Xl=2pifl

Solutions:

Please see answer in bold letters.

Note pi = 3.1415….

1. The voltage across a 15Ω is as indicated. Find the sinusoidal expression for the current. In addition, sketch the v and i waveform on the same axis.

Note: For the graph of a and b please see attached jpg photo with filename 1ab.jpg and for c and d please see attached photo with filename 1cd.jpg.

a. 15sin20t

v= 15sin20t

By ohms law,

i = v/r

i = 15sin20t / 15

i = sin20t A

Computation of period for graphing:

v= 15sin20t

i = sin20t

w = 20 = 2pi*f

f = 3.183 Hz

Period =1/f = 0.314 seconds

b. 300sin (377t+20)

v = 300sin (377t+20)

i = 300sin (377t+20) /15

i = 20 sin (377t+20) A

Computation of period for graphing:

v = 300sin (377t+20)

i = 20 sin (377t+20)

w = 377 = 2pi*f

f = 60 Hz

Period = 1/60 = 0.017 seconds

shift to the left by:

2pi/0.017 = (20/180*pi)/x

x = 9.44x10-4 seconds

c. 60cos (wt+10)

v = 60cos (wt+10)

i = 60cos (wt+10)/15

i = 4cos (wt+10) A

Computation of period for graphing:

let’s denote the period as w sifted to the left by:

10/180*pi = pi/18

d. -45sin (wt+45)

v = -45sin (wt+45)

i = -45sin (wt+45) / 15

i = -3 sin (wt+45) A

Computation of period for graphing:

let’s denote the period as w sifted to the left by:

45/180 * pi = 1/4*pi

2. Determine the inductive reactance (in ohms) of a 5mH coil for

a. dc

Note at dc, frequency (f) = 0

Formula: XL = 2*pi*fL

XL = 2*pi* (0) (5m)

XL = 0 Ω

b. 60 Hz

Formula: XL = 2*pi*fL

XL = 2 (60) (5m)

XL = 1.885 Ω

c. 4kHz

Formula: XL = 2*pi*fL

XL = = 2*pi* (4k)(5m)

XL = 125.664 Ω

d. 1.2 MHz

Formula: XL = 2*pi*fL

XL = 2*pi* (1.2 M) (5m)

XL = 37.7 kΩ

3. Determine the frequency at which a 10 mH inductance has the following inductive reactance.

a. XL = 10 Ω

Formula: XL = 2*pi*fL

Express in terms in f:

f = XL/2 pi*L

f = 10 / (2pi*10m)

f = 159.155 Hz

b. XL = 4 kΩ

f = XL/2pi*L

f = 4k / (2pi*10m)

f = 63.662 kHz

c. XL = 12 kΩ

f = XL/2piL

f = 12k / (2pi*10m)

f = 190.99 kHz

d. XL = 0.5 kΩ

f = XL/2piL

f = 0.5k / (2pi*10m)

f = 7.958 kHz

4. Determine the frequency at which a 1.3uF capacitor has the following capacitive reactance.

a. 10 Ω

Formula: XC = 1/ (2pifC)

Expressing in terms of f:

f = 1/ (2pi*XC*C)

f = 1/ (2pi*10*1.3u)

f = 12.243 kΩ

b. 1.2 kΩ

f = 1/ (2pi*XC*C)

f = 1/ (2pi*1.2k*1.3u)

f = 102.022 Ω

c. 0.1 Ω

f = 1/ (2pi*XC*C)

f = 1/ (2pi*0.1*1.3u)

f = 1.224 MΩ

d. 2000 Ω

f = 1/ (2pi*XC*C)

f = 1/ (2pi*2000*1.3u)

f = 61.213 Ω

5. For the following pairs of voltage and current, indicate whether the element is a capacitor, an inductor and a capacitor, an inductor, or a resistor and find the value of C, L, or R if insufficient data are given.

a. v = 55 sin (377t + 50)

i = 11 sin (377t -40)

Element is inductor

In this case voltage leads current (ELI) by exactly 90 degrees so that means the circuit is inductive and the element is inductor.

XL = 55/11 = 5 Ω

we know the w=2pif so

w= 377=2pif

f= 60 Hz

To compute for the value of L,

XL= 2pifL

L = 5/ (2pi*60)

L = 0.013 H

b.

Element is inductor

In this case voltage leads current (ELI) by exactly 90 degrees so that means the circuit is inductive and the element is inductor.

