Inductive Circuits Fill in all the missing values. Refer to the following formulas: X L = 2 πfL L= X L 2 πf f = X L 2 πL Inductance (H) Frequency (Hz) Inductive Reactance ( Ω ) 1.2 60 0.085 213.628 1000 4712.389 0.65 600 3.6 678.584 25 411.459 0.5 60 0.85 6408.849 20 201.062 0.45 400 4.8 2412.743 1000 40.841

Question

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Answer 1

Textbook Question

Chapter 16, Problem 1PP

Inductive Circuits

Fill in all the missing values. Refer to the following formulas:

X L = 2 πfL L= X L 2 πf f = X L 2 πL

Inductance (H)	Frequency (Hz)	Inductive Reactance ( Ω )
1.2	60
0.085		213.628
	1000	4712.389
0.65	600
3.6		678.584
	25	411.459
0.5	60
0.85		6408.849
	20	201.062
0.45	400
4.8		2412.743
	1000	40.841

Expert Solution & Answer

To determine

The missing values in the table.

Answer to Problem 1PP

Inductance (H)	Frequency(Hz)	Inductive Reactance(Ω)
1.2	60	452.39
0.085	400	213.628
0.75	1000	4712.389
0.65	600	2450.44
3.6	30	678.584
2.61	25	411.459
0.5	60	188.50
0.85	1200	6408.849
1.6	20	201.062
0.45	400	1130
4.8	80	2412.743
6.44m	1000	40.841

Explanation of Solution

We are given the following formulae,

XL=2πfL ... (1)L=XL2πf ... (2)f=XL2πL ... (3)

(1) XL=2πfL =2π×60×1.2 =452.39 Ω(2) f=XL2πL =213.6282π×0.085 =400 Hz(3) L=XL2πf =4712.3892π×1000 =0.75 H

(4) XL=2πfL =2π×600×0.65 =2450.44 Ω(5) f=XL2πL =678.5842π×3.6 =30 Hz(6) L=XL2πf =411.4592π×25 =2.61 H

(7) XL=2πfL =2π×60×0.5 =188.50 Ω(8) f=XL2πL =6408.8492π×0.85 =1200 Hz(9) L=XL2πf =201.0622π×20 =1.6 H

(10) XL=2πfL =2π×400×0.45 =1130 Ω(11) f=XL2πL =2412.7432π×4.8 =80 Hz(12) L=XL2πf =40.4812π×1000 =6.44 mH

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Answer 2

Textbook Question

Chapter 16, Problem 1PP

Inductive Circuits

Fill in all the missing values. Refer to the following formulas:

X L = 2 πfL L= X L 2 πf f = X L 2 πL

Inductance (H)	Frequency (Hz)	Inductive Reactance ( Ω )
1.2	60
0.085		213.628
	1000	4712.389
0.65	600
3.6		678.584
	25	411.459
0.5	60
0.85		6408.849
	20	201.062
0.45	400
4.8		2412.743
	1000	40.841

Expert Solution & Answer

To determine

The missing values in the table.

Answer to Problem 1PP

Inductance (H)	Frequency(Hz)	Inductive Reactance(Ω)
1.2	60	452.39
0.085	400	213.628
0.75	1000	4712.389
0.65	600	2450.44
3.6	30	678.584
2.61	25	411.459
0.5	60	188.50
0.85	1200	6408.849
1.6	20	201.062
0.45	400	1130
4.8	80	2412.743
6.44m	1000	40.841

Explanation of Solution

We are given the following formulae,

XL=2πfL ... (1)L=XL2πf ... (2)f=XL2πL ... (3)

(1) XL=2πfL =2π×60×1.2 =452.39 Ω(2) f=XL2πL =213.6282π×0.085 =400 Hz(3) L=XL2πf =4712.3892π×1000 =0.75 H

(4) XL=2πfL =2π×600×0.65 =2450.44 Ω(5) f=XL2πL =678.5842π×3.6 =30 Hz(6) L=XL2πf =411.4592π×25 =2.61 H

(7) XL=2πfL =2π×60×0.5 =188.50 Ω(8) f=XL2πL =6408.8492π×0.85 =1200 Hz(9) L=XL2πf =201.0622π×20 =1.6 H

(10) XL=2πfL =2π×400×0.45 =1130 Ω(11) f=XL2πL =2412.7432π×4.8 =80 Hz(12) L=XL2πf =40.4812π×1000 =6.44 mH

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Answer 3

Textbook Question

Chapter 16, Problem 1PP

Inductive Circuits

Fill in all the missing values. Refer to the following formulas:

X L = 2 πfL L= X L 2 πf f = X L 2 πL

Inductance (H)	Frequency (Hz)	Inductive Reactance ( Ω )
1.2	60
0.085		213.628
	1000	4712.389
0.65	600
3.6		678.584
	25	411.459
0.5	60
0.85		6408.849
	20	201.062
0.45	400
4.8		2412.743
	1000	40.841

Expert Solution & Answer

To determine

The missing values in the table.

