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.
Let’s see how this method works on the same example problem:
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 B_{1}, R_{1}, and R_{2} will be the first while the loop formed by B_{2}, R_{2}, and R_{3} 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 I_{1} and I_{2} are both going “up” through resistor R_{2}, 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.
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, 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 I_{1} and I_{2} as the current through resistor R_{2}. This is because mesh currents I_{1} and I_{2} are going the same direction through R_{2}, and thus complement each other. Distributing the coefficient of 2 to the I_{1} and I_{2} terms, and then combining I_{1} 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:
Now, with two equations, we can use one of several methods to mathematically solve for the unknown currents I_{1} and I_{2}:
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 I_{2} means that we initially assumed the direction of current was incorrect. In actuality, I_{2} 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 R_{2} and R_{3} due to current I_{2}. From here, we can say that the current through R_{1} is 5 amps, with the voltage drop across R_{1} 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 R_{3} 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 R_{2}?
Mesh current I_{1} is going “down” through R_{2}, while mesh current I_{2} is going “up” through R_{2}. To determine the actual current through R_{2}, we must see how mesh currents I_{1} and I_{2} interact (in this case they’re in opposition), and algebraically add them to arrive at a final value. Since I_{1} is going “down” at 5 amps, and I_{2} is going “up” at 1 amp, the real current through R_{2} must be a value of 4 amps, going “down”:
A current of 4 amps through R_{2}‘s resistance of 2 Ω gives us a voltage drop of 8 volts (E=IR), positive on the top and negative on the bottom.
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 (I_{1} through I_{5}). 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).
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 R_{1}/R_{4} and R_{2}/R_{5} are unequal, we know that there will be the voltage across resistor R_{3}, 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 (I_{1} through I_{6}), 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.
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 I_{1} nor I_{2} goes through the battery. So, we must add a third mesh current, I_{3}:
Here, I have chosen I_{3} to loop from the bottom side of the battery, through R_{4}, through R_{1}, and back to the top side of the battery. This is not the only path I could have chosen for I_{3}, but it seems the simplest.
Now, we must label the resistor voltage drop polarities, following each of the assumed currents’ directions:
Notice something very important here: at resistor R_{4}, the polarities for the respective mesh currents do not agree. This is because those mesh currents (I_{2} and I_{3}) are going through R_{4} 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 R_{4} current clash. (See Example below)
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 R_{3}, with a value of 100 Ω, has its voltage drop represented in the above KVL equation by the expression 100(I_{1} + I_{2}), since both currents I_{1} and I_{2} go through R_{3} from right to left. The same may be said for resistor R_{1}, with its voltage drop expression shown as 150(I_{1} + I_{3}), since both I_{1} and I_{3} 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 R_{4}. 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 R_{4}‘s value of 300 Ω multiplied by the difference between I_{2} and I_{3} (I_{2} - I_{3}). 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(I_{2} - I_{3}) does not mean the same thing as 300(I_{3} - I_{2}). I chose to write 300(I_{2} - I_{3}) because I was thinking first of I_{2}‘s effect (creating a positive voltage drop, measuring with an imaginary voltmeter across R_{4}, red lead on the bottom and black lead on the top), and secondarily of I_{3}‘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 I_{3}‘s effect first and I_{2}‘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(I_{3} - I_{2}). Note that this expression is mathematically equivalent to the first one: +300(I_{2} - I_{3}).
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 I_{1}, I_{2}, and I_{3} using whatever simultaneous equation method we prefer:
Example: Use Octave to find the solution for I_{1}, I_{2}, and I_{3} 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: I_{1}, _{2}, and I_{3} 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 I_{1} 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 I_{3} that does not produce a conflicting polarity on any resistor compared to I_{1} or I_{2}. R_{4} was the offending component. (b) Find values for I_{1}, I_{2}, and I_{3}. (c) Find the five resistor currents and compare them to the previous values.
Solution:
(a) Route I_{3} through R_{5}, R_{3, }and R_{1} as shown:
Note that the conflicting polarity on R_{4} 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 I_{1}, I_{2}, and I_{3} are the same (I_{2}) 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:
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:
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 I_{3}clockwise 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: I_{1}, I_{2}, and I_{3} 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 I_{1} and I_{2} in the pair of equations. Tracing the assumed clockwise loop 1 current through B_{1} 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 B_{2}, a voltage drop. Thus, the sign of B_{2} 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 I_{1} = 5 A, and I_{2} = 1 A are both positive. They both flow in the direction of the clockwise loops. This compares with previous results.
Summary:
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