07_Superposition

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Physics 302/328: Superposition Objectives :  Discuss the concept of superposition for circuits with more than one independent power supply o This is section 5.2 in the book Basic Engineering Circuit Analysis, 11 th ed. 1) The basic idea for using this as a problem solving strategy is this: a) Suppose that we have a circuit with more than one independent power supply. For any resistor in that circuit, R x , the current going through R x is determined by all of the sources in the circuit. What we are trying to show here is that the current through any resistor equals the addition of the current produced by each power supply, when all the other power supplies are removed from the circuit . 2) A specific example: Consider the following problem where a probe is powered by its own battery and it is hooked up to a calibrator and a display which are all modeled as resistors. Find I 3 and I 1 and I 2 (the current thru the respective resistors, R 3 , R 1 and R 2 ). R 1 = 200  R 2 = 100  R 3 = 150  V 1 = 6 V V 2 = 4 V There are three main equations that describe this full circuit : (full eqn 1) V 2 + I 3 R 3 – I 2 R 2 = 0 (from KVL & Ohm’s law) (full eqn 2) I 1 = I 2 + I 3 (from KCL) (full eqn 3) V 1 – I 1 R 1 – I 3 R 3 = 0 (from KVL & Ohm’s law) And we can go through and solve for all the currents (3 eqns and 3 unknowns). But let’s suppose we only had V 1 in the circuit and V 2 = 0 (we “short out” V 2 ). Then the equations are: (V1 only eqn 1) 0+ I V1_3 R 3 – I V1_2 R 2 = 0 (from KVL & Ohm’s law) (V1 only eqn 2) I V1_1 = I V1_2 + I V1_3 (from KCL) (V1 only eqn 3) V 1 – I V1_1 R 1 – I V1_3 R 3 = 0 (from KVL & Ohm’s law) Where the “V1_” subscript (like on I V1_3 ) means it is the current when only V1 is present Page 1 of 10

Physics 302/328: Superposition But let’s suppose we only had V 2 in the circuit and V 1 = 0 (we “short out” V 1 ). Then the equations are: (V2 only eqn 1) (from KVL & Ohm’s law, small loop) (V2 only eqn 2) (from KCL) (V2 only eqn 3) (from KVL & Ohm’s law; bigger loop) Surely, there must be some relation between the full circuit (with V 1 and V 2 not zero) and the circuits when one is zero and the other is not. Let’s make a Hypothesis: Is it something as simple as I 1 = ? I V 1 1 + I V 2 1 and I 2 = ? I V 1 2 + I V 2 2 and I 3 = ? I V 1 3 + I V 2 3 Well, let’s check. How can we check? (1) One way is to solve the “V2 only eqn x” for all the currents when only V2 is present. Then do the same for the “V1 only eqn x” for all currents and finally do it for the full circuit (i.e. “full eqn x”). Sub them in to the questionable equations above and see if the LHS = RHS (2) Another way to do it: Given: Check: Well, let’s suppose that I V2_x is the solution to the set of equations when V 1 = 0 and suppose that I V1_x is the solution to the set of equations when V 2 = 0. Then, let’s sub these into the main equations and see if they work In other words, V 2 + I V2_3 R 3 – I V2_2 R 2 = 0 I V2_1 = I V2_2 + I V2_3 0 – I V2_1 R 1 – I V2_3 R 3 = 0 Are all taken to be true In other words, 0+ I V1_3 R 3 – I V1_2 R 2 = 0 I V1_1 = I V1_2 + I V1_3 V 1 – I V1_1 R 1 – I V1_3 R 3 = 0 Are all taken to be true Can we show that this is true: V 2 + I 3 R 3 – I 2 R 2 = 0 I 1 = I 2 + I 3 V 1 – I 1 R 1 – I 3 R 3 = 0 Where Page 2 of 10

Physics 302/328: Superposition I 1 = I V 1 1 + I V 2 1 I 2 = I V 1 2 + I V 2 2 I 3 = I V 1 3 + I V 2 3 To check we will sub in our currents into the questionable equations: V 2 + I 3 R 3 – I 2 R 2 = 0 becomes V 2 + (I V2_3 + I V1_3 )R 3 – (I V2_2 + I V1_2 )R 2 = 0 I 1 = I 2 + I 3 becomes I V1_1 + I V2_1 = (I V1_2 + I V2_2 ) + (I V1_3 + I V2_3 ) V 1 – I 1 R 1 – I 3 R 3 = 0 becomes V 1 – (I V2_1 + I V1_1 )R 1 – (I V2_3 + I V1_3 )R 3 = 0 And now we need to check if each one of these is true. We’ll start with the first one: V 2 + I 3 R 3 – I 2 R 2 = 0 (from the full circuit) V 2 + I V2_3 R 3 – I V2_2 R 2 = 0 (from when V 1 = 0) and I V1_3 R 3 – I V1_2 R 2 = 0 (from when V 2 = 0) are both true, that implies the first equation must be true too. Can you see why? Let’s check the 2 nd equation (KCL for the full circuit; I 1 = I 2 + I 3 ) I V2_1 = I V2_2 + I V2_3 and I V1_1 = I V1_2 + I V1_3 are true, then we know that I V1_1 + I V2_1 = (I V1_2 + I V2_2 ) + (I V1_3 + I V2_3 ) is also true. Can you see why? The last equation with V 1 it is also true for similar reasons. 3) What does this all mean? Well, it is basically telling us that if we have more than one independent source in our circuit, then we can find the solution to the total circuit by a) Removing all sources except for one b) Find the current with that one source c) Repeat with all other sources removed except for a different one d) Repeat e) After doing this for each source, add up all the currents to get the total current. i) This is what it means for a “linear circuit”…the total is the addition of each individual piece. Page 3 of 10

