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3.4 Electrical networks One application of linear equations is to electrical networks containing voltage sources and resistors. Here is a simple electrical network. Example 1. This network consists of three lines that are connected at two nodes. The positive direction in each line is indicated by an arrow and the currents in the lines are labeled i1, i2 and i3. Each line has a node #1 voltage source with the positive and negative terminals indicated by + i1 i2 i3 and -. The voltage of each voltage + sources are labeled E1, E2 and E3. E1 = 5v E2 = 10v E3 = 8v Each line has a resistor with + + resistances labeled by R1, R2 and R3. line #1 line #2 line #3 In this network values are given for R1 = 4Ω R2 = 2Ω R3 = 6Ω the voltages and resistances and the problem is to find the currents. node #2 By way of background, here is some information on these three types of electrical components. i. Lines: conducting paths through which current can flow. Current is the flow of charged particles in a line, mostly electrons. The current i in the line is the rate at which charge is crossing a cross section in the line. Positive charges moving in the positive direction in the line count positively and those moving in the opposite direction count negatively. For negative charges it is just the reverse. To determine the sign of the current, it is necessary to choose a positive direction in the line. Charge is measured in coulombs and current is measured in amperes. One ampere of current flowing in a wire means there is a net flow of one coulomb of charge past any cross section in one second. The different lines are connected at nodes. ii. Voltage sources: these include batteries, power supplies and generators. They create electrical fields that cause charged particles in a line to move and hence produce a current in the line. Voltage sources have two terminals, one is the positive terminal designated by + and the other the negative terminal designated by -. The voltage E of a voltage source is a measure of the strength of the voltage source. More precisely, it is the change in electrical potential as one goes from the negative terminal to the positive terminal. More generally, for every pair of points p and q in space there is a potential difference as one goes from p to q. Typically, as p or q moves along lines, the potential difference between p and q doesn’t change except when p or q move through a circuit element such as a 3.4 - 1 resistor. Voltage is measured in volts. If V is the potential difference between two points, then the electric field transfers an amount of energy equal to QV to a particle with charge Q that moves between these two points. iii. Resistors: these impede the flow of current in a line. We shall only consider linear resistors that obey Ohm’s law which says V = iR where V is the drop in potential as one goes through the resistor in the direction of the current, i is the current in the line and R is a constant called the resistance of the resistor. Resistance is measured in Ohms. A resistor with resistance of R ohms has the property that in order to make a current of i amperes flow through the resistor it is necessary to supply a potential difference of iR to the two ends of the resistor. In order to get equations that allow us to find the currents in the lines, we use two electrical laws, namely Kirchoff’s current law and Kirchoff’s voltage law. Kirchoff’s current law states the following. At any node, the sum of the currents going into the node is the sum of the currents going out of the node. or in’s = out’s Kirchoff’s current law is just a restatement of the principle of conservation of charge at the node. If we apply this to node #1 of the above network we get the equation i3 = i1 + i2 or i1 + i2 - i3 = 0 We get the same equation if we apply the current law at node #2. Kirchoff’s voltage law states the following. In any loop of the network, the sum of the voltage rises in the loop is the sum of the voltage drops in the loop. or up’s = down’s 3.4 - 2 Kirchoff’s voltage law is a consequence of the fact that the sum of the voltage changes as you go around the loop is zero. To apply this, we have to pick out some loops in the network and pick out directions to traverse the loops. One loop is at the right. If we go around this loop counterclockwise and apply the voltage law to this loop we get the equation i1 i2 - E1 = 5v + E2 = 10v + line #1 E1 + i2R2 = i1R1 + E2 or line #2 R1 = 4Ω R2 = 2Ω 5 + 2i2 = 4i1 + 10 or 4i1 - 2i2 = -5 The second loop is at the right. If we go around this loop counterclockwise and apply the voltage law to this loop we get the equation i2 i3 - + E2 = 10v E2 + E3 = i2R2 + i2R2 + line #2 or 10 + 8 = 2i2 + 6i3 R2 = 2Ω or E3 = 8v line #3 R3 = 6Ω 2i2 + 6i3 = 18 There is another loop where we go around the entire circuit first going along line 1 and then line 3. However the equation we get from applying the voltage law to this loop is the sum of the other two loop equations, so it doesn’t yield any new information. We now have the following three equations for i1, i2 and i3. i1 + i2 - i3 = 0 4i1 - 2i2 = -5 2i2 + 6i3 = 18 1 1 -1 | 0 The augmented matrix is 4 -2 0 | -5. Subtracting 4 times row 1 from row 2 gives 0 2 6 | 18 1 1 -1 | 0 0 -6 4 | -5. Interchanging rows 2 and 3 gives 01 21 -16 || 180 . Dividing row 2 by 2 gives 0 2 6 | 18 0 -6 4 | -5 1 1 -1 | 0 0 1 3 | 9 . Subtracting row 2 from row 1 and adding 6 times row 2 to row 3 gives 0 -6 4 | -5 10 01 -43 || 9-9 . Dividing row 3 by 22 gives 10 01 -43 || 9-9 . Adding 4 times row 3 to row 1 0 0 22 | 49 0 0 1 | 49/22 1 0 0 | -2/22 and subtracting 3 times row 3 from row 2 gives 0 1 0 | 51/22 . So the solution is 0 0 1 | 49/22 i1 = -2/22, i2 = 51/22 and i3 = 49/22. 3.4 - 3 If one or more of the voltage source values is not given, then we have more unknowns than equations. node #1 Example 2. In Example 1 suppose the value of E3 is not given, so one has the network at the right. i1 i2 - + E1 = 5v Now the equation that one obtains by applying the voltage law to the loop at the right is i3 E2 = 10v + E3 + - line #1 R1 = 4Ω line #2 line #3 R2 = 2Ω R3 = 6Ω 10 + E3 = 2i2 + 6i3 or 2i2 + 6i3 - E3 = 10 node #2 So one now has the following set of three equations for i1, i2, i3 and E3. i1 + i2 - i3 = 0 4i1 - 2i2 = -5 2i2 + 6i3 - E3 = 10 1 Let's solve for i1, i2 and i3 in terms of E3. 4 0 R2 R3 10 0 R3/22 R3 1 -1 0 | 0 2 6 -1 | 10 -6 4 0 | -5 10 0 R2/2 R2 0 -4 1/2 | -5 1 3 -1/2 | 5 0 1 -3/22 | 25/22 10 0 1 -1 0 | 0 -2 0 0 | -5 2 6 -1 | 10 1 -1 0 | 0 1 3 -1/2 | 5 -6 4 0 | -5 R1 + 4R3 R1 R2 - 3R3 R2 10 0 R2 - 4R1 R2 R1 - R2 R1 R3 + 6R2 R3 01 0 10 0 1 -1 0 | 0 -6 4 0 | -5 2 6 -1 | 10 0 -4 1/2 | -5 1 3 -1/2 | 5 0 22 -3 | 25 0 0 -1/22 | -10/22 1 0 2/22 | 35/22 . 0 1 -3/22 | 25/22 The corresponding equations are i1 - (1/22)E3 = -10/22 i2 - (2/22)E3 = 35/22 i3 - (3/22)E3 = 25/22 So the formulas for i1, i2 and i3 in terms of E3 are i1 = (1/22)E3 - 10/22 i2 = (2/22)E3 + 35/22 i3 = (3/22)E3 + 25/22 or i1 1/22 -10/22 i2 = 2/22 E3 + 35/22 i3 3/22 25/22 These formulas would be useful if we set E3 to different values at different times. We could plug the current value of E3 into these formulas and find i1, i2 and i3. 3.4 - 4