The required z parameters for the given condition.

Question

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Answer 1

Question

Chapter 16, Problem 39E

To determine

The required z parameters for the given condition.

Expert Solution & Answer

Answer to Problem 39E

The required z parameters is z=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω.

Explanation of Solution

Given data:

The angular frequency is ω=108 rad/s.

Calculation:

The given diagram is shown in Figure 1.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 1

The conversion from pF to F is given by,

1 pF=10−12 F

The conversion from 5 pF to F is given by,

5 pF=5×10−12 5 F

The conversion from kΩ to Ω is given by,

1 kΩ=103 Ω

The conversion from 100 kΩ to Ω is given by,

100 kΩ=100×103 Ω=100000 Ω

The conversion from 10 kΩ to Ω is given by,

10 kΩ=10×103 Ω=10000 Ω

The capacitive reactance of 5 pF capacitor is given by,

XC5=1jωC5

Substitute 108 rad/s for ω and 5×10−12 F for C5 in the above equation.

XC5=1j(108 rad/s)(5×10−12 F)=−j2000 Ω

The capacitive reactance of 1 pF capacitor is given by,

XC1=1jωC1

Substitute 108 rad/s for ω and 1×10−12 F for C1 in the above equation.

XC1=1j(108 rad/s)(1×10−12 F)=−j10000 Ω

The modified diagram is shown in Figure 2.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 2

Apply KCL at node VA.

−I1+VA100000+VA−j2000+VA−VB−j10000=0I1=10−5VA+j(5×10−4)VA+j(1×10−4)(VA−VB)I1=(10−5+j(6×10−4))VA−j(1×10−4)VB (1)

Apply KCL at node VB.

−I2+VB10000+0.01VA+VB−VA−j10000=0I2=(1×10−4)VB+0.01VA+j(1×10−4)(VB−VA)I1=(0.01−j(1×10−4))VA−((1×10−4)+j(1×10−4))VB (2)

The standard equation for admittance parameters is,

I1=y11VA+y12VA (3)

I2=y21VA+y22VB (4)

Write equation (1) and equation (2) in matrix form.

[I1I2]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)][VAVB] (5)

Write equation (3) and equation (4) in matrix form

[I1I2]=[y11y12y21y22][VAVB] (6)

Compare equation (5) with equation (6).

[y11y12y21y22]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]y=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]

The voltage expression V1 by Crammer’s rule is given by,

V1=|I1y12I2y22||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|I1−j(1×10−4)I2(1×10−4)+j(1×10−4)||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=(10−4+j10−4)I1−(−j10−4)I2(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−4+j10−4)I1+j10−4I2−(4.9×10−8)+j(1.061×10−6)=1.414×10−4∠45°I1+10−4∠90°I21.062×10−6∠92.64°

V1=(89.71−j98.38)I1+(94.06−j4.34)I2 (7)

The voltage expression V2 by Crammer’s rule is given by,

V2=|y11I1y21I2||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|10−5+j(6×10−4)I10.01−j(1×10−4)I2I2||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=10−5+j(6×10−4)I2−(0.01−j(1×10−4))I1(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−5+j(6×10−4))I2−(0.01−j(1×10−4))I1−(4.9×10−8)+j(1.061×10−6)=6×10−4∠89.04°I2+0.01∠−0.57°I11.062×10−6∠92.64°

V2=(527.27+j9401.42)I1+(563.86−j35.47)I2 (8)

The standard equation for impedance parameter are,

V1=z11I1+z12I2 (9)

V2=z21I1+z22I2 (10)

Compare equation (7) with equation (9).

z11=(89.71−j98.38) Ω

z12=(94.06−j4.34) Ω

Compare equation (8) with equation (10).

z21=(527.27+j9401.42) Ω

z22=(563.86−j35.47) Ω

The z matrix can be written as,

z=[(89.71−j98.38) Ω(94.06−j4.34) Ω(527.27+j9401.42) Ω(563.86−j35.47) Ω]=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω

Conclusion:

Therefore, the required z parameters is z=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω.

