Source Transformations 2

Other Useful Transformations – 1 As with all equivalent circuits, these two are equivalent only with respect to the things connected to the equivalents. Voltage sources in series can be replaced by a single voltage source, where the value of the equivalent v2 source is equal to the algebraic sum of the voltage sources it is replacing. An example is shown here with two v1 sources with random polarities. A + - A - ~ + B v2-v1 + B Other Useful Transformations – 2 Current sources in parallel can be replaced by a single current source, where the value of the equivalent source is equal to the algebraic sum of the current sources it is i2 replacing. An example is shown here with two sources with random polarities. As with all equivalent circuits, these two are equivalent only with respect to the things connected to the equivalents. A A ~ i1 B i2-i1 B Other Useful Transformations – 3 A voltage source in parallel with anything can be replaced by that voltage source. The “anything” can be a resistor, a current source, or any other combination of elements. If the “anything” is a voltage source, the two voltage sources must be equal for KVL to hold. As with all equivalent circuits, these two are equivalent only with respect to the things connected to the equivalents. A vX + Anything A ~ B vX + B Other Useful Transformations – 4 A current source in series with anything can be replaced by that current source. The “anything” can be a resistor, a voltage source, or any other combination of elements. If the “anything” is a current source, the two current sources must be equal for KCL to hold. As with all equivalent circuits, these two are equivalent only with respect to the things connected to the equivalents. Anything A A ~ iX B iX B Go back to Overview slide. Notes 1. These equivalent circuits can go in either direction. That is, we can replace the circuit on the right with the one on the left, or the other way around. 2. The polarities of the sources with respect to the terminals are important. This is one good reason for naming the terminals of these equivalents. 3. As with all equivalent circuits, these are equivalent only with respect to the things connected to the equivalent circuits. 4. These equivalent circuits hold for dependent sources as well as independent sources. The key is that the variable, which the dependent sources depend on, must remain intact. That is, the voltage or current that the dependent sources use must be outside of the circuit being replaced. Example Problem We wish to solve for the voltage vX in the circuit given below. While we could certainly solve this by writing a series of KVL and KCL equations, we are going to solve it instead by using a series of equivalent circuits and simplify the circuit down step by step. R1= 22[W] R2= 33[W] + vX + vS = 100[V] iS= 4[A] R3= 10[W] R4= 15[W] - Example Problem – Step 1 We wish to solve for the voltage vX in the circuit given below. We note that we have a voltage source, vS, in series with a resistor, R1. We can replace them with a current source in parallel with a resistor. When we do, we will have current sources in parallel and resistors in parallel, which we can simplify further. So, let’s take the first step. R1= 22[W] R2= 33[W] + vX + vS = 100[V] iS= 4[A] R3= 10[W] R4= 15[W] - Example Problem – Step 2 We want to replace the voltage source in series with a resistor, with a current source in parallel with a resistor. Here, we have made this replacement. Note that we now have two current sources in parallel, and two resistors in parallel. Since the voltage we are looking for is outside these combinations, we can replace them with their equivalents. That is our next step. R2= 33[W] + vX vS/R1= i1= 4.546[A] R1= 22[W] iS= 4[A] R3= 10[W] R4= 15[W] - Example Problem – Step 3 We have replaced the parallel current sources and parallel resistors with their equivalents. Now, we can note that we have a current source in parallel with a resistor. We could replace this with a voltage source in series with a resistor, and then we could simplify the circuit further. Let’s do this. R2=33[W] + vX i2= 8.546[A] R5= 6.875[W] R4= 15[W] - Example Problem – Step 4 We have replaced the current source in parallel with a resistor with a voltage source in series with a resistor. At this point, we have three resistors in series, and we want the voltage across one of them. This means that we use the voltage divider rule, and write, R4 v X  v3  R4  R5  R2 R5=6.875[W] R2=33[W] 15[W] v X  58.75[V] 15[W]  6.875[W]  33[W] v X  16.06[V]. + + - vX v3= i2R5= 58.75[V] R4= 15[W] - Go to Comments slide. Go back to Overview slide. Example Problem – Solution The solution for this problem then is v X  16.06[V]. R1= 22[W] R2= 33[W] + vX + vS = 100[V] iS= 4[A] R3= 10[W] R4= 15[W] - Go to Comments slide. Go back to Overview slide. Why Are Source Transformations Needed? • This is a good question. In fact, it should be pointed out that these transformations are not necessary. Rather, they are like many other aspects of circuit analysis in that they allow us to solve circuits more quickly and more easily. They are used in cases where we want to be able to combine elements using other equivalent circuit simplifications. • One key in applying these equivalents is to get the proper sources in the proper polarities in the equivalents and equations. We recommend that you name the terminals each time, on the circuit diagrams, to help you get these polarities right. Go back to Overview slide. Example Problem 68[W] + Find iQ. 3.3[kW] 4[mA] + + 75[V] - 1.5[kW] 22[V] 7[V] - 1[kW] iQ