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