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Answer 2

Question

Chapter 16, Problem 39E

To determine

The required z parameters for the given condition.

Expert Solution & Answer

Answer to Problem 39E

The required z parameters is z=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω.

Explanation of Solution

Given data:

The angular frequency is ω=108 rad/s.

Calculation:

The given diagram is shown in Figure 1.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 1

The conversion from pF to F is given by,

1 pF=10−12 F

The conversion from 5 pF to F is given by,

5 pF=5×10−12 5 F

The conversion from kΩ to Ω is given by,

1 kΩ=103 Ω

The conversion from 100 kΩ to Ω is given by,

100 kΩ=100×103 Ω=100000 Ω

The conversion from 10 kΩ to Ω is given by,

10 kΩ=10×103 Ω=10000 Ω

The capacitive reactance of 5 pF capacitor is given by,

XC5=1jωC5

Substitute 108 rad/s for ω and 5×10−12 F for C5 in the above equation.

XC5=1j(108 rad/s)(5×10−12 F)=−j2000 Ω

The capacitive reactance of 1 pF capacitor is given by,

XC1=1jωC1

Substitute 108 rad/s for ω and 1×10−12 F for C1 in the above equation.

XC1=1j(108 rad/s)(1×10−12 F)=−j10000 Ω

The modified diagram is shown in Figure 2.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 2

Apply KCL at node VA.

−I1+VA100000+VA−j2000+VA−VB−j10000=0I1=10−5VA+j(5×10−4)VA+j(1×10−4)(VA−VB)I1=(10−5+j(6×10−4))VA−j(1×10−4)VB (1)

Apply KCL at node VB.

−I2+VB10000+0.01VA+VB−VA−j10000=0I2=(1×10−4)VB+0.01VA+j(1×10−4)(VB−VA)I1=(0.01−j(1×10−4))VA−((1×10−4)+j(1×10−4))VB (2)

The standard equation for admittance parameters is,

I1=y11VA+y12VA (3)

I2=y21VA+y22VB (4)

Write equation (1) and equation (2) in matrix form.

[I1I2]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)][VAVB] (5)

Write equation (3) and equation (4) in matrix form

[I1I2]=[y11y12y21y22][VAVB] (6)

Compare equation (5) with equation (6).

[y11y12y21y22]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]y=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]

The voltage expression V1 by Crammer’s rule is given by,

V1=|I1y12I2y22||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|I1−j(1×10−4)I2(1×10−4)+j(1×10−4)||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=(10−4+j10−4)I1−(−j10−4)I2(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−4+j10−4)I1+j10−4I2−(4.9×10−8)+j(1.061×10−6)=1.414×10−4∠45°I1+10−4∠90°I21.062×10−6∠92.64°

V1=(89.71−j98.38)I1+(94.06−j4.34)I2 (7)

The voltage expression V2 by Crammer’s rule is given by,

V2=|y11I1y21I2||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|10−5+j(6×10−4)I10.01−j(1×10−4)I2I2||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=10−5+j(6×10−4)I2−(0.01−j(1×10−4))I1(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−5+j(6×10−4))I2−(0.01−j(1×10−4))I1−(4.9×10−8)+j(1.061×10−6)=6×10−4∠89.04°I2+0.01∠−0.57°I11.062×10−6∠92.64°

V2=(527.27+j9401.42)I1+(563.86−j35.47)I2 (8)

The standard equation for impedance parameter are,

V1=z11I1+z12I2 (9)

V2=z21I1+z22I2 (10)

Compare equation (7) with equation (9).

z11=(89.71−j98.38) Ω

z12=(94.06−j4.34) Ω

Compare equation (8) with equation (10).

z21=(527.27+j9401.42) Ω

z22=(563.86−j35.47) Ω

The z matrix can be written as,

z=[(89.71−j98.38) Ω(94.06−j4.34) Ω(527.27+j9401.42) Ω(563.86−j35.47) Ω]=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω

Conclusion:

Therefore, the required z parameters is z=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω.