Answer to Problem 1PP

Inductance (H)	Frequency(Hz)	Inductive Reactance(Ω)
1.2	60	452.39
0.085	400	213.628
0.75	1000	4712.389
0.65	600	2450.44
3.6	30	678.584
2.61	25	411.459
0.5	60	188.50
0.85	1200	6408.849
1.6	20	201.062
0.45	400	1130
4.8	80	2412.743
6.44m	1000	40.841

Explanation of Solution

We are given the following formulae,

XL=2πfL ... (1)L=XL2πf ... (2)f=XL2πL ... (3)

(1) XL=2πfL =2π×60×1.2 =452.39 Ω(2) f=XL2πL =213.6282π×0.085 =400 Hz(3) L=XL2πf =4712.3892π×1000 =0.75 H

(4) XL=2πfL =2π×600×0.65 =2450.44 Ω(5) f=XL2πL =678.5842π×3.6 =30 Hz(6) L=XL2πf =411.4592π×25 =2.61 H

(7) XL=2πfL =2π×60×0.5 =188.50 Ω(8) f=XL2πL =6408.8492π×0.85 =1200 Hz(9) L=XL2πf =201.0622π×20 =1.6 H

(10) XL=2πfL =2π×400×0.45 =1130 Ω(11) f=XL2πL =2412.7432π×4.8 =80 Hz(12) L=XL2πf =40.4812π×1000 =6.44 mH

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Answer 4

Textbook Question

Chapter 16, Problem 1PP

Inductive Circuits

Fill in all the missing values. Refer to the following formulas:

X L = 2 πfL L= X L 2 πf f = X L 2 πL

Inductance (H)	Frequency (Hz)	Inductive Reactance ( Ω )
1.2	60
0.085		213.628
	1000	4712.389
0.65	600
3.6		678.584
	25	411.459
0.5	60
0.85		6408.849
	20	201.062
0.45	400
4.8		2412.743
	1000	40.841

Expert Solution & Answer

To determine

The missing values in the table.

Answer to Problem 1PP

Inductance (H)	Frequency(Hz)	Inductive Reactance(Ω)
1.2	60	452.39
0.085	400	213.628
0.75	1000	4712.389
0.65	600	2450.44
3.6	30	678.584
2.61	25	411.459
0.5	60	188.50
0.85	1200	6408.849
1.6	20	201.062
0.45	400	1130
4.8	80	2412.743
6.44m	1000	40.841

Explanation of Solution

We are given the following formulae,

XL=2πfL ... (1)L=XL2πf ... (2)f=XL2πL ... (3)

(1) XL=2πfL =2π×60×1.2 =452.39 Ω(2) f=XL2πL =213.6282π×0.085 =400 Hz(3) L=XL2πf =4712.3892π×1000 =0.75 H

(4) XL=2πfL =2π×600×0.65 =2450.44 Ω(5) f=XL2πL =678.5842π×3.6 =30 Hz(6) L=XL2πf =411.4592π×25 =2.61 H

(7) XL=2πfL =2π×60×0.5 =188.50 Ω(8) f=XL2πL =6408.8492π×0.85 =1200 Hz(9) L=XL2πf =201.0622π×20 =1.6 H

(10) XL=2πfL =2π×400×0.45 =1130 Ω(11) f=XL2πL =2412.7432π×4.8 =80 Hz(12) L=XL2πf =40.4812π×1000 =6.44 mH

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The missing values in the table.

