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Mesh current methodAnalysis, Mesh Current method Mesh Current analysisThe Mesh 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. Voltage polarityThe 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 upperleft corner and moving counterclockwise (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 upperleft hand corner of the right loop and trace counterclockwise: 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 our initially assumed direction of current was incorrect. In actuality, I_{2} is flowing in a counterclockwise 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 "up" through R_{2}, while mesh current I_{2} is going "down" 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 "up" at 5 amps, and I_{2} is going "down" at 1 amp, the real current through R_{2} must be a value of 4 amps, going "up:" 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 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 simultaneous equation solution by hand (without a calculator). Wheatstone bridge, unbalancedAnother 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 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 seriesparallel 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 is 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. Normally, we try to avoid this when establishing our mesh current directions, but in a bridge circuit it is unavoidable: two of the mesh currents will inevitably clash through a component. This does not preclude the use of the Mesh Current method of analysis, but it does complicate it a bit. 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 righthand node, and tracing counterclockwise): 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 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: 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 is as such: Calculating voltage drops across each resistor: A SPICE simulation will confirm 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


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