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Answer 3

Question

Chapter 16, Problem 39E

To determine

The required z parameters for the given condition.

Expert Solution & Answer

Answer to Problem 39E

The required z parameters is z=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω.

Explanation of Solution

Given data:

The angular frequency is ω=108 rad/s.

Calculation:

The given diagram is shown in Figure 1.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 1

The conversion from pF to F is given by,

1 pF=10−12 F

The conversion from 5 pF to F is given by,

5 pF=5×10−12 5 F

The conversion from kΩ to Ω is given by,

1 kΩ=103 Ω

The conversion from 100 kΩ to Ω is given by,

100 kΩ=100×103 Ω=100000 Ω

The conversion from 10 kΩ to Ω is given by,

10 kΩ=10×103 Ω=10000 Ω

The capacitive reactance of 5 pF capacitor is given by,

XC5=1jωC5

Substitute 108 rad/s for ω and 5×10−12 F for C5 in the above equation.

XC5=1j(108 rad/s)(5×10−12 F)=−j2000 Ω

The capacitive reactance of 1 pF capacitor is given by,

XC1=1jωC1

Substitute 108 rad/s for ω and 1×10−12 F for C1 in the above equation.

XC1=1j(108 rad/s)(1×10−12 F)=−j10000 Ω

The modified diagram is shown in Figure 2.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 2

Apply KCL at node VA.

−I1+VA100000+VA−j2000+VA−VB−j10000=0I1=10−5VA+j(5×10−4)VA+j(1×10−4)(VA−VB)I1=(10−5+j(6×10−4))VA−j(1×10−4)VB (1)

Apply KCL at node VB.

−I2+VB10000+0.01VA+VB−VA−j10000=0I2=(1×10−4)VB+0.01VA+j(1×10−4)(VB−VA)I1=(0.01−j(1×10−4))VA−((1×10−4)+j(1×10−4))VB (2)

The standard equation for admittance parameters is,

I1=y11VA+y12VA (3)

I2=y21VA+y22VB (4)

Write equation (1) and equation (2) in matrix form.

[I1I2]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)][VAVB] (5)

Write equation (3) and equation (4) in matrix form

[I1I2]=[y11y12y21y22][VAVB] (6)

Compare equation (5) with equation (6).

[y11y12y21y22]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]y=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]

The voltage expression V1 by Crammer’s rule is given by,

V1=|I1y12I2y22||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|I1−j(1×10−4)I2(1×10−4)+j(1×10−4)||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=(10−4+j10−4)I1−(−j10−4)I2(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−4+j10−4)I1+j10−4I2−(4.9×10−8)+j(1.061×10−6)=1.414×10−4∠45°I1+10−4∠90°I21.062×10−6∠92.64°

V1=(89.71−j98.38)I1+(94.06−j4.34)I2 (7)

The voltage expression V2 by Crammer’s rule is given by,

V2=|y11I1y21I2||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|10−5+j(6×10−4)I10.01−j(1×10−4)I2I2||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=10−5+j(6×10−4)I2−(0.01−j(1×10−4))I1(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−5+j(6×10−4))I2−(0.01−j(1×10−4))I1−(4.9×10−8)+j(1.061×10−6)=6×10−4∠89.04°I2+0.01∠−0.57°I11.062×10−6∠92.64°

V2=(527.27+j9401.42)I1+(563.86−j35.47)I2 (8)

The standard equation for impedance parameter are,

V1=z11I1+z12I2 (9)

V2=z21I1+z22I2 (10)

Compare equation (7) with equation (9).

z11=(89.71−j98.38) Ω

z12=(94.06−j4.34) Ω

Compare equation (8) with equation (10).

z21=(527.27+j9401.42) Ω

z22=(563.86−j35.47) Ω

The z matrix can be written as,

z=[(89.71−j98.38) Ω(94.06−j4.34) Ω(527.27+j9401.42) Ω(563.86−j35.47) Ω]=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω

Conclusion:

Therefore, the required z parameters is z=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω.

