theorem are the steps used in Norton’s Theorem to calculate the Norton source current (I_Norton) and Norton resistance (R_Norton).

Identify The Load Resistance

As before, the first step is to identify the load resistance and remove it from the original circuit:

Find The Norton Current

Then, to find the Norton current (for the current source in the Norton equivalent circuit), place a direct wire (short) connection between the load points and determine the resultant current. Note that this step is exactly opposite the respective step in Thevenin’s Theorem, where we replaced the load resistor with a break (open circuit):

With zero voltage dropped between the load resistor connection points, the current through R₁ is strictly a function of B1‘s voltage and R₁‘s resistance: 7 amps (I=E/R). Likewise, the current through R₃ is now strictly a function of B₂‘s voltage and R₃‘s resistance: 7 amps (I=E/R). The total current through the short between the load connection points is the sum of these two currents: 7 amps + 7 amps = 14 amps. This figure of 14 amps becomes the Norton source current (I_Norton) in our equivalent circuit:

Find Norton Resistance

Remember, the arrow notation for current source points in the direction of conventional current flow. To calculate the Norton resistance (R_Norton), we do the exact same thing as we did for calculating Thevenin resistance (R_Thevenin): take the original circuit (with the load resistor still removed), remove the power sources (in the same style as we did with the Superposition Theorem: voltage sources replaced with wires and current sources replaced with breaks), and figure total resistance from one load connection point to the other:

Now our Norton equivalent circuit looks like this:

Determine The Voltage Across The Load Resistor

If we re-connect our original load resistance of 2 Ω, we can analyze the Norton circuit as a simple parallel arrangement:

As with the Thevenin equivalent circuit, the only useful information from this analysis is the voltage and current values for R₂; the rest of the information is irrelevant to the original circuit. However, the same advantages seen with Thevenin’s Theorem apply to Norton’s as well: if we wish to analyze load resistor voltage and current over several different values of load resistance, we can use the Norton equivalent circuit, again and again, applying nothing more complex than simple parallel circuit analysis to determine what’s happening with each trial load.

REVIEW:

• Norton’s Theorem is a way to reduce a network to an equivalent circuit composed of a single current source, parallel resistance, and parallel load.
• Steps to follow for Norton’s Theorem:
  • Find the Norton source current by removing the load resistor from the original circuit and calculating the current through a short (wire) jumping across the open connection points where the load resistor used to be.
  • 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.
  • 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.
  • Analyze voltage and current for the load resistor following the rules for parallel circuits.

Superposition theorem is one of those strokes of genius that takes a complex subject and simplifies it in a way that makes perfect sense. A theorem like Millman’s certainly works well, but it is not quite obvious why it works so well. Superposition, on the other hand, is obvious.

Series/Parallel Analysis

The strategy used in the Superposition Theorem is to eliminate all but one source of power within a network at a time, using series/parallel analysis to determine voltage drops (and/or currents) within the modified network for each power source separately. Then, once voltage drops and/or currents have been determined for each power source working separately, the values are all “superimposed” on top of each other (added algebraically) to find the actual voltage drops/currents with all sources active. Let’s look at our example circuit again and apply Superposition Theorem to it:

Since we have two sources of power in this circuit, we will have to calculate two sets of values for voltage drops and/or currents, one for the circuit with only the 28-volt battery in effect. . .

. . . and one for the circuit with only the 7-volt battery in effect:

When re-drawing the circuit for series/parallel analysis with one source, all other voltage sources are replaced by wires (shorts), and all current sources with open circuits (breaks). Since we only have voltage sources (batteries) in our example circuit, we will replace every inactive source during analysis with a wire.

Analyzing the circuit with only the 28-volt battery, we obtain the following values for voltage and current:

Analyzing the circuit with only the 7-volt battery, we obtain another set of values for voltage and current:

By Superimposing

When superimposing these values of voltage and current, we have to be very careful to consider polarity (of the voltage drop) and direction (of the current flow), as the values have to be added algebraically.

