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Matthews Engineering 17 Fall 2020 Introduction to Kirchhoff's Laws

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Matthews Engineering 17 Fall 2020 Introduction to Kirchhoff's Laws Matthews Engineering 17 Fall 2020 Introduction to Kirchhoff’s laws Gustav Robert Kirchhoff, a German physicist, expressed two laws that characterize the behavior of electric circuits. Kirchoff’s current law (KCL): We have so far used the assumption that the current flowing into one terminal of a two-terminal element must be equal to the current flowing out of the other terminal of that element. Consider the hydraulic analogy of water in a pipe to current in a wire. For a single piece of pipe, the flow rate of water entering the pipe must equal the flow rate of water leaving that pipe. Suppose that the flow rate entering is 2 ft3 per minute (2 cfm). Then if the pipe is rigid and the water does not compress, the flow rate leaving must be 2cfm. Water in Pipe Water out 2cfm 2cfm Now, if there is a “Tee” connector in the pipe, how do we extend the principle above? Matthews Engineering 17 Fall 2020 Water in 1 cfm Water in Water out 2 cfm 3 cfm From the figure above, we see that the algebraic sum of all the flow rates is zero. Suppose we define flow entering positive and flow leaving negative. This gives 2+ 1 + ( − 3) = 0. Or we could define flow entering as negative and flow leaving as positive. This gives (− 2) + ( − 1) + (3) = 0. Kirchoff’s current law (KCL) states that the algebraic sum of currents entering a node is equal to zero. Example: iR 1A 2A 3A 2 vR For the circuit above, what is vR ? Matthews Engineering 17 Fall 2020 First, identify the nodes in the circuit. There are two nodes. Node 1 is inside the dashed box and is at the top of the drawing. It should be noted that any two points connected by an ideal conductor are the same electrical Node 1 iR iR vR 1A 2A 3A 2 vR Reminder Node 2 node (the same electrical “location”) in the circuit. Note that Node 2 has been marked as the “reference node” by use of the ground symbol . All voltages can then be expressed in terms of their potential difference from the reference node. This allows us to refer to the voltage “at” Node 1 with a single variable, which in this case isvR (because the + terminal of vR is Node 1 and the − terminal of vR is the reference node). Sum currents leaving Node 1; that is, define positive currents as leaving the node: v +1 − 2 + 3 +i = 0 which implies that +1 − 2 + 3 +R = 0 . R 2 Now we have one equation with one unknown. The leading “+” sign could be suppressed. For the sake of simplicity, we omitted units such as the “A” symbol that stands for “Amps”. Solving, we get that vR =−4V . Matthews Engineering 17 Fall 2020 A more complicated example follows. Note that the use of KCL leads to nodal analysis, in which nodal equations (which sum currents) are written in terms of node voltages. The goal is to solve for the node voltages. Here is an example of nodal analysis using KCL: v 12 1 6 6V 4 12V Draw the same circuit. Label a reference node. Discuss “reference node” and “ground” (important) Label unknown node voltage (in this case, “ v1 ”). Matthews Engineering 17 Fall 2020 Sum currents into node: v 12 1 6 iA iB 6V iC 4 12V NOTE: The labels iA , iB , and iC are used for clarity in the figure, but in this type of analysis the unknowns are the node voltages. The labels for the currents are used here only as placeholders and are not needed after some practice. Principle (KCL): iABC+ i + i = 0 6−v 0 − v 12 − v Write: 1+ 1 + 1 = 0 and 6−v − 3 v + 24 − 2 v = 0 12 4 6 1 1 1 ii i AB C −6v1 = − 30 and v1 = 5 . Note: Not all quiz and exam problems will result in answers which are round numbers! Matthews Engineering 17 Fall 2020 Kirchhoff’s voltage law (KVL): This law states that the sum of voltage drops going around a closed loop must equal zero. It is also true that the sum of voltage rises going around a closed loop must equal zero. A familiar example is that of going for a hike in the mountains. If the hike brings you back to the same location (i.e. it is a closed loop), it must be true that the sum of your elevation changes around the loop is equal