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Answer 4

Question

Chapter 16, Problem 39E

To determine

The required z parameters for the given condition.

Expert Solution & Answer

Answer to Problem 39E

The required z parameters is z=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω.

Explanation of Solution

Given data:

The angular frequency is ω=108 rad/s.

Calculation:

The given diagram is shown in Figure 1.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 1

The conversion from pF to F is given by,

1 pF=10−12 F

The conversion from 5 pF to F is given by,

5 pF=5×10−12 5 F

The conversion from kΩ to Ω is given by,

1 kΩ=103 Ω

The conversion from 100 kΩ to Ω is given by,

100 kΩ=100×103 Ω=100000 Ω

The conversion from 10 kΩ to Ω is given by,

10 kΩ=10×103 Ω=10000 Ω

The capacitive reactance of 5 pF capacitor is given by,

XC5=1jωC5

Substitute 108 rad/s for ω and 5×10−12 F for C5 in the above equation.

XC5=1j(108 rad/s)(5×10−12 F)=−j2000 Ω

The capacitive reactance of 1 pF capacitor is given by,

XC1=1jωC1

Substitute 108 rad/s for ω and 1×10−12 F for C1 in the above equation.

XC1=1j(108 rad/s)(1×10−12 F)=−j10000 Ω

The modified diagram is shown in Figure 2.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 2

Apply KCL at node VA.

−I1+VA100000+VA−j2000+VA−VB−j10000=0I1=10−5VA+j(5×10−4)VA+j(1×10−4)(VA−VB)I1=(10−5+j(6×10−4))VA−j(1×10−4)VB (1)

Apply KCL at node VB.

−I2+VB10000+0.01VA+VB−VA−j10000=0I2=(1×10−4)VB+0.01VA+j(1×10−4)(VB−VA)I1=(0.01−j(1×10−4))VA−((1×10−4)+j(1×10−4))VB (2)

The standard equation for admittance parameters is,

I1=y11VA+y12VA (3)

I2=y21VA+y22VB (4)

Write equation (1) and equation (2) in matrix form.

[I1I2]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)][VAVB] (5)

Write equation (3) and equation (4) in matrix form

[I1I2]=[y11y12y21y22][VAVB] (6)

Compare equation (5) with equation (6).

[y11y12y21y22]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]y=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]

The voltage expression V1 by Crammer’s rule is given by,

V1=|I1y12I2y22||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|I1−j(1×10−4)I2(1×10−4)+j(1×10−4)||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=(10−4+j10−4)I1−(−j10−4)I2(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−4+j10−4)I1+j10−4I2−(4.9×10−8)+j(1.061×10−6)=1.414×10−4∠45°I1+10−4∠90°I21.062×10−6∠92.64°

V1=(89.71−j98.38)I1+(94.06−j4.34)I2 (7)

The voltage expression V2 by Crammer’s rule is given by,

V2=|y11I1y21I2||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|10−5+j(6×10−4)I10.01−j(1×10−4)I2I2||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=10−5+j(6×10−4)I2−(0.01−j(1×10−4))I1(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−5+j(6×10−4))I2−(0.01−j(1×10−4))I1−(4.9×10−8)+j(1.061×10−6)=6×10−4∠89.04°I2+0.01∠−0.57°I11.062×10−6∠92.64°

V2=(527.27+j9401.42)I1+(563.86−j35.47)I2 (8)

The standard equation for impedance parameter are,

V1=z11I1+z12I2 (9)

V2=z21I1+z22I2 (10)

Compare equation (7) with equation (9).

z11=(89.71−j98.38) Ω

z12=(94.06−j4.34) Ω

Compare equation (8) with equation (10).

z21=(527.27+j9401.42) Ω

z22=(563.86−j35.47) Ω

The z matrix can be written as,

z=[(89.71−j98.38) Ω(94.06−j4.34) Ω(527.27+j9401.42) Ω(563.86−j35.47) Ω]=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω

Conclusion:

Therefore, the required z parameters is z=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω.

