Thévenin’s and Norton’s Theorems Notes Ch4

Example Problem – Step 8 We have applied a test current source to the two terminals. A test voltage source would have been just as good. We chose a current source because we thought it might make the solution a little easier, since we can find vC so easily now. But it really does not matter. Don’t worry about which one to choose. Let’s solve. 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 9 Let’s solve for vT. We note that we can write an expression for vC using Ohm’s Law, and get vC  1[A]R2  5600[V]. This voltage may seem very large. Don’t let this bother you. We do not actually have this voltage; it is just for calculating the resistance. R 1= 2.2[kW] 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 10 Next, let’s write KCL for the top center node. We get vD  5vD vD  iT   iS  0, or by substituting, R1 R3 4vD vD  1[A]   25[mS]  5600[V]  0. 2.2[kW] 10[kW] R 1= 2.2[kW] 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 11 Solving for vD yields 4vD vD   139[A], or 2.2[kW] 10[kW]  1.72[mS] vD  139[A], or vD  80,900[V]. R 1= 2.2[kW] 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 12 Taking KVL, we get vD  vC  vT  0, or vT  vD  vC  80,900[V]   5600[V]  86,500[V]. R 1= 2.2[kW] 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 13 So, we can find the equivalent resistance by finding REQ vT 86,500[V]    86.5[kW]. iT 1[A] R 1= 2.2[kW] 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 14 So, the Thévenin equivalent is given in the circuit below. Note that the Thévenin voltage is zero, and so we don’t even show the voltage source at all. The Thévenin resistance is shown, and in this case, it is the Thévenin equivalent. A REQ= 86.5[kW] B Is the Test-Source Method Really That Important? • This is a good question. Basically, the answer is yes. There are many cases where we have dependent sources present, and wish to use a Thévenin equivalent. While in some cases we can get the Thévenin resistance from the open-circuit voltage and the short-circuit current, there are others where we cannot. The Test-Source Method is also quicker in some cases. • Some students go to great lengths to avoid learning this method. This seems like a waste of energy. Just learn it Go back to and use it. Overview slide. Example Problem #1 For the circuit given below, find the Norton equivalent as seen by the current source. Find the power delivered by the current source in this circuit. 30[W] + 10[V] 37[W] 13[A] 56[W] 40[W] + - 125[W]iQ 32[W] iQ Sample Problem #2 a) For the circuit shown below, find the Norton equivalent as seen by the 22[kW] resistor. b) Use this equivalent circuit to solve for iQ. iX R1= 10[kW] + - vS1= 100[V] iS1= 10iX R2= 33[kW] iQ R3= 22[kW] R4= 18[kW] iS2= 12[mA] Sample Problem #2 a) For the circuit shown below, find the Norton equivalent as seen by the 22[kW] resistor. b) Use this equivalent circuit to solve for iQ. Soln: a) iN = -102[mA], RN = -1.228[kW] b) -5.988[mA] iX R1= 10[kW] + - vS1= 100[V] iS1= 10iX R2= 33[kW] iQ R3= 22[kW] R4= 18[kW] iS2= 12[mA]