Introductory Mathematics for Engineering Applications
Introductory Mathematics for Engineering Applications
1st Edition
ISBN: 9781118141809
Author: Nathan Klingbeil
Publisher: WILEY
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Textbook Question
Chapter 5, Problem 1P

In the series RL circuit shown in Fig. P5.1, voltage V L Leads voltage V R by 90 ° (i.e. if the angle of V R is 0 ° . The angle of V L is 90 ° ). Assume V R = 1 0 ° V and V L = 1 90 ° V .

  1. Write V R and V L
  2. in rectangular form.
  3. Determine V = V R + V L
  4. in both its rectangular and polar forms.
  5. Write the real and imaginary parts of V .  

Chapter 5, Problem 1P, In the series RL circuit shown in Fig. P5.1, voltage VL Leads voltage VR by 90 (i.e. if the angle of

Figure PS.1 A series RL circuit for problem P5-1.

Expert Solution
Check Mark
To determine

(a)

The rectangular representation of VR and VL.

Answer to Problem 1P

The rectangular representation of given voltages,

  VR=1+j0 VVL=0+j1 V

Explanation of Solution

Given Information:

The series RL circuit is given,

Introductory Mathematics for Engineering Applications, Chapter 5, Problem 1P , additional homework tip  1

  VR=10° VVL=190° V

Calculation:

Consider voltage across resistor,

  VR=10° VVR=1cos(0°)+j1sin(0°) =1+j0 V

Consider voltage across inductor,

  VL=190° VVL=1cos(90°)+j1sin(90°) =0+j1 V

Conclusion:

The rectangular representation of given voltages,

  VR=1+j0 VVL=0+j1 V

Expert Solution
Check Mark
To determine

(b)

Total voltage in rectangular and polar forms.

Answer to Problem 1P

Total voltage in rectangular form is V=1+j1 V, and in polar form is V=1.41445° V.

Explanation of Solution

Given Information:

The series RL circuit is given,

Introductory Mathematics for Engineering Applications, Chapter 5, Problem 1P , additional homework tip  2

  VR=10° V=1+j0 VVL=190° V=0+j1 V

Calculation:

Total voltage in rectangular form,

  V=VR+VL =(1+j0)+(0+j1) =1+j1 V

In polar form,

  V=12+12tan1(11) =1.41445° V

Conclusion:

Total voltage in rectangular form is V=1+j1 V, and in polar form is V=1.41445° V.

Expert Solution
Check Mark
To determine

(c)

The real and imaginary part of total voltage V.

Answer to Problem 1P

Real part and imaginary part of total voltage are,

  Re(V)=1Im(V)=1

Explanation of Solution

Given Information:

The series RL circuit is given,

Introductory Mathematics for Engineering Applications, Chapter 5, Problem 1P , additional homework tip  3

  VR=10° VVL=190° VV=1+j1 V

Calculation:

The total voltage is, V=1+j1 V.

From this rectangular form, real part,

  Re(V)=1.

Imaginary part,

  Im(V)=1.

Conclusion:

Real part and imaginary part of total voltage are,

  Re(V)=1Im(V)=1

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Chapter 5 Solutions

Introductory Mathematics for Engineering Applications

Ch. 5 - Repeat problem P5-10 if IR=0.50mA and IC=0.290mA.Ch. 5 - Repeat problem P5-10 if IR=0.92921.8mA and...Ch. 5 - The output voltage across the capacitor in a...Ch. 5 - The current flowing through the resistor in a...Ch. 5 - A resistor, capacitor, and an inductor are...Ch. 5 - Two circuit elements are connected in series as...Ch. 5 - An RC circuit is subjected to an alternating...Ch. 5 - A series-parallel electric circuit consists of the...Ch. 5 - The circuit shown in Fig. P5.19 consist of a...Ch. 5 - An electric circuit consists of two components as...Ch. 5 - A sinusoidal voltage source V = 110 V of frequency...Ch. 5 - A sinusoidal voltage source V=110223.2V is applied...Ch. 5 - An electric circuit consists of a resistor R, an...Ch. 5 - In the circuit shown in Fig. P5.24, the impedances...Ch. 5 - In the RC circuit shown in Fig. P5.25, the...Ch. 5 - In the RL circuit shown in Fig. P5.26, the...Ch. 5 - A resistor, capacitor, and inductor are connected...Ch. 5 - In the current divider circuit shown in Fig....Ch. 5 - In the current divider circuit shown in Fig....Ch. 5 - In the current divider circuit shown in Fig....Ch. 5 - In the current divider circuit shown in Fig....Ch. 5 - In the Op-Amp circuit shown in Fig. P5.32. the...Ch. 5 - In the Op-Amp circuit shown in Fig. P5.33, the...Ch. 5 - In the Op-Amp circuit shown in Fig. P534, the...Ch. 5 - In the Op-Amp circuit shown in Fig. P5.35. the...Ch. 5 - In the Op-Amp circuit shown in Fig. P5.36. the...Ch. 5 - Prob. 37PCh. 5 - Repeat problem P5-37 if Za=10, Zb=j20 and...Ch. 5 - In the circuit shown in Fig. 5.37, the impedances...Ch. 5 - Repeat problem P5-39 ¡f Z1=3.33+j3.33,...

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