barcode fonts for ssrs VTh and ZTh are connected as in Fig. 1-4(b) to produce the Thevenin equivalent circuit. in Software

Make Code 3 of 9 in Software VTh and ZTh are connected as in Fig. 1-4(b) to produce the Thevenin equivalent circuit.

VTh and ZTh are connected as in Fig. 1-4(b) to produce the Thevenin equivalent circuit.
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For the circuit and values of Problem 1.7, nd the Norton equivalent for the network to the left of terminals a; b.
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With terminals a; b shorted, the component of current Iab due to V1 alone is
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0 Iab
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V1 10 2:5 A R1 4 V2 15 2:5 A 6 R2
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Similarly, the component due to V2 alone is
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00 Iab
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Then, by superposition,
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0 00 IN Iab Iab Iab 2:5 2:5 5 A
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Now, with RTh as found in Problem 1.7, YN 1 1 0:4167 A RTh 2:4
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IN and YN are connected as in Fig. 1-4(c) to produce the Norton equivalent circuit.
CIRCUIT ANALYSIS: PORT POINT OF VIEW
[CHAP. 1
For the circuit and values of Problems 1.7 and 1.8, nd the Thevenin impedance as the ratio of open-circuit voltage to short-circuit current to illustrate the equivalence of the results.
The open-circuit voltage is VTh as found in Problem 1.7, and the short-circuit current is IN from Problem 1.8. Thus, ZTh which checks with the result of Problem 1.7. VTh 12 2:4  5 IN
Thevenin s and Norton s theorems are applicable to other than dc steady-state circuits. For the frequency-domain circuit of Fig. 1-18 (where s is frequency), nd (a) the Thevenin equivalent and (b) the Norton equivalent of the circuit to the right of terminals a; b.
a IL(s) sL Load + V1(s) _ b _ I(s) + V2(s) 1 sC
Fig. 1-18 (a) With terminals a; b open-circuited, only loop current I s ows; by KVL and Ohm s law, with all currents and voltages understood to be functions of s, we have I Now KVL gives VTh Vab V1 sLI V1 sL V2 V1 V1 s2 LCV2 sL 1=sC s2 LC 1 V2 V1 sL 1=sC
With the independent sources deactivated, the Thevenin impedance can be determined as ZTh sLk (b) The Norton current can be found as V IN Th ZTh and the Norton admittance as YN 1 s2 LC 1 ZTh sL V1 s2 LCV2 2 s2 LC 1 V1 s LCV2 sL sL s2 LC 1 1 sL 1=sC sL sC sL 1=sC s2 LC 1
Determine the z parameters for the two-port network of Fig. 1-19.
For I2 0, by Ohm s law, Ia V1 V 1 10 6 16
CHAP. 1]
CIRCUIT ANALYSIS: PORT POINT OF VIEW
10 W 3
I2 + _ VB = 0 +
V1 _ I1 0
0.3Ia
6W Ia I2
V2 _
Fig. 1-19 Also, at node b, KCL gives I1 0:3Ia Ia 1:3Ia 1:3 Thus, by (1.10), z11 Further, again by Ohm s law, Ia Substitution of (2) into (1) yields I1 1:3 so that, by (1.12), z21 V2 6 V2 6 2 V1 16 1
V1 16 12:308  I1 I2 0 1:3
V2 6 4:615  I1 I2 0 1:3
Now with I1 0, applying KCL at node a gives us I2 Ia 0:3Ia 1:3Ia The application of KVL then leads to V1 V2 10 0:3Ia 6Ia 3Ia 3Ia so that, by (1.11), z12 Now, substitution of (2) in (3) gives I2 1:3Ia 1:3 Hence, from (1.13), z22 V2 6 V1 3 2:308  I2 I1 0 1:3 3I2 1:3 3
V2 6 4:615  I2 I1 0 1:3
Solve Problem 1.11 using a SPICE method similar to that of Example 1.9.
The SPICE netlist code is
CIRCUIT ANALYSIS: PORT POINT OF VIEW
[CHAP. 1
Prbl_12.CIR z-parameter evaluation .PARAM I1value=1mA I2value=0mA I1 0 1 AC {I1value} F 1 0 VB 0.3 R1 1 2 10ohm VB 2 3 0V ; Current sense R2 3 0 6ohm I2 0 2 AC {I2value} .DC I1 0 1mA 1mA I2 1mA 0 1mA ; Nested loop .PRINT DC V(1) I(I1) V(2) I(I2) .END
A nested loop is used in the .DC statement to eliminate the need for two separate executions. As a consequence, data is generated for I1 I2 1mA and I1 I2 0, which is extraneous to the problem. Execute <Prb1_12.CIR> and poll the output le to obtain data to evaluate the z parameters by use of (1.10) to (1.13). V V 1 1:231 10 2 z11 1 12:31  I1 I2 0 I I1 I I2 0 1 10 3 V V 1 2:308 10 3 2:308  z12 1 I2 I1 0 I I2 I I1 0 1 10 3 V V 2 4:615 10 3 4:615  z21 2 I1 I2 0 I I1 I I1 0 1 10 3 V V 2 4:615 10 3 4:615  z22 2 I2 I1 0 I I2 I I1 0 1 10 3
Determine the h parameters for the two-port network of Fig. 1-19.
For V2 0; Ia  0; thus, I1 V1 =10 and, by (1.16), V 10  h11 1 I1 V2 0 Further, I2 I1 and, by (1.18), h21 Now, Ia V2 =6. With I1 0, KVL yields V1 V2 10 0:3Ia V2 10 0:3 and, from (1.17), h12 Finally, applying KCL at node a gives I2 Ia 0:3Ia 1:3 so that, by (1.19), h22 V2 6 V1 0:5 V2 I1 0 V2 1 V2 2 6 I2 1 I1 V2 0
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