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Solutions:
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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 and for c and d please see attached photo with filename 1cd .
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 = 9.44×10-4 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 = 8.485×10-3
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 = 8.485×10-3 ∠225o
d.
Z = 200
+ j0.02
Z = sqrt ((200)2 + (0.02)2)
Z = 200
θ = tan-1 (0.02/200)
θ = 5.73×10-3 degrees
In polar form:
Z = 200 ∠5.72×10-3
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
