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ECE 2300 Circuit Analysis Dr. Dave Shattuck Example Equivalent Circuit Problem 37[k] 2.2[k] 4.7[k] + 8.2[k] 5[mA] 6.2[k] iX 27[k] 3.3[k] vW - Find iX and vW. Solution: We begin this problem by replacing the 3.3[k] resistor and the 4.7[k] resistor with their equivalent resistor value, which is 8.0[k]. Then, we take that 8.0[k] resistor, which is in parallel with the 6.2[k] resistor, and replace them by their equivalent resistance, which is 3.49[k]. We redraw the circuit that results, and get the following circuit diagram. 37[k] 2.2[k] + 8.2[k] 5[mA] 3.49[k] vX 27[k] There are several things to notice about this diagram. First, we note that vW is no longer present in this circuit. There is no place to label vW in this diagram. The voltage vW is inside the equivalent circuit, and cannot be labeled outside the equivalent circuit in the original circuit. We redraw the original circuit in the figure that follows, enclosing the part of the circuit that we have replaced within a dashed line. 37[k] 2.2[k] 4.7[k] + 8.2[k] + 6.2[k] 5[mA] vX iX 27[k] + - vX - 3.3[k] vW - Note that because the upper terminal for the voltage vW is inside the equivalent circuit, and not present outside the equivalent circuit, it is not possible to label vW on the diagram after replacing the equivalent circuit with the 3.49[kW] resistor. Note, however, that since the voltage vX can be labeled outside the equivalent circuit, it can be labeled both before and after replacing the equivalent circuit with the 3.49[kW] resistor. We have shown vX both inside and outside the equivalent circuit, in this diagram. It should be clear that both of these versions of vX are the same. Let’s again show the simplified circuit, and work from there. 37[k] 2.2[k] + 8.2[k] 5[mA] 3.49[k] vX 27[k] We continue to simplify the circuit, by replacing the 3.49[k] resistor and the 2.2[k] resistor with their equivalent resistor, which is 5.69[k]. Then, we take that 5.69[k] resistor, which is in parallel with the 8.2[k] resistor, and replace them by their equivalent resistance, which is 3.36[k]. We redraw the circuit that results, and get the following circuit diagram. 37[k] + 5[mA] 3.36[k] vZ 27[k] - It is now pretty straightforward to solve the voltage vZ, using Ohm’s Law. Note that the 5[mA] current is now flowing up through the 3.36[k] resistor, in the active sign convention with respect to the voltage vZ. Thus, using Ohm’s Law, we can write that vZ 5 mA 3.36 k 16.8 V . Now that we have vZ, we go back to our earlier diagram, and noting that vZ is also present in this diagram, we use it to solve for vX with the VDR. For this example, we reproduce the diagram following this paragraph, adding the definition of vZ. Of course, if we were solving this by hand, we would simply have added vZ to our existing diagram, from earlier in the problem. 37[k] 2.2[k] + 8.2[k] 5[mA] 27[k] vZ - + 3.49[k] vX - In this circuit, the 2.2[k] resistor and the 3.49[k] resistor are in series, and so we can use VDR to write 3.49 k v X vZ 3.49 k 2.2 k 3.49 k v X 16.8 V 3.49 k 2.2 k v X 10.3 V . Next, we can use the original diagram to find iX, using Ohm’s Law. We reproduce the original diagram here, just for clarity. 37[k] 2.2[k] 4.7[k] + 8.2[k] + 6.2[k] 5[mA] iX 27[k] vX - 3.3[k] vW - Using Ohm’s Law, we write vX 10.3[V] iX 6.2[k] 6.2[k] iX 1.66 mA. In addition, we can use VDR to solve for vW, and get 3.3 k vW v X 3.3 k 4.7 k 3.3 k vW 10.3 V 3.3 k 4.7 k vW 4.25 V. Alternative Solution: There are several good ways to solve this problem. Another choice would have been to have defined the current through the 2.2[k] resistor. We will define it as iQ, as shown in the circuit that follows. 37[k] 2.2[k] + 8.2[k] iQ 5[mA] 27[k] vZ - + 3.49[k] vX - Again, note that we did not have to redraw the circuit, as we have done here. This definition could have been completed in the diagrams already given. In this diagram, iQ can be found from the CDR. Note that the series combination of the 2.2[k] resistor and the 3.49[k] resistor is in parallel with the 8.2[k] resistor. Note also that the 5[mA] current feeds this parallel combination. Thus, we can use CDR, to get 8.2 k iQ 5[mA] 8.2 k 5.69 k iQ 2.95[mA]. We can now use this value of iQ to find iX, using the CDR again. The diagram is reproduced here for clarity. 37[k] 2.2[k] 4.7[k] + 8.2[k] iQ 5[mA] 27[k] + 6.2[k] iX vX - 3.3[k] vW - Using CDR again, we can write 8.0 k iX iQ 8.0 k 6.2 k 8.0 k iX 2.95[mA] 8.0 k 6.2 k iX 1.66[mA] . This is the same result that we got earlier, which should be no surprise. It is simply a different way to get the same answer. It is left as an exercise for the student to solve for the current through the 3.3[k] resistor, and find vW in this way. The answer should not change.