to zero. Matthews Engineering 17 Fall 2020 Example: The circuit to the right is to be analyzed using KVL. This type of analysis 3V is called mesh analysis. d iR c A systematic approach is to first define the v loop current (or all loop currents in the case 2V 4 R of multiple loops) including the direction of b each current. The figure shows a circuit with a single loop, so only one loop current 1V is required. a Nodes are labeled in the figure (a,b,c,d). The labels for nodes are used here to clarify the process and are not needed after some practice. Start at node a and proceed to node b (in the direction of the current). Going from a to b, there is a voltage rise of +1V. In this case we are using the rule of adding voltage drops, so the voltage drop is equal to −1V . Following this rule we have: −1V + 2V − 3V + 4iR = 0 a→ b b → c c → d da→ Notice that in this example, the first three voltage drops are due to voltage sources, so the terms in the equation can be written by inspection (including the polarity due to a voltage drop or a voltage rise). The last term is a voltage that depends on the current, so we write an equation for that voltage in terms of the current. Note that mesh analysis intrinsically finds currents, and all terms in the equations are voltages. So now we find 42iR = . Re- write that equation showing units, for clarity: 4iR A = 2V . Finally, V iR = 0.5A . From this result, we can find vR = 2V. Matthews Engineering 17 Fall 2020 Another Example of KVL iX 4 10V 6 Find iX . Show how to obtain: −10 + 4iiXX + 6 = 0 Solve iX =1 This will lead to the equivalent value of resistors in series, but that is a later topic. Matthews Engineering 17 Fall 2020 Another KVL problem. Find iX . 2 4 iX 6V 9V Show how to get −6 + 2iiXX + 4 + 9 = 0 Note that voltage source is not always negative. When it is a drop, sign of voltage source in loop equation is positive. Note that when we go around the loop in the direction of the loop current, and we add voltage drops, we have a positive sign associated with each term that is due to a resistor. Find: iX =−0.5A Note: When we assigned a direction to iX , we did not assume that the current would indeed be in the direction of the arrow; we were not “wrong”. In fact, all we did was assign a direction to the variable iX , and for our assumed direction the value is expressed as −0.5A. The following two statements are equivalent (and both true): 1) The current is -0.5A in a clockwise direction 2) The current is +0.5A in a counter-clockwise direction Matthews Engineering 17 Fall 2020 i 2 X 4 v 6V X 9V Now, same problem (no changes made to circuit), but find vX . We can write another loop equation including vX , even though we are not going around a loop for which the current is the same through every element, as before. −6 + 2ivXX + = 0 . Since the circuit has not changed from before, and we know iX =−0.5A , we get −6 + 2( − 0.5) +vX = 0 and vX = 7V . We can go around the other loop to get vX = 7V also. (You should try it!) Matthews Engineering 17 Fall 2020 Another problem: 2 4 iX 6V 3A We know the value of iX , but let’s go ahead and write a loop equation: −6 + 2iiXX + 2 + ? = 0 We use ? because we cannot express the voltage across the current source as a function of the current through it (why?) Just define the voltage across the current source as a variable, vCS as shown below: 2 4 iX 6V vCS 3A Now we get: −6 + 2iX + 2 i X + v CS = 0 Have two unknowns but one equation. But wait! We know iX = 3 so vCS =−12 Matthews Engineering 17 Fall 2020 Another problem: 12 6 i i 6V 1 4 2 12V Now we need two loops. For loop of i1 : −6 + 12i1 + 4( i 1 − i 2 ) = 0 --> 16ii12−= 4 6 For loop of i2 : 4(i2− i 1 ) + 6 i 2 + 12 = 0 --> −4ii12 + 10 = − 12 Solve to get i2 =−1.167 and i1 = 0.083 Note that ii12−=1.25 and 12i1 = 1; substitute into first equation to confirm What if we added the loop equations immediately? For loop of i1 : −6 + 12i1 + 4( i 1 − i 2 ) = 0 For loop of i2 : 4(i2− i 1 ) + 6 i 2 + 12 = 0 note that 4(ii21− ) cancels 4(ii12− ) Sum: −6 + 12ii12 + + 6 + 12 = 0 This is the equation of the outside loop!!! Only two independent equations because there are only two loops. .
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