Answer to Problem 1PP

Inductance (H)	Frequency(Hz)	Inductive Reactance(Ω)
1.2	60	452.39
0.085	400	213.628
0.75	1000	4712.389
0.65	600	2450.44
3.6	30	678.584
2.61	25	411.459
0.5	60	188.50
0.85	1200	6408.849
1.6	20	201.062
0.45	400	1130
4.8	80	2412.743
6.44m	1000	40.841

Explanation of Solution

We are given the following formulae,

XL=2πfL ... (1)L=XL2πf ... (2)f=XL2πL ... (3)

(1) XL=2πfL =2π×60×1.2 =452.39 Ω(2) f=XL2πL =213.6282π×0.085 =400 Hz(3) L=XL2πf =4712.3892π×1000 =0.75 H

(4) XL=2πfL =2π×600×0.65 =2450.44 Ω(5) f=XL2πL =678.5842π×3.6 =30 Hz(6) L=XL2πf =411.4592π×25 =2.61 H

(7) XL=2πfL =2π×60×0.5 =188.50 Ω(8) f=XL2πL =6408.8492π×0.85 =1200 Hz(9) L=XL2πf =201.0622π×20 =1.6 H

(10) XL=2πfL =2π×400×0.45 =1130 Ω(11) f=XL2πL =2412.7432π×4.8 =80 Hz(12) L=XL2πf =40.4812π×1000 =6.44 mH

Answer 8

Expert Solution & Answer

To determine

The missing values in the table.

Answer to Problem 1PP

Inductance (H)	Frequency(Hz)	Inductive Reactance(Ω)
1.2	60	452.39
0.085	400	213.628
0.75	1000	4712.389
0.65	600	2450.44
3.6	30	678.584
2.61	25	411.459
0.5	60	188.50
0.85	1200	6408.849
1.6	20	201.062
0.45	400	1130
4.8	80	2412.743
6.44m	1000	40.841

Explanation of Solution

We are given the following formulae,

XL=2πfL ... (1)L=XL2πf ... (2)f=XL2πL ... (3)

(1) XL=2πfL =2π×60×1.2 =452.39 Ω(2) f=XL2πL =213.6282π×0.085 =400 Hz(3) L=XL2πf =4712.3892π×1000 =0.75 H

(4) XL=2πfL =2π×600×0.65 =2450.44 Ω(5) f=XL2πL =678.5842π×3.6 =30 Hz(6) L=XL2πf =411.4592π×25 =2.61 H

(7) XL=2πfL =2π×60×0.5 =188.50 Ω(8) f=XL2πL =6408.8492π×0.85 =1200 Hz(9) L=XL2πf =201.0622π×20 =1.6 H

(10) XL=2πfL =2π×400×0.45 =1130 Ω(11) f=XL2πL =2412.7432π×4.8 =80 Hz(12) L=XL2πf =40.4812π×1000 =6.44 mH

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The missing values in the table.

Answer 12

Answer to Problem 1PP

Inductance (H)	Frequency(Hz)	Inductive Reactance(Ω)
1.2	60	452.39
0.085	400	213.628
0.75	1000	4712.389
0.65	600	2450.44
3.6	30	678.584
2.61	25	411.459
0.5	60	188.50
0.85	1200	6408.849
1.6	20	201.062
0.45	400	1130
4.8	80	2412.743
6.44m	1000	40.841

Answer 13

Explanation of Solution

We are given the following formulae,

XL=2πfL ... (1)L=XL2πf ... (2)f=XL2πL ... (3)

(1) XL=2πfL =2π×60×1.2 =452.39 Ω(2) f=XL2πL =213.6282π×0.085 =400 Hz(3) L=XL2πf =4712.3892π×1000 =0.75 H

(4) XL=2πfL =2π×600×0.65 =2450.44 Ω(5) f=XL2πL =678.5842π×3.6 =30 Hz(6) L=XL2πf =411.4592π×25 =2.61 H

(7) XL=2πfL =2π×60×0.5 =188.50 Ω(8) f=XL2πL =6408.8492π×0.85 =1200 Hz(9) L=XL2πf =201.0622π×20 =1.6 H

(10) XL=2πfL =2π×400×0.45 =1130 Ω(11) f=XL2πL =2412.7432π×4.8 =80 Hz(12) L=XL2πf =40.4812π×1000 =6.44 mH

Answer 14

Ch. 16 - 1. How many degrees are the current and voltage...Ch. 16 - 2. How many degrees are the current and voltage...Ch. 16 - To what is inductive reactance proportional?Ch. 16 - Four inductors, each having an inductance of 0.6...Ch. 16 - Three inductors are connected in parallel....Ch. 16 - 6. If the three inductors in Question 5 were...Ch. 16 - 7. An inductor is connected to a 240-V, 1000-Hz...Ch. 16 - 8. An inductor with an inductance of 3.6 H is...Ch. 16 - If the frequency in Question 8 is reduced to 50...Ch. 16 - 10. An inductor has an inductive reactance of 250...

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Answer to Problem 1PP

Explanation of Solution

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Chapter 16 Solutions