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Power Supply 123: Direct Current (DC) Chapter 10DC NETWORK ANALYSISMesh current by inspectionWe take a second look at the “mesh current method” with all the currents runing counterclockwise (ccw). 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 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.
+(sum of R's loop 1)I_{1}  (common R loop 12)I_{2}  (common R loop 13)I_{3} = E_{1} (common R loop 12)I_{1} + (sum of R's loop 2)I_{2}  (common R loop 23)I_{3} = E_{2} (common R loop 13)I_{1}  (common R loop 23)I_{2} + (sum of R's loop 3)I_{3} = E_{3}
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, counterclockwise. 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 current direction. The I_{3}counterclockwise current traverses the 24V source from (+) to (). This is a voltage rise for electron 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 counterclockwise 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 loopcurrent, meshcurrent method in those text will run the assumed mesh currents clockwise.[aef] 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 counterclockwise 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 counter clockwise 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 counterclockwise loops. This compares with previous results.


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