XL = 36/4 = 9 Ω

We know the w=2pif so

w= 754=2pif

f= 120 Hz

To compute for the value of L,

XL= 2pifL

L = 9/ (2pi*120)

L = 0.012 H

c. v=10.5sin(wt-13)

i = 1.5sin (wt-13)

In this case, the voltage and current are in phase which means that the circuit is resistive only. So

R = 10.5/1.5 = 7 Ω

6. For the network in the figure and the applied source:

i = 12sin (102t + 45)

a. Determine the sinusoidal expression for the source voltage vs

w= 102 = 100

100 = 2pif

f = 15.915 Hz

First find individual inductive reactances:

XL1 = 2pifL1

XL1 = 2pi(15.915 )(30 m)

XL1 = 3 Ω

XL2 = 2pifL2

XL2 = 2pi(15.915 )(90 m)

XL1 = 9 Ω

Total reactance = XL = 9(3)/ (9+3)

XL = 2.25 Ω

so vs = 2.25*12sin(102t + 45) and knowing that voltage leads the current by 90 degrees in an inductor, the equation is

vs = 27sin(100t +135o) volts

b. Find the sinusoidal expression for i1 and i2

By Current Division Theorem:

i1 = is [XL2/ (XL1+XL2)]

i1 = 12sin (102t + 45) (9/ (9+3))

i1 = 9 sin (102t + 45) A

i2 = is [XL1 /(XL1+XL2)]

i2 = 12sin (102t + 45) (12/ (9+3))

i2 = 3 sin (102t + 45) A

7. Convert the following from rectangular to polar

a. Z = -8-j16

let rectangular form Z= a + jb and polar form Z = |Z| ∠ θ

where |Z| = sqrt ( a2 + b2 ) and θ = tan-1 (b/a)

Z = sqrt [(-8)2 + (-16)2 ]

Z = 17.89

θ = tan-1 (b/a)

θ = tan-1 (-16/-8)

θ = 64.43 degrees

Since it is in third quadrant θ = 64.43 + 180 = 244.43 degrees

In polar form:

Z = 17.89∠ 244.43o

b. Z = 0.02 – j0.003

Z = sqrt ((0.02)2 + (-0.003)2 )

Z = 0.0202

θ = tan-1 (b/a)

θ = tan-1 (-0.003/0.02)

θ = -8.53 degrees

In polar form:

Z = 0.0202∠- 8.53o

c. Z = -6*10-3 – j6*10-3

Z = sqrt ((-6*10-3)2 + (-6*10-3)2)

Z = 8.485x10-3

θ = tan-1 (b/a)

θ = tan-1 (-6*10-3/-6*10-3)

θ = 45 degrees

Since it is in third quadrant θ = 45 + 180 = 225 degrees

In polar form:

Z = 8.485x10-3 ∠225o

d. Z = 200 + j0.02

Z = sqrt ((200)2 + (0.02)2)

Z = 200

θ = tan-1 (0.02/200)

θ = 5.73x10-3 degrees

In polar form:

Z = 200 ∠5.72x10-3

e. . Z = -1000+j20

Z = sqrt ((-1000)2 + (20)2 )

Z = 1000.20

θ = tan-1 (20/-1000)

θ = -1.146 degrees

Since it is in third quadrant θ = 180 -1.146 = 178.854 degrees

In polar form:

Z = 1000.20 ∠178.854 o

8. Perform the following operations in their respective forms

a. (142 + j7) + (9.8+j42) + (0.1 + j0.9)

Note: Add all real and imaginary numbers separately so

= (142 + 9.8 + 0.1) + j (7+42+0.9)

= 151.9 + j49.9

b. (167 + j243) – (-42.3 – j68)

= (167 + 42.3) +j (243 + 68)

= 209.3+j311

c. (7.8+j1)(4+j2)(7+j6)

First two factors: (7.8+j1)(4+j2) = (31.2+15.6j +4j -2)

= 29.2 +j19.6

Multiply this with the 3rd factor:

(29.2 +j19.6)( (7+j6) = 204.4 + j175.2 +j137.2 -117.6

= 86.8 + j312.4 This is the final answer.

d. (6.9∠8)(7.2∠72)

We note that to multiply in polar form, magnitudes will be multiplied and angles will be added so

(6.9∠8)(7.2∠72) = 6.9*7.2∠ (8+72)

= 49.68∠ 80

e. (8 + j8)/(2+j2)

Convert first to polar form

let Z1 = (8 + j8)

Z1 = sqrt ((8)2 + (8)2)

Z1 = 11.31

θ1 = tan-1 (8/8)

θ1 = 45 degrees

Z1 = 11.31 ∠45

let Z2 = (2 + j2)

Z2 = sqrt ((2)2 + (2)2)

Z2 = 2.83

θ2 = tan-1 (2/2)

θ2 = 45 degrees

Z2 = 2.83 ∠45

so

(8 + j8)/(2+j2) = 11.31 ∠45/2.83 ∠45

= 4∠0

Converting it to rectangular form

4∠0 = 4 + j0 This is the final answer

Applied Sciences

Architecture and Design

Biology

Business & Finance

Chemistry

Computer Science

Geography

Geology

Education

Engineering

English

Environmental science

Spanish

Government

History

Human Resource Management

Information Systems

Law

Literature

Mathematics

Nursing

Physics

Political Science

Psychology

Reading

Science

Social Science

Home

Blog

Archive

Contact

google+twitterfacebook

Copyright © 2019 HomeworkMarket.com