Applying these superimposed voltage figures to the circuit, the end result looks something like this:

Currents add up algebraically as well and can either be superimposed as done with the resistor voltage drops or simply calculated from the final voltage drops and respective resistances (I=E/R). Either way, the answers will be the same. Here I will show the superposition method applied to current:

Once again applying these superimposed figures to our circuit:

Prerequisites for the Superposition Theorem

Quite simple and elegant, don’t you think? It must be noted, though, that the Superposition Theorem works only for circuits that are reducible to series/parallel combinations for each of the power sources at a time (thus, this theorem is useless for analyzing an unbalanced bridge circuit), and it only works where the underlying equations are linear (no mathematical powers or roots). The requisite of linearity means that Superposition Theorem is only applicable for determining voltage and current, not power!!! Power dissipations, being nonlinear functions, do not algebraically add to an accurate total when only one source is considered at a time. The need for linearity also means this Theorem cannot be applied in circuits where the resistance of a component changes with voltage or current. Hence, networks containing components like lamps (incandescent or gas-discharge) or varistors could not be analyzed.

Another prerequisite for Superposition Theorem is that all components must be “bilateral,” meaning that they behave the same with electrons flowing in either direction through them. Resistors have no polarity-specific behavior, and so the circuits we’ve been studying so far all meet this criterion.

The Superposition Theorem finds use in the study of alternating current (AC) circuits, and semiconductor (amplifier) circuits, where sometimes AC is often mixed (superimposed) with DC. Because AC voltage and current equations (Ohm’s Law) are linear just like DC, we can use Superposition to analyze the circuit with just the DC power source, then just the AC power source, combining the results to tell what will happen with both AC and DC sources in effect. For now, though, Superposition will suffice as a break from having to do simultaneous equations to analyze a circuit.

REVIEW:

• The Superposition Theorem states that a circuit can be analyzed with only one source of power at a time, the corresponding component voltages and currents algebraically added to find out what they’ll do with all power sources in effect.
• To negate all but one power source for analysis, replace any source of voltage (batteries) with a wire; replace any current source with an open (break).

The Mesh-Current Method, also known as the Loop Current Method, is quite similar to the Branch Current method in that it uses simultaneous equations, Kirchhoff’s Voltage Law, and Ohm’s Law to determine unknown currents in a network. It differs from the Branch Current method in that it does not use Kirchhoff’s Current Law, and it is usually able to solve a circuit with less unknown variables and less simultaneous equations, which is especially nice if you’re forced to solve without a calculator.

Mesh Current, conventional method

Let’s see how this method works on the same example problem:

Identify Loops

The first step in the Mesh Current method is to identify “loops” within the circuit encompassing all components. In our example circuit, the loop formed by B₁, R₁, and R₂ will be the first while the loop formed by B₂, R₂, and R₃ will be the second. The strangest part of the Mesh Current method is envisioning circulating currents in each of the loops. In fact, this method gets its name from the idea of these currents meshing together between loops like sets of spinning gears:

The choice of each current’s direction is entirely arbitrary, just as in the Branch Current method, but the resulting equations are easier to solve if the currents are going the same direction through intersecting components (note how currents I₁ and I₂ are both going “up” through resistor R₂, where they “mesh,” or intersect). If the assumed direction of a mesh current is wrong, the answer for that current will have a negative value.

Label the Voltage Drop Polarities

The next step is to label all voltage drop polarities across resistors according to the assumed directions of the mesh currents. Remember that the “upstream” end of a resistor will always be negative, and the “downstream” end of a resistor positive with respect to each other, since electrons are negatively charged. The battery polarities, of course, are dictated by their symbol orientations in the diagram, and may or may not “agree” with the resistor polarities (assumed current directions):

Using Kirchhoff’s Voltage Law, we can now step around each of these loops, generating equations representative of the component voltage drops and polarities. As with the Branch Current method, we will denote a resistor’s voltage drop as the product of the resistance (in ohms) and its respective mesh current (that quantity being unknown at this point). Where two currents mesh together, we will write that term in the equation with resistor current being the sum of the two meshing currents.

