1、附录 A A few examples will refresh your memory about the content of Chapter 8 and the general approach to a nodal-analysis solution. EXAMPLE 17.12 Determine the voltage across the inductor for the network of Fig. Solution:Steps 1 and 2 are as ndicated in Fig.17.22. Step 3:Note Fig.17.23 for the applic
2、ation of Kirchhoff s current law to node V1: Fig.17.22 Fig.17.23 Ii= I0 0=I1+I2+I3V1-E/Z1+(V1/Z2)+(V1-V2)/Z3=0 Rearranging terms: V11/Z1+1/Z2+1/Z3-V21/Z3=E1/Z1 (17.1) Note Fig.17.24 for the application of Kirchhoff s current law to node V2: 0=I3+I4+I V2-V1/Z3+V2/Z4+I=0 Rearranging terms: V21/Z3+1/Z4
3、-V11/Z3=-I (17.2) Fig.17.24 Grouping equations: V11/Z1+1/Z2+1/Z3-V21/Z3=E1/Z1 V11/Z3-V21/Z3+1/Z4 =I 1/Z1+1/Z2+1/Z3=1/0.5k +1/10 k +1/2k =2.5mS -2.29 1/Z3+1/Z4=1/2k +1/-5k =0.539mS 21.80 and V12.5ms -2.29 -V20.5mS 0 =24m 0 V10.5mS 0 -V20.539mS 21.80 =4m 0 with 24m 0 -0.5mS 0 4m 0 -0.539mS 21.80 V1= 2
4、.5ms -2.29 -0.5mS 0 0.5mS 0 -0.539mS 21.80 =(24m 0 )(-0.539mS 21.80 )+(0.5mS 0 )(4m 0 )/(2.5ms -2.29 )(-0.539mS 21.80 )+(0.5mS 0 )(0.5mS 0 ) =-10.01 -j4.81 /-1.021-j0.45=11.106 -154.33 /1.116 -156.21 V1=9.95 1.88 MathCad The length and complexity of the above mathematical development strongly sugges
5、t the use of an alternative approach such as MathCad.Note in MathCad 17.2 that the equations are entered in the same format as Eqs.(17.1) and (17.2).Both V1 and V2 were generated,but because only V1 was asked for,it was the only solution converted to the polar form.In the lower solution the complexi
6、ty was significantly reduced by simply recognizing that the current is in milliamperes and the impedances in kilohms. The result will then be in volts. K :=10 m :=0.01 rad :=1 V1 :=1+j V2 :=1+j deg := /180 Given V1 1/5 k+1/10j k+1/2 k-V2 1/2 k 24 m V1 1/2 k-V21/2 k+1/-5j k 4 m Find(V1,V2)= 9.944 +0.
7、319j Volts 1.786 -0.396j Volts V1 :=9.944+0.319j V1 =9.949 arg(V1)=1.837 deg Recognizing that current in mA results ehen Z is in kilohmns,an alternative format follows: Given V1 1/5+1/10j+1/2-V2 1/2 24 V1 1/2-V21/2+1/-5j 4 Find(V1,V2)= 9.944 +0.319j Volts 1.786 -0.396j Volts V1 :=9.944+0.319j V1 =9.
8、949 arg(V1)=1.837 deg MATHCAD 17.2 Dependent Current Sources For dependent current sources,the procedure is modified as follows: Steps 1 and 2 are the same as those applied for independent sources. Step 3 is modified as follows:Treat each dependent current source like an independent source when Kirc
9、hhoffs current law applied to each defined node.However,once the equations are established,substitute the equation for the controlling quantity to ensure that the unknowns are limited solely to the chosen nodal voltages. 1. Step 4 is as before. EXAMPLE 17.13 Write the nodal equations for the network
10、 of Fig.17.25 having a dependent current source. Solution: Steps 1 and 2 are as defined in Fig.17.25. Fig.17.25. Step 3: At node V1, I=I1+I2 V1/Z1+V1-V2/Z2-I=0 and V11/Z1+1/Z2-V21/Z2=I At node V2, I2+I3+ I=0 V2-V1/Z2+V2/Z3+ V1-V2/Z2=0 and V11- /Z2-V21- /Z2+1/Z3=0 resulting in two equations and two u
11、nknowns. Independent Voltage Sources between Assigned Nodes For independent voltage sources between assigned nodes,the procedure is modified as follows: 1. Steps 1 and 2 are the same as those applied for independent sources. 2. Step 3 is modefied as follows:Treat each source betwwen defined nodes as
12、 a short circuit(recall the supernode classification of Chapter 8),and write the nodal equations for each remaining independent node.Then relate the chosen nodal voltages to the independent voltage source to ensure that the unknowns of the ginal equations are limited solely to the nodal voltages. 3.
13、 Step 4 is as before. EXAMPLE 17.14 Write the nodal equations for the network of Fig.17.26 having an independent source between two assigned nodes. Solution: Steps 1 and 2 defined in Fig.17.26. Step 3:Replacing the independent source E with a short-circuit equivalent results in a supernode that will generate the following equation when Kirchhoffs current law is applied to node V1: I1=V1/Z1+V2/Z2+I2 with V2-V1=E