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Answer 5

Chapter 16, Problem 39E

To determine

The required z parameters for the given condition.

Expert Solution & Answer

Answer to Problem 39E

The required z parameters is z=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω.

Explanation of Solution

Given data:

The angular frequency is ω=108 rad/s.

Calculation:

The given diagram is shown in Figure 1.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 1

The conversion from pF to F is given by,

1 pF=10−12 F

The conversion from 5 pF to F is given by,

5 pF=5×10−12 5 F

The conversion from kΩ to Ω is given by,

1 kΩ=103 Ω

The conversion from 100 kΩ to Ω is given by,

100 kΩ=100×103 Ω=100000 Ω

The conversion from 10 kΩ to Ω is given by,

10 kΩ=10×103 Ω=10000 Ω

The capacitive reactance of 5 pF capacitor is given by,

XC5=1jωC5

Substitute 108 rad/s for ω and 5×10−12 F for C5 in the above equation.

XC5=1j(108 rad/s)(5×10−12 F)=−j2000 Ω

The capacitive reactance of 1 pF capacitor is given by,

XC1=1jωC1

Substitute 108 rad/s for ω and 1×10−12 F for C1 in the above equation.

XC1=1j(108 rad/s)(1×10−12 F)=−j10000 Ω

The modified diagram is shown in Figure 2.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 2

Apply KCL at node VA.

−I1+VA100000+VA−j2000+VA−VB−j10000=0I1=10−5VA+j(5×10−4)VA+j(1×10−4)(VA−VB)I1=(10−5+j(6×10−4))VA−j(1×10−4)VB (1)

Apply KCL at node VB.

−I2+VB10000+0.01VA+VB−VA−j10000=0I2=(1×10−4)VB+0.01VA+j(1×10−4)(VB−VA)I1=(0.01−j(1×10−4))VA−((1×10−4)+j(1×10−4))VB (2)

The standard equation for admittance parameters is,

I1=y11VA+y12VA (3)

I2=y21VA+y22VB (4)

Write equation (1) and equation (2) in matrix form.

[I1I2]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)][VAVB] (5)

Write equation (3) and equation (4) in matrix form

[I1I2]=[y11y12y21y22][VAVB] (6)

Compare equation (5) with equation (6).

[y11y12y21y22]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]y=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]

The voltage expression V1 by Crammer’s rule is given by,

V1=|I1y12I2y22||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|I1−j(1×10−4)I2(1×10−4)+j(1×10−4)||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=(10−4+j10−4)I1−(−j10−4)I2(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−4+j10−4)I1+j10−4I2−(4.9×10−8)+j(1.061×10−6)=1.414×10−4∠45°I1+10−4∠90°I21.062×10−6∠92.64°

V1=(89.71−j98.38)I1+(94.06−j4.34)I2 (7)

The voltage expression V2 by Crammer’s rule is given by,

V2=|y11I1y21I2||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|10−5+j(6×10−4)I10.01−j(1×10−4)I2I2||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=10−5+j(6×10−4)I2−(0.01−j(1×10−4))I1(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−5+j(6×10−4))I2−(0.01−j(1×10−4))I1−(4.9×10−8)+j(1.061×10−6)=6×10−4∠89.04°I2+0.01∠−0.57°I11.062×10−6∠92.64°

V2=(527.27+j9401.42)I1+(563.86−j35.47)I2 (8)

The standard equation for impedance parameter are,

V1=z11I1+z12I2 (9)

V2=z21I1+z22I2 (10)

Compare equation (7) with equation (9).

z11=(89.71−j98.38) Ω

z12=(94.06−j4.34) Ω

Compare equation (8) with equation (10).

z21=(527.27+j9401.42) Ω

z22=(563.86−j35.47) Ω

The z matrix can be written as,

z=[(89.71−j98.38) Ω(94.06−j4.34) Ω(527.27+j9401.42) Ω(563.86−j35.47) Ω]=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω

Conclusion:

Therefore, the required z parameters is z=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω.