Tracing the Left Loop of the Circuit with Equations

Tracing the left loop of the circuit, starting from the upper-left corner and moving counter-clockwise (the choice of starting points and directions is ultimately irrelevant), counting polarity as if we had a voltmeter in hand, red lead on the point ahead and black lead on the point behind, we get this equation:

Notice that the middle term of the equation uses the sum of mesh currents I₁ and I₂ as the current through resistor R₂. This is because mesh currents I₁ and I₂ are going the same direction through R₂, and thus complement each other. Distributing the coefficient of 2 to the I₁ and I₂ terms, and then combining I₁ terms in the equation, we can simplify as such:

At this time we have one equation with two unknowns. To be able to solve for two unknown mesh currents, we must have two equations. If we trace the other loop of the circuit, we can obtain another KVL equation and have enough data to solve for the two currents. Creature of habit that I am, I’ll start at the upper-left-hand corner of the right loop and trace counter-clockwise:

Simplifying the equation as before, we end up with:

Solving For The Unknown

Now, with two equations, we can use one of several methods to mathematically solve for the unknown currents I₁ and I₂:

Redraw Circuit

Knowing that these solutions are values for mesh currents, not branch currents, we must go back to our diagram to see how they fit together to give currents through all components:

The solution of -1 amp for I₂ means that we initially assumed the direction of current was incorrect. In actuality, I₂ is flowing in a counter-clockwise direction at a value of (positive) 1 amp:

This change of current direction from what was first assumed will alter the polarity of the voltage drops across R₂ and R₃ due to current I₂. From here, we can say that the current through R₁ is 5 amps, with the voltage drop across R₁ being the product of current and resistance (E=IR), 20 volts (positive on the left and negative on the right).

Also, we can safely say that the current through R₃ is 1 amp, with a voltage drop of 1 volt (E=IR), positive on the left and negative on the right. But what is happening at R₂?

Mesh current I₁ is going “down” through R₂, while mesh current I₂ is going “up” through R₂. To determine the actual current through R₂, we must see how mesh currents I₁ and I₂ interact (in this case they’re in opposition), and algebraically add them to arrive at a final value. Since I₁ is going “down” at 5 amps, and I₂ is going “up” at 1 amp, the real current through R₂ must be a value of 4 amps, going “down”:

A current of 4 amps through R₂‘s resistance of 2 Ω gives us a voltage drop of 8 volts (E=IR), positive on the top and negative on the bottom.

Advantage of Mesh Current Analysis

The primary advantage of Mesh Current analysis is that it generally allows for the solution of a large network with fewer unknown values and fewer simultaneous equations. Our example problem took three equations to solve the Branch Current method and only two equations using the Mesh Current method. This advantage is much greater as networks increase in complexity:

To solve this network using Branch Currents, we’d have to establish five variables to account for each and every unique current in the circuit (I₁ through I₅). This would require five equations for the solution, in the form of two KCL equations and three KVL equations (two equations for KCL at the nodes, and three equations for KVL in each loop):

I suppose if you have nothing better to do with your time than to solve for five unknown variables with five equations, you might not mind using the Branch Current method of analysis for this circuit. For those of us who have better things to do with our time, the Mesh Current method is a whole lot easier, requiring only three unknowns and three equations to solve:

Less equations to work with is a decided advantage, especially when performing a simultaneous equation solution by hand (without a calculator).

Unbalanced Wheatstone Bridge

Another type of circuit that lends itself well to Mesh Current is the unbalanced Wheatstone Bridge. Take this circuit, for example:

Since the ratios of R₁/R₄ and R₂/R₅ are unequal, we know that there will be the voltage across resistor R₃, and some amount of current through it. As discussed at the beginning of this chapter, this type of circuit is irreducible by normal series-parallel analysis, and may only be analyzed by some other method.

We could apply the Branch Current method to this circuit, but it would require six currents (I₁ through I₆), leading to a very large set of simultaneous equations to solve. Using the Mesh Current method, though, we may solve for all currents and voltages with much fewer variables.

