Thévenin’s and Norton’s Theorems Notes Ch3

Test-Source Method – Note 4 To get the equivalent resistance of a circuit, as seen by two terminals of that circuit, we follow these steps. When you apply these voltages and currents, 1) Set all independent sources equal to zero. apply them in the active sign relationship for 2) Find the equivalent resistance. the source. This gives the “proper” sign. a) If there are no dependent sources, find this equivalent resistance using the equivalent resistance rules that have been used before. These include series combinations, parallel combinations, and delta-to-wye equivalents. b) If there are dependent sources present, apply a test source to the two terminals. It can be either a voltage source or a current source. 1) If you apply a voltage source, find the current through that voltage source. iT A 2) If you apply a current source, find the voltage across that + Any circuit current source. vT made up of 3) Then, find the ratio of the voltage to the Test resistors and current, which will be the Source equivalent resistance. dependent sources - B Test-Source Method – Note 5 To get the equivalent resistance of a circuit, as seen by two terminals of that circuit, we follow these steps. The active sign relationship for the test source 1) Set all independent sources equal to zero. gives the passive sign relationship for the 2) Find the equivalent resistance. circuit, which is the resistance, by Ohm’s Law. a) If there are no dependent sources, find this equivalent resistance using the equivalent resistance rules that have been used before. These include series combinations, parallel combinations, and delta-to-wye equivalents. b) If there are dependent sources present, apply a test source to the two terminals. It can be either a voltage source or a current source. 1) If you apply a voltage source, find the current through that voltage source. iT A 2) If you apply a current source, find the voltage across that + Any circuit current source. vT made up of 3) Then, find the ratio of the voltage to the Test resistors and current, which will be the Source equivalent resistance. dependent sources - B Go back to Overview slide. Notes 1. The Test-Source Method usually requires some practice before it becomes natural for students. It is important to work several problems to get this practice in. 2. There is a tendency to assume that one could just ignore the Test-Source Method, and just find the open-circuit voltage and short-circuit current whenever a dependent source is present. However, sometimes this does not work. In particular, when the open-circuit voltage and short-circuit current are zero, we must use the Test-Source Method. Learn how to use it. Example Problem We wish to find the Thévenin equivalent of the circuit below, as seen from terminals A and B. R 1= 2.2[kW] R2= 5.6[kW] A + + vS= 5vD + - iS= 25[mS]vC vC - vD R 3= 10[kW] B Example Problem – Step 1 We wish to find the Thévenin equivalent of the circuit below, as seen from terminals A and B. We will start by find the open-circuit voltage at the terminals, as defined below. R1= 2.2[kW] R2 = 5.6[kW] A + + + vS= 5vD + - i S= 25[mS]vC vC - vD vOC R3= 10[kW] - B Example Problem – Step 2 To find vOC, we will first find vD, by writing KCL at the top center node. We have vD  5vD vD  0   iS  0. R1 R3 Note that we recognize that the current through R2 must be zero since R2 is in series with an open circuit. R1= 2.2[kW] R2 = 5.6[kW] A + + + vS= 5vD + - i S= 25[mS]vC vC - vD vOC R3= 10[kW] - B Example Problem – Step 3 We can substitute in the value for iS, 25[mS]vC. We note that since the current through R2 is zero, the voltage across it is zero, so vC is zero. So, we write vD  5vD  0  vD  25[mS]v  0, or C R1 R3 vD  5vD vD   0. R1 R3 R1= 2.2[kW] R2 = 5.6[kW] A + + + vS= 5vD + - i S= 25[mS]vC vC - vD vOC R3= 10[kW] - B Example Problem – Step 4 Next, we substitute in values and solve for vD. We write 4vD vD   0. With some math, we find 2.2[kW] 10[kW] vD  0. R1= 2.2[kW] R2 = 5.6[kW] A + + + vS= 5vD + - i S= 25[mS]vC vC - vD vOC R3= 10[kW] - B Example Problem – Step 5 Now, we can take KVL around the loop, and we write vD  vC  vOC  0, and so vOC  0. The Thévenin voltage is equal to this open-circuit voltage, so the Thévenin voltage must be zero. The short-circuit current will also be zero. To get the resistance, we need to use the R1= R2 = Test-Source 2.2[kW] 5.6[kW] Method. A + + + vS= 5vD + - i S= 25[mS]vC vC - vD vOC R3= 10[kW] - B Example Problem – Step 6 We have applied a test current source to the two terminals. We have also labeled a voltage across this current source, vT. This voltage has been defined in the active sign relationship for the current source. As noted earlier, this will give us the passive sign relationship for vT and iT for the circuit that we are finding the equivalent resistance of. Thus, we will have R 1= 2.2[kW] vT  REQ . iT R2= 5.6[kW] A + + vS= 5vD + - iS= 25[mS]vC vC vD R 3= 10[kW] - - + vT B iT = 1[A] Example Problem – Step 7 We have applied a test current source to the two terminals. We don’t need to do this, but doing so makes it clear that we are now just solving another circuit, like the many that we have solved before. We have even given the source a value, in this case, 1[A]. This is just a convenience. Many people choose to leave this as an arbitrary source. We choose to use a value, an easy value like 1[A], to allow us to find an actual value for vT. v T R 1= 2.2[kW] R2= 5.6[kW] A + + vS= 5vD + - iS= 25[mS]vC vC vD R 3= 10[kW] - - iT  REQ . + vT B iT = 1[A]