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Physics 302/328: Superposition How do you “remove” a source? i) If it is a voltage source, the one way to get V = 0 is to “short out” the voltage source (i.e. replace with a single wire). ii) If it is a current source, the way to get I = 0, is to cut open the circuit (open circuit) by the current source so no current can get out. 4) Example: What is V o ? a) Educated Guess: i) If R 2 is really big, then there is no current through it; nor is there current through R 4 . Thus the V o is due to the voltage drop across the 2mA source. And that, from KVL is: 12 – R 3 (2 mA) – V 2mA = 0. Thus V o = 12 - R 3 (2 mA) ii) If R 3 = R 4 = 0, then V o = 12 Volts iii) If R 3 = 0, then the R 4 and R 2 serve as a voltage divider for the 12 V supply and V o = R 2 *12/(R 2 + R 4 ) b) Concepts: Try this with Superposition . Find I 2 (the current thru the 2k  ) when the 2mA is open circuited (call it I 2_V ) and then find I 2 when the 12 volts is short circuited (call it I 2_I ). Then I 2 = I 2_V + I 2_I . c) Outline solution i) Find I 2_V : Open circuit the 2 mA then the 12 V is in series with the resistors…that is easy to solve (eqn 1) I 2 V = ( 12 V ) R 3 + R 2 + R 4 ii) Find I 2_I : Short Circuit 12 V Then the 2mA will go in parallel loops through R 2 and R 4 (in series) and through R 3 . Need to use KVL and KCL to get this one (eqn 2; KCL) 2mA = I 3 + I 2_I (eqn 3: KVL) R 3 I 3 – ( I 2_I R 4 + I 2_I R 2 ) = 0  R 3 I 3 = I 2_I ( R 4 + R 2 ) Solve 3 for I 3 . Sub it into eqn 2 to get: I 2 I = ( 2 mA ) R 3 R 3 + R 2 + R 4 Page 4 of 10

Physics 302/328: Superposition Thus the total current is (since the two go in opposite directions; we subtract) I 2 = ( 12 V ) R 3 + R 2 + R 4 − ( 2 mA ) R 3 R 3 + R 2 + R 4 = 12 V − ( 2 mA ) R 3 R 3 + R 2 + R 4 Hence V o is: V o = I 2 R 2 = ( 12 V − ( 2 mA ) R 3 ) R 2 R 3 + R 2 + R 4 iii) Check answer: (1) Units (2) Educated Guess iv) Compare it to the answer from the old method of solving using KVL and KCL. ======================== 5) Why does this work? It is because in any resistive circuit V and I are linear. In other words, we can always write I = mV + b (i.e. just like y = mx + b). a) Here is a specific example for what I mean. Consider the simple circuit shown here. Imagine V bat and R 1 are constants and we measure I as we change the load resistor, R L . In other words, we choose some value for R L and measure V L and I. Then we do it again. Question: What should we expect for a graph of V L versus I? Note that we can’t use V L = R L I since R L is changing for each data point. Instead, we want an equation where only V L and I are the variables and everything else is constant (which in the case are R 1 and V bat ). The KVL equation can be written as V bat – IR 1 – V L = 0 (don’t sub in for V L because we want to solve for it) i) Thus we find that V L = V bat – IR 1 Page 5 of 10

Physics 302/328: Superposition ii) Conclusion: V L and I are linearly related. The slope is downward with a value of R 1 and the y-intercept is the voltage of the battery. (1) When R L = infinity…. (2) When R L = 0…. b) Example 2: A more complex circuit. Show that the current through R var is linearly related to the voltage V var as the value of R var changes. Our final equation can depend on R 2 , R 3 V bat2 and V bat3 . It is possible to show (remember we want to not have R var in the answer since it changes along with I var and V var ) V var = ( V bat 2 + V bat 3 R 2 R 3 ) R 3 R 2 + R 3 − I var R 3 R 2 R 3 + R 2 Once again, we see it is linear for I var and V var . And, in comparing it to the previous circuit where we got V L = V bat – IR 1 it would suggest that the two circuits would be exactly equivalent if we had: V bat = ( V bat 2 + V bat 3 R 2 R 3 ) R 3 R 2 + R 3 and R 1 = R 3 R 2 R 3 + R 2 c) Extending this idea i) Any linear circuit is related by this relationship (using y = mx + b): V = mI + b. Page 6 of 10