Draw Mesh

The first step in the Mesh Current method is to draw just enough mesh currents to account for all components in the circuit. Looking at our bridge circuit, it should be obvious where to place two of these currents:

The directions of these mesh currents, of course, is arbitrary. However, two mesh currents are not enough in this circuit, because neither I₁ nor I₂ goes through the battery. So, we must add a third mesh current, I₃:

Here, I have chosen I₃ to loop from the bottom side of the battery, through R₄, through R₁, and back to the top side of the battery. This is not the only path I could have chosen for I₃, but it seems the simplest.

Label The Resistor Voltage Drop Polarities

Now, we must label the resistor voltage drop polarities, following each of the assumed currents’ directions:

Notice something very important here: at resistor R₄, the polarities for the respective mesh currents do not agree. This is because those mesh currents (I₂ and I₃) are going through R₄ in different directions. This does not preclude the use of the Mesh Current method of analysis, but it does complicate it a bit. Though later, we will show how to avoid the R₄ current clash. (See Example below)

Using KVL

Generating a KVL equation for the top loop of the bridge, starting from the top node and tracing in a clockwise direction:

In this equation, we represent the common directions of currents by their sums through common resistors. For example, resistor R₃, with a value of 100 Ω, has its voltage drop represented in the above KVL equation by the expression 100(I₁ + I₂), since both currents I₁ and I₂ go through R₃ from right to left. The same may be said for resistor R₁, with its voltage drop expression shown as 150(I₁ + I₃), since both I₁ and I₃ go from bottom to top through that resistor, and thus work together to generate its voltage drop.

Generating a KVL equation for the bottom loop of the bridge will not be so easy since we have two currents going against each other through resistor R₄. Here is how I do it (starting at the right-hand node, and tracing counter-clockwise):

Note how the second term in the equation’s original form has resistor R₄‘s value of 300 Ω multiplied by the difference between I₂ and I₃ (I₂ – I₃). This is how we represent the combined effect of two mesh currents going in opposite directions through the same component. Choosing the appropriate mathematical signs is very important here: 300(I₂ – I₃) does not mean the same thing as 300(I₃ – I₂). I chose to write 300(I₂ – I₃) because I was thinking first of I₂‘s effect (creating a positive voltage drop, measuring with an imaginary voltmeter across R₄, red lead on the bottom and black lead on the top), and secondarily of I₃‘s effect (creating a negative voltage drop, red lead on the bottom and black lead on the top). If I had thought in terms of I₃‘s effect first and I₂‘s effect secondarily, holding my imaginary voltmeter leads in the same positions (red on the bottom and black on top), the expression would have been -300(I₃ – I₂). Note that this expression is mathematically equivalent to the first one: +300(I₂ – I₃).

Well, that takes care of two equations, but I still need a third equation to complete my simultaneous equation set of three variables, three equations. This third equation must also include the battery’s voltage, which up to this point does not appear in either two of the previous KVL equations. To generate this equation, I will trace a loop again with my imaginary voltmeter starting from the battery’s bottom (negative) terminal, stepping clockwise (again, the direction in which I step is arbitrary, and does not need to be the same as the direction of the mesh current in that loop):

Solving for the Currents

Solving for I₁, I₂, and I₃ using whatever simultaneous equation method we prefer:

Example: Use Octave to find the solution for I₁, I₂, and I₃ from the above-simplified form of equations.

Solution: In Octave, an open-source Matlab® clone, enter the coefficients into the A matrix between square brackets with column elements comma-separated, and rows semicolon-separated. Enter the voltages into the column vector: b. The unknown currents: I₁, ₂, and I₃ are calculated by the command: x=A\b. These are contained within the x column vector.

 
octave:1>A = [300,100,150;100,650,-300;-150,300,-450]
        A =
          300  100  150
          100  650  -300
          -150 300  -450
 
        octave:2> b = [0;0;-24]
        b =
          0
          0
          -24
               
        octave:3> x = A\b
        x =
          -0.093793
           0.077241
           0.136092
 