Answer 6

Chapter 16, Problem 39E

To determine

The required z parameters for the given condition.

Expert Solution & Answer

Answer to Problem 39E

The required z parameters is z=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω.

Explanation of Solution

Given data:

The angular frequency is ω=108 rad/s.

Calculation:

The given diagram is shown in Figure 1.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 1

The conversion from pF to F is given by,

1 pF=10−12 F

The conversion from 5 pF to F is given by,

5 pF=5×10−12 5 F

The conversion from kΩ to Ω is given by,

1 kΩ=103 Ω

The conversion from 100 kΩ to Ω is given by,

100 kΩ=100×103 Ω=100000 Ω

The conversion from 10 kΩ to Ω is given by,

10 kΩ=10×103 Ω=10000 Ω

The capacitive reactance of 5 pF capacitor is given by,

XC5=1jωC5

Substitute 108 rad/s for ω and 5×10−12 F for C5 in the above equation.

XC5=1j(108 rad/s)(5×10−12 F)=−j2000 Ω

The capacitive reactance of 1 pF capacitor is given by,

XC1=1jωC1

Substitute 108 rad/s for ω and 1×10−12 F for C1 in the above equation.

XC1=1j(108 rad/s)(1×10−12 F)=−j10000 Ω

The modified diagram is shown in Figure 2.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 2

Apply KCL at node VA.

−I1+VA100000+VA−j2000+VA−VB−j10000=0I1=10−5VA+j(5×10−4)VA+j(1×10−4)(VA−VB)I1=(10−5+j(6×10−4))VA−j(1×10−4)VB (1)

Apply KCL at node VB.

−I2+VB10000+0.01VA+VB−VA−j10000=0I2=(1×10−4)VB+0.01VA+j(1×10−4)(VB−VA)I1=(0.01−j(1×10−4))VA−((1×10−4)+j(1×10−4))VB (2)

The standard equation for admittance parameters is,

I1=y11VA+y12VA (3)

I2=y21VA+y22VB (4)

Write equation (1) and equation (2) in matrix form.

[I1I2]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)][VAVB] (5)

Write equation (3) and equation (4) in matrix form

[I1I2]=[y11y12y21y22][VAVB] (6)

Compare equation (5) with equation (6).

[y11y12y21y22]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]y=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]

The voltage expression V1 by Crammer’s rule is given by,

V1=|I1y12I2y22||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|I1−j(1×10−4)I2(1×10−4)+j(1×10−4)||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=(10−4+j10−4)I1−(−j10−4)I2(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−4+j10−4)I1+j10−4I2−(4.9×10−8)+j(1.061×10−6)=1.414×10−4∠45°I1+10−4∠90°I21.062×10−6∠92.64°

V1=(89.71−j98.38)I1+(94.06−j4.34)I2 (7)

The voltage expression V2 by Crammer’s rule is given by,

V2=|y11I1y21I2||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|10−5+j(6×10−4)I10.01−j(1×10−4)I2I2||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=10−5+j(6×10−4)I2−(0.01−j(1×10−4))I1(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−5+j(6×10−4))I2−(0.01−j(1×10−4))I1−(4.9×10−8)+j(1.061×10−6)=6×10−4∠89.04°I2+0.01∠−0.57°I11.062×10−6∠92.64°

V2=(527.27+j9401.42)I1+(563.86−j35.47)I2 (8)

The standard equation for impedance parameter are,

V1=z11I1+z12I2 (9)

V2=z21I1+z22I2 (10)

Compare equation (7) with equation (9).

z11=(89.71−j98.38) Ω

z12=(94.06−j4.34) Ω

Compare equation (8) with equation (10).

z21=(527.27+j9401.42) Ω

z22=(563.86−j35.47) Ω

The z matrix can be written as,

z=[(89.71−j98.38) Ω(94.06−j4.34) Ω(527.27+j9401.42) Ω(563.86−j35.47) Ω]=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω

Conclusion:

Therefore, the required z parameters is z=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω.

