Engineering Circuit Analysis
Engineering Circuit Analysis
9th Edition
ISBN: 9780073545516
Author: Hayt, William H. (william Hart), Jr, Kemmerly, Jack E. (jack Ellsworth), Durbin, Steven M.
Publisher: Mcgraw-hill Education,
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Nodal Analysis
In the analysis of any electric circuit, nodal analysis is the method of determining the potential difference or voltage between different nodes in an electric circuit. A node is a point where two or more branches connect each other. In the nodal analysi…
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Chapter 10, Problem 43E

(a)

To determine

Find the equivalent impedance of the network at f=1 Hz.

(a)

Expert Solution
Check Mark

Answer to Problem 43E

The equivalent impedance of the network at f=1 Hz is 30∠−0.35° Ω_.

Explanation of Solution

Given data:

f=1 Hz.

Formula used:

Consider the general expression for inductive impedance.

ZL=jωL        (1)

Here,

ω is the frequency.

L is the inductance.

Consider the general expression for capacitive impedance.

ZC=−j1ωC        (2)

Here,

ω is the frequency.

C is the capacitance.

Consider the general expression for angular frequency.

ω=2πf        (3)

Here,

f is the frequency.

Calculation:

Refer to Figure in the respective question.

Substitute 1 Hz for f in equation (3) as follows.

ω=2π(1 Hz)=6.28 rad/s

From Figure, write the expression for equivalent impedance.

Zeq=[(ZC||60 Ω)+(60 Ω||ZL)]||60 Ω        (4)

Substitute 10 mH for L and 6.28 rad/s for ω in equation (1) as follows.

ZL=j(6.28 rad/s)(10 mH)=j(6.28)(10×10−3) {∵1 mH=1×10−3 H}=j0.0628  Ω

Substitute 30 μF for C and 6.28 rad/s for ω in equation (2) as follows.

ZC=−j1(6.28 rad/s)(30 μF)=−j1(6.28)(30×10−6)=−j11.884×10−4=−j5307.8 Ω

Substitute j0.0628  Ω for ZL and −j5307.8 Ω for ZC in equation (4) as follows.

Zeq=[(−j5307.8 Ω||60 Ω)+(60 Ω||j0.0628  Ω)]||60 Ω=[(60(−j5307.8)60−j5307.8)+(60(j0.0628)60+j0.0628)]||60=[(−j31846860−j5307.8)+(j3.76860+j0.0628)]||60=[(59.9−j0.678)+(6.57×10−5−j0.062)]||60

Simplify the equation as follows.

Zeq=(59.9−j0.74)||60=(59.9−j0.74)60(59.9−j0.74)+60=3594−j44.4119.9−j0.74=29.9−j0.185

Simplify the expression as follows:

Zeq=29.9∠−0.35° Ω≅30∠−0.35° Ω

Conclusion:

Thus, the equivalent impedance of the network at f=1 Hz is 30∠−0.35° Ω_.

(b)

To determine

Find the equivalent impedance of the network at f=1 kHz.

(b)

Expert Solution
Check Mark

Answer to Problem 43E

The equivalent impedance of the network at f=1 kHz is 25.38∠52.7° Ω_.

Explanation of Solution

Given data:

f=1 kHz.

Calculation:

Substitute 1 kHz for f in equation (3) as follows.

ω=2π(1 kHz)=6283.1 rad/s

Substitute 10 mH for L and 6283.1 rad/s for ω in equation (1) as follows.

ZL=j(6283.1 rad/s)(10 mH)=j(6283.1)(10×10−3) {∵1 mH=1×10−3 H}=j62.83 Ω

Substitute 30 μF for C and 6283.1 rad/s for ω in equation (2) as follows.

ZC=−j1(6283.1 rad/s)(30 μF)=−j1(6283.1)(30×10−6)=−j10.1884=−j5.307 Ω

Substitute j62.83 Ω for ZL and −j5.307 Ω for ZC in equation (4) as follows.

Zeq=[(−j5.307 Ω||60 Ω)+(60 Ω||j62.83 Ω)]||60 Ω=[(60(−j5.307 Ω)60−j5.307 Ω)+(60(j62.83)60+j62.83)]||60=[(−j318.4260−j5.307 Ω)+(j3769.860+j62.83)]||60=[(0.465−j5.265)+(31.38+j29.9)]||60

Simplify the equation as follows.