The negative value arrived at for I₁ tells us that the assumed direction for that mesh current was incorrect. Thus, the actual current values through each resistor are as such:

Calculating voltage drops across each resistor:

A SPICE simulation confirms the accuracy of our voltage calculations:

unbalanced wheatstone bridge  
v1 1 0 
r1 1 2 150     
r2 1 3 50      
r3 2 3 100     
r4 2 0 300     
r5 3 0 250     
.dc v1 24 24 1 
.print dc v(1,2) v(1,3) v(3,2) v(2,0) v(3,0)   
.end
v1          v(1,2)     v(1,3)      v(3,2)     v(2)       v(3)           
2.400E+01   6.345E+00  4.690E+00   1.655E+00  1.766E+01  1.931E+01

Example:

(a) Find a new path for current I₃ that does not produce a conflicting polarity on any resistor compared to I₁ or I₂. R₄ was the offending component. (b) Find values for I₁, I₂, and I₃. (c) Find the five resistor currents and compare them to the previous values.

Solution:

(a) Route I₃ through R₅, R_3, and R₁ as shown:

Note that the conflicting polarity on R₄ has been removed. Moreover, none of the other resistors have conflicting polarities.

(b) Octave, an open source (free) Matlab clone, yields a mesh current vector at “x”:

 
octave:1> A = [300,100,250;100,650,350;-250,-350,-500]
        A =
          300  100  250
          100  650  350
          -250  -350  -500
      
        octave:2> b = [0;0;-24]
        b =
          0
          0
        -24
              
        octave:3> x = A\b
        x =
          -0.093793
          -0.058851
           0.136092

Not all currents I₁, I₂, and I₃ are the same (I₂) as the previous bridge because of different loop paths However, the resistor currents compare to the previous values:

 
        IR1 = I1 + I3 = -93.793 ma + 136.092 ma = 42.299 ma
        IR2 = I1 = -93.793 ma
        IR3 = I1 + I2 + I3 = -93.793 ma -58.851 ma  + 136.092 ma =  -16.552 ma
        IR4 = I2 = -58.851 ma
        IR5 = I2 + I3 = -58.851 ma  + 136.092 ma = 77.241 ma

Since the resistor currents are the same as the previous values, the resistor voltages will be identical and need not be calculated again.

REVIEW:

• Steps to follow for the “Mesh Current” method of analysis:
• (1) Draw mesh currents in loops of the circuit, enough to account for all components.
• (2) Label resistor voltage drop polarities based on assumed directions of mesh currents.
• (3) Write KVL equations for each loop of the circuit, substituting the product IR for E in each resistor term of the equation. Where two mesh currents intersect through a component, express the current as of the algebraic sum of those two mesh currents (i.e. I₁ + I₂) if the currents go in the same direction through that component. If not, express the current as of the difference (i.e. I₁ – I₂).
• (4) Solve for unknown mesh currents (simultaneous equations).
• (5) If any solution is negative, then the assumed current direction is wrong!
• (6) Algebraically add mesh currents to find current components sharing multiple mesh currents.
• (7) Solve for voltage drops across all resistors (E=IR).

Mesh current by inspection

We take a second look at the “mesh current method” with all the currents running clockwise (cw). The motivation is to simplify the writing of mesh equations by ignoring the resistor voltage drop polarity. Though, we must pay attention to the polarity of voltage sources with respect to the assumed current direction. The sign of the resistor voltage drops will follow a fixed pattern.

If we write a set of conventional mesh-current equations for the circuit below, where we do pay attention to the signs of the voltage drop across the resistors, we may rearrange the coefficients into a fixed pattern:

Once rearranged, we may write equations by inspection. The signs of the coefficients follow a fixed pattern in the pair above or the set of three in the rules below.