Answer 7

Chapter 16, Problem 39E

To determine

The required z parameters for the given condition.

Answer 8

Expert Solution & Answer

Answer to Problem 39E

The required z parameters is z=[(89.71−j98.38)(94.06−j4.34)(527.27+j9401.42)(563.86−j35.47)] Ω.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given data:

The angular frequency is ω=108 rad/s.

Calculation:

The given diagram is shown in Figure 1.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 1

The conversion from pF to F is given by,

1 pF=10−12 F

The conversion from 5 pF to F is given by,

5 pF=5×10−12 5 F

The conversion from kΩ to Ω is given by,

1 kΩ=103 Ω

The conversion from 100 kΩ to Ω is given by,

100 kΩ=100×103 Ω=100000 Ω

The conversion from 10 kΩ to Ω is given by,

10 kΩ=10×103 Ω=10000 Ω

The capacitive reactance of 5 pF capacitor is given by,

XC5=1jωC5

Substitute 108 rad/s for ω and 5×10−12 F for C5 in the above equation.

XC5=1j(108 rad/s)(5×10−12 F)=−j2000 Ω

The capacitive reactance of 1 pF capacitor is given by,

XC1=1jωC1

Substitute 108 rad/s for ω and 1×10−12 F for C1 in the above equation.

XC1=1j(108 rad/s)(1×10−12 F)=−j10000 Ω

The modified diagram is shown in Figure 2.

ENGINEERING CIRCUIT ANALYSIS ACCESS >I<, Chapter 16, Problem 39E , additional homework tip 2

Apply KCL at node VA.

−I1+VA100000+VA−j2000+VA−VB−j10000=0I1=10−5VA+j(5×10−4)VA+j(1×10−4)(VA−VB)I1=(10−5+j(6×10−4))VA−j(1×10−4)VB (1)

Apply KCL at node VB.

−I2+VB10000+0.01VA+VB−VA−j10000=0I2=(1×10−4)VB+0.01VA+j(1×10−4)(VB−VA)I1=(0.01−j(1×10−4))VA−((1×10−4)+j(1×10−4))VB (2)

The standard equation for admittance parameters is,

I1=y11VA+y12VA (3)

I2=y21VA+y22VB (4)

Write equation (1) and equation (2) in matrix form.

[I1I2]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)][VAVB] (5)

Write equation (3) and equation (4) in matrix form

[I1I2]=[y11y12y21y22][VAVB] (6)

Compare equation (5) with equation (6).

[y11y12y21y22]=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]y=[10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)]

The voltage expression V1 by Crammer’s rule is given by,

V1=|I1y12I2y22||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|I1−j(1×10−4)I2(1×10−4)+j(1×10−4)||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=(10−4+j10−4)I1−(−j10−4)I2(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−4+j10−4)I1+j10−4I2−(4.9×10−8)+j(1.061×10−6)=1.414×10−4∠45°I1+10−4∠90°I21.062×10−6∠92.64°

V1=(89.71−j98.38)I1+(94.06−j4.34)I2 (7)

The voltage expression V2 by Crammer’s rule is given by,

V2=|y11I1y21I2||y11y12y21y22|

Substitute 10−5+j(6×10−4) for y11, −j(1×10−4) for y12, 0.01−j(1×10−4) for y21 and (1×10−4)+j(1×10−4) for y22 in above equation.

V1=|10−5+j(6×10−4)I10.01−j(1×10−4)I2I2||10−5+j(6×10−4)−j(1×10−4)0.01−j(1×10−4)(1×10−4)+j(1×10−4)|=10−5+j(6×10−4)I2−(0.01−j(1×10−4))I1(−5.9×10−8+j6.1×10−8)−(−1×10−8−j10−6)=(10−5+j(6×10−4))I2−(0.01−j(1×10−4))I1−(4.9×10−8)+j(1.061×10−6)=6×10−4∠89.04°I2+0.01∠−0.57°I11.062×10−6∠92.64°