Zeq=(31.845−j24.635)||60=(31.845−j24.635)60(31.845−j24.635)+60=1910.7+j1478.191.845−j24.635=15.38+j20.2

Zeq=25.38∠52.7° Ω

Conclusion:

Thus, the equivalent impedance of the network at f=1 kHz is 25.38∠52.7° Ω_.

(c)

To determine

Find the equivalent impedance of the network at f=1 MHz.

(c)

Expert Solution
Check Mark

Answer to Problem 43E

The equivalent impedance of the network at f=1 MHz is 29.97∠0.042° Ω_.

Explanation of Solution

Given data:

f=1 MHz.

Calculation:

Substitute 1 MHz for f in equation (3) as follows.

ω=2π(1 MHz)=6.283 Mrad/s

Substitute 10 mH for L and 6.283 Mrad/s for ω in equation (1) as follows.

ZL=j(6.283 Mrad/s)(10 mH)=j(6.283×106)(10×10−3) {∵1 mH=1×10−3 H}=j62,830 Ω

Substitute 30 μF for C and 6.2831 Mrad/s for ω in equation (2) as follows.

ZC=−j1(6.2831 Mrad/s)(30 μF)=−j1(6.2831×106)(30×10−6)=−j1188.493=−j5.305×10−3 Ω

Substitute j62,830 Ω for ZL and −j5.305×10−3 Ω for ZC in equation (4) as follows.

Zeq=[(−j5.305×10−3 Ω||60 Ω)+(60 Ω||j62,830 Ω)]||60 Ω=[(60(−j5.305×10−3 Ω)60−j5.305×10−3 Ω)+(60(j62,830 Ω)60+j62,830 Ω)]||60=[(−j0.318360−j5.305×10−3 Ω)+(j376980060+j62,830 Ω)]||60=[(4.69×10−7−j5.3×10−3)+(59.9+j0.057)]||60

Simplify the equation as follows.

Zeq=(59.9+j0.01517)||60=(59.9+j0.01517)60(59.9+j0.01517)+60=3594+j3.102119.9+j0.01517=29.97+j0.022

Zeq=29.97∠0.042° Ω

Conclusion:

Thus, the equivalent impedance of the network at f=1 MHz is 29.97∠0.042° Ω_.

(d)

To determine

Find the equivalent impedance of the network at f=1 GHz.

(d)

Expert Solution
Check Mark

Answer to Problem 43E

The equivalent impedance of the network at f=1 GHz is 30∠0° Ω_.

Explanation of Solution

Given data:

f=1 GHz.

Calculation:

Substitute 1 GHz for f in equation (3) as follows.

ω=2π(1 GHz)=6.283 Grad/s

Substitute 10 mH for L and 6.283 Grad/s for ω in equation (1) as follows.

ZL=j(6.283 Grad/s)(10 mH)=j(6.283×109)(10×10−3) {∵1 mH=1×10−3 H}=j62830000 Ω

Substitute 30 μF for C and 6.283 Grad/s for ω in equation (2) as follows.

ZC=−j1(6.283 Grad/s)(30 μF)=−j1(6.2831×109)(30×10−6)=−j1188493=−j5.305×10−6 Ω

Substitute j62830000 Ω for ZL and −j5.305×10−6 Ω for ZC in equation (4) as follows.

Zeq=[(−j5.305×10−6 Ω||60 Ω)+(60 Ω||j62830000 Ω)]||60 Ω=[(60(−j5.305×10−6 Ω)60−j5.305×10−6 Ω)+(60(j62830000 Ω)60+j62830000 Ω)]||60=[(−j3.183×10−460−j5.305×10−3 Ω)+(j376980000060+j62830000 Ω)]||60=[(4.69×10−10−j5.3×10−6)+(60+j5.72×10−5)]||60

Simplify the equation as follows.

Zeq=(60+j5.19×10−5)||60=(60+j5.19×10−5)60(60+j5.19×10−5)+60=3600+j3.114×10−3120+j5.19×10−5=30+j1.2975×10−5

Simplify the equation as follows.

Zeq=30∠2.47×10−5° Ω≅30∠0° Ω

Conclusion:

Thus, the equivalent impedance of the network at f=1 GHz is 30∠0° Ω_.

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