Mesh current rules:

• This method assumes conventional current flow voltage sources. Replace any current source in parallel with a resistor with an equivalent voltage source in series with an equivalent resistance.
• Ignoring current direction or voltage polarity on resistors, draw counterclockwise current loops traversing all components. Avoid nested loops.
• Write voltage-law equations in terms of unknown currents: I₁, I₂, and I₃. Equation 1 coefficient 1, equation 2, coefficient 2, and equation 3 coefficient 3 are the positive sums of resistors around the respective loops.
• All other coefficients are negative, representative of the resistance common to a pair of loops. Equation 1 coefficient 2 is the resistor common to loops 1 and 2, coefficient 3 the resistor common to loops 1 an 3. Repeat for other equations and coefficients.
• +(sum of R’s loop 1)I1 – (common R loop 1-2)I2 – (common R loop 1-3)I3   = E1
  -(common R loop 1-2)I1 + (sum of R’s loop 2)I2 – (common R loop 2-3)I3   = E2
  -(common R loop 1-3)I1 – (common R loop 2-3)I2 + (sum of R’s loop 3)I3   = E3
• The right-hand side of the equations is equal to an electron current flow voltage source. A voltage rise with respect to the counterclockwise assumed current is positive, and 0 for no voltage source.
• Solve equations for mesh currents: I₁, I₂, and I3. Solve for currents through individual resistors with KCL. Solve for voltages with Ohms Law and KVL.

While the above rules are specific for a three mesh circuit, the rules may be extended to smaller or larger meshes. The figure below illustrates the application of the rules. The three currents are all drawn in the same direction, clockwise. One KVL equation is written for each of the three loops. Note that there is no polarity drawn on the resistors. We do not need it to determine the signs of the coefficients. Though we do need to pay attention to the polarity of the voltage source with respect to the current direction. The I₃clockwise current flows out from the (+) positive terminal of the l24V source then returns to the (-) terminal. This is a voltage rise for conventional current flow. Therefore, the third equation right-hand side is -24V.

In Octave, enter the coefficients into the A matrix with column elements comma-separated, and rows semicolon-separated. Enter the voltages into the column vector b. Solve for the unknown currents: I₁, I₂, and I₃ with the command: x=A\b. These currents are contained within the x column vector. The positive values indicate that the three mesh currents all flow in the assumed clockwise direction.

 
octave:2> A=[300,-100,-150;-100,650,-300;-150,-300,450]
           A =
             300  -100  -150
             -100  650  -300
             -150  -300  450

           octave:3> b=[0;0;24]
           b =
              0
              0
             24

           octave:4> x=A\b
           x =
             0.093793
             0.077241
             0.136092
 

The mesh currents match the previous solution by a different mesh current method. The calculation of resistor voltages and currents will be identical to the previous solution. No need to repeat here.

Note that electrical engineering texts are based on conventional current flow. The loop-current, mesh-current method in those texts will run the assumed mesh currents clockwise. The conventional current flows out the (+) terminal of the battery through the circuit, returning to the (-) terminal. A conventional current-voltage rise corresponds to tracing the assumed current from (-) to (+) through any voltage sources.

One more example of a previous circuit follows. The resistance around loop 1 is 6 Ω, around loop 2: 3 Ω. The resistance common to both loops is 2 Ω. Note the coefficients of I₁ and I₂ in the pair of equations. Tracing the assumed clockwise loop 1 current through B₁ from (+) to (-) corresponds to an electron current flow voltage rise.

Thus, the sign of the 28 V is positive. The loop 2 counterclockwise assumed current traces (-) to (+) through B₂, a voltage drop. Thus, the sign of B₂ is negative, -7 in the 2nd mesh equation. Once again, there are no polarity markings on the resistors. Nor do they figure into the equations.

The currents I₁ = 5 A, and I₂ = 1 A are both positive. They both flow in the direction of the clockwise loops. This compares with previous results.

Summary:

• The modified mesh-current method avoids having to determine the signs of the equation coefficients by drawing all mesh currents clockwise for conventional current flow.
• However, we do need to determine the sign of any voltage sources in the loop. The voltage source is positive if the assumed ccw current flows with the battery (source). The sign is negative if the assumed ccw current flows against the battery.
• See the rules above for details.