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Physics 302/328: Superposition ii) Since any linear circuit can be written this way, then if we can just find “m” and “b”, then we would know what V and I are no matter what the R L is. What this implies is that we can model any circuit as one resistor in series with one voltage supply since that type of simple circuit also has an “m” and a “b” (it is exactly the same equation). Use the last example to compare. The question is “what is m for a complex circuit and what is b?” Namely, how do we find these values? Well, we need to have two equations….so go to the extremes… Any linear circuit V = mI + b one resistor, one voltage supply and a load resistor all in series Complex circuit When open circuited , that means, I = 0 so b = V oc From the circuit V oc = V bat when I = 0 When short circuited , that means V = 0, so m =− b i sc We know b already. This will give us m From the circuit, when V L = 0 we get: R 1 = V bat i sc Conclusion : (subbing in for the m and b for each circuit) Does this match with what we got? V L = V bat – IR 1 6) Try this circuit: Imagine we want to know how the current across the variable resistor depends on the voltage across it. The answer, of course, is V = mI + b. For this circuit, what is the value of “m” and “b”? Your answer should be in terms of the constants given (R 1 , R 2 and  V bat ) Page 7 of 10

Physics 302/328: Superposition 7) Summary a) Circuits with resistors and power supplies (dependent and independent) are linear. So, for any given resistor, there is a relation between the voltage and the current as that resistor’s value changes: V = mI + b. b) Because circuits are linear, we can use superposition to find the current or voltage through a resistor by considering one independent source at a time. The total current is then the sum of the currents from each of the supplies. I = I 1 + I 2 + … i) Short out voltage supplies ii) Open circuit current supplies ======================================================= Appendix : Looking at this same problem in matrices. The main equations can be written as the following in matrix form: Page 8 of 10

Physics 302/328: Superposition [ − 1 1 1 0 R 2 − R 3 R 1 0 R 3 ] [ I 1 I 2 I 3 ] = [ 0 V 2 V 1 ] Which is the same as (eqn 1) [ − 1 1 1 0 R 2 − R 3 R 1 0 R 3 ] [ I 1 I 2 I 3 ] = [ 0 0 V 1 ] + [ 0 V 2 0 ] which can also be written as (although this is not as important but it is interesting because it shows the “scaling feature” of solutions…namely doubling the voltages will double the current). [ − 1 1 1 0 R 2 − R 3 R 1 0 R 3 ] [ I 1 I 2 I 3 ] = V 1 [ 0 0 1 ] + V 2 [ 0 1 0 ] So what? Well, if we set V 2 = 0  in other words, we short circuit the independent voltages, then we get a solution of (notice that the notation on the currents has changed because it is a different circuit) (eqn 2) [ − 1 1 1 0 R 2 − R 3 R 1 0 R 3 ] [ I V 1 1 I V 1 2 I V 1 3 ] = [ 0 0 V 1 ] Also, if V 1 is zero but not V 2 we get: (eqn 3) [ − 1 1 1 0 R 2 − R 3 R 1 0 R 3 ] [ I V 2 1 I V 2 2 I V 2 3 ] = [ 0 V 2 0 ] Now, the question is this, does I 1 = I V 1 1 + I V 2 1 ? Where I 1 is for the total circuit. The same is said about I 2 and I 3 . Well, let’s sub this into the lhs of eqn 1 and see if it equals the rhs Lhs (eqn 1)= [ − 1 1 1 0 R 2 − R 3 R 1 0 R 3 ] [ I V 1 1 + I V 2 1 I V 1 2 + I V 2 2 I V 1 3 + I V 2 3 ] Page 9 of 10

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Physics 302/328: Superposition = [ − 1 1 1 0 R 2 − R 3 R 1 0 R 3 ] [ I V 2 1 I V 2 2 I V 2 3 ] + [ − 1 1 1 0 R 2 − R 3 R 1 0 R 3 ] [ I V 1 1 I V 1 2 I V 1 3 ] Using eqn 2 and 3 this becomes = [ 0 V 2 0 ] + [ 0 0 V 1 ] Which is the same as the RHS of eqn 1. Thus we just proved the superposition theorem for this specific example. Now let’s be more generic (this will include impedance and dependent power sources as well as current sources  everything that is linear): [ X 11 X 12 X 13 X 21 X 22 X 23 X 31 X 32 X 33 ][ I 1 I 2 I 3 ] = [ S 1 S 2 S 3 ] where X’s could be 1 or -1 or 0s (like above). Or X’s could be the  for dependent sources or they could be impedances (like R, jwL, 1/jwC) and S 1 and S 2 and S 3 are the sources (current or voltage sources). Can you see the “linearity” here? Page 10 of 10

07_Superposition

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