07_Superposition

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