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 . Write V R and V L in rectangular form. Determine V = V R + V L in both its rectangular and polar forms. Write the real and imaginary parts of V . Â Figure PS.1 A series RL circuit for problem P5-1.

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Answer 1

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 .

Write V R and V L
in rectangular form.
Determine V = V R + V L
in both its rectangular and polar forms.
Write the real and imaginary parts of V . Â

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

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=1∠0° VVL=1∠90° V

Calculation:

Consider voltage across resistor,

VR=1∠0° V⇒VR=1cos(0°)+j1sin(0°) =1+j0 V

Consider voltage across inductor,

VL=1∠90° V⇒VL=1cos(90°)+j1sin(90°) =0+j1 V

Conclusion:

The rectangular representation of given voltages,

VR=1+j0 VVL=0+j1 V

Expert Solution

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=1.414∠45° 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=1∠0° V=1+j0 VVL=1∠90° 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+12∠tan−1(11) =1.414∠45° V

Conclusion:

Total voltage in rectangular form is V=1+j1 V, and in polar form is V=1.414∠45° V.

Expert Solution

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=1∠0° VVL=1∠90° 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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Answer 2

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 .

Write V R and V L
in rectangular form.
Determine V = V R + V L
in both its rectangular and polar forms.
Write the real and imaginary parts of V . Â

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

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=1∠0° VVL=1∠90° V

Calculation:

Consider voltage across resistor,

VR=1∠0° V⇒VR=1cos(0°)+j1sin(0°) =1+j0 V

Consider voltage across inductor,

VL=1∠90° V⇒VL=1cos(90°)+j1sin(90°) =0+j1 V

Conclusion:

The rectangular representation of given voltages,

VR=1+j0 VVL=0+j1 V

Expert Solution

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=1.414∠45° 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=1∠0° V=1+j0 VVL=1∠90° 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+12∠tan−1(11) =1.414∠45° V

Conclusion:

Total voltage in rectangular form is V=1+j1 V, and in polar form is V=1.414∠45° V.

Expert Solution

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=1∠0° VVL=1∠90° 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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Answer 3

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 .

Write V R and V L
in rectangular form.
Determine V = V R + V L
in both its rectangular and polar forms.
Write the real and imaginary parts of V . Â

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

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=1∠0° VVL=1∠90° V

Calculation:

Consider voltage across resistor,

VR=1∠0° V⇒VR=1cos(0°)+j1sin(0°) =1+j0 V

Consider voltage across inductor,

VL=1∠90° V⇒VL=1cos(90°)+j1sin(90°) =0+j1 V

Conclusion:

The rectangular representation of given voltages,

VR=1+j0 VVL=0+j1 V

Expert Solution

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=1.414∠45° 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=1∠0° V=1+j0 VVL=1∠90° 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+12∠tan−1(11) =1.414∠45° V

Conclusion:

Total voltage in rectangular form is V=1+j1 V, and in polar form is V=1.414∠45° V.

Expert Solution

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=1∠0° VVL=1∠90° 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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Answer 4

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 .

Write V R and V L
in rectangular form.
Determine V = V R + V L
in both its rectangular and polar forms.
Write the real and imaginary parts of V . Â

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

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=1∠0° VVL=1∠90° V

Calculation:

Consider voltage across resistor,

VR=1∠0° V⇒VR=1cos(0°)+j1sin(0°) =1+j0 V

Consider voltage across inductor,

VL=1∠90° V⇒VL=1cos(90°)+j1sin(90°) =0+j1 V

Conclusion:

The rectangular representation of given voltages,

VR=1+j0 VVL=0+j1 V

Expert Solution

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=1.414∠45° 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=1∠0° V=1+j0 VVL=1∠90° 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+12∠tan−1(11) =1.414∠45° V

Conclusion:

Total voltage in rectangular form is V=1+j1 V, and in polar form is V=1.414∠45° V.

Expert Solution

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=1∠0° VVL=1∠90° 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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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

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=1∠0° VVL=1∠90° V

Calculation:

Consider voltage across resistor,

VR=1∠0° V⇒VR=1cos(0°)+j1sin(0°) =1+j0 V

Consider voltage across inductor,

VL=1∠90° V⇒VL=1cos(90°)+j1sin(90°) =0+j1 V

Conclusion:

The rectangular representation of given voltages,

VR=1+j0 VVL=0+j1 V

Expert Solution

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=1.414∠45° 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=1∠0° V=1+j0 VVL=1∠90° 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+12∠tan−1(11) =1.414∠45° V

Conclusion:

Total voltage in rectangular form is V=1+j1 V, and in polar form is V=1.414∠45° V.

Expert Solution

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=1∠0° VVL=1∠90° 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

Answer 8

Expert Solution

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=1∠0° VVL=1∠90° V

Calculation:

Consider voltage across resistor,

VR=1∠0° V⇒VR=1cos(0°)+j1sin(0°) =1+j0 V

Consider voltage across inductor,

VL=1∠90° V⇒VL=1cos(90°)+j1sin(90°) =0+j1 V

Conclusion:

The rectangular representation of given voltages,

VR=1+j0 VVL=0+j1 V

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

The rectangular representation of VR and VL.

Answer 13

Answer to Problem 1P

The rectangular representation of given voltages,

VR=1+j0 VVL=0+j1 V

Answer 14

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=1∠0° VVL=1∠90° V

Calculation:

Consider voltage across resistor,

VR=1∠0° V⇒VR=1cos(0°)+j1sin(0°) =1+j0 V

Consider voltage across inductor,

VL=1∠90° V⇒VL=1cos(90°)+j1sin(90°) =0+j1 V

Conclusion:

The rectangular representation of given voltages,

VR=1+j0 VVL=0+j1 V

Answer 15

Expert Solution

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=1.414∠45° 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=1∠0° V=1+j0 VVL=1∠90° 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+12∠tan−1(11) =1.414∠45° V

Conclusion:

Total voltage in rectangular form is V=1+j1 V, and in polar form is V=1.414∠45° V.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

Total voltage in rectangular and polar forms.

Answer 20

Answer to Problem 1P

Total voltage in rectangular form is V=1+j1 V, and in polar form is V=1.414∠45° V.

Answer 21

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=1∠0° V=1+j0 VVL=1∠90° 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+12∠tan−1(11) =1.414∠45° V

Conclusion:

Total voltage in rectangular form is V=1+j1 V, and in polar form is V=1.414∠45° V.

Answer 22

Expert Solution

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=1∠0° VVL=1∠90° 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

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

(c)

The real and imaginary part of total voltage V.

Answer 27

Answer to Problem 1P

Real part and imaginary part of total voltage are,

Re(V)=1Im(V)=1

Answer 28

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=1∠0° VVL=1∠90° 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

Answer 29

Ch. 5 - In the series RL circuit shown in Fig. P5.1,...Ch. 5 - Repeat problem P5-1 if VR=1045V and VL=545V.Ch. 5 - Repeat problem P5-1 if VR=926.6V and Ch. 5 - Prob. 4P Ch. 5 - Repeat problem P5-4 if VR=9.518.44V and...Ch. 5 - Repeat problem P5-4 if VR=1060V and VC=17.3230V.Ch. 5 - Prob. 7P Ch. 5 - Repeat problem P5-7 if IR=0.70745A and...Ch. 5 - Repeat problem P5-7 if IR=86.630A and IL=5060A.Ch. 5 - In the parallel RC circuit shown in Fig. P5.10....

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