The document provides detailed calculations and sinusoidal expressions for current and voltage in various circuits based on given parameters, including various values of resistance, inductance, and capacitance. It covers methods for determining inductive and capacitive reactance, frequency, and converting complex numbers between rectangular and polar forms. The document also includes graphing instructions and specific numerical examples for illustrating the calculations.

1. Solution
s:
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

2. 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)

3. 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)

4. 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:

5. 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

6. 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

7. 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

8. 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

9. 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)

10. 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

11. 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.

12. 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

13. 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)

14. 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))

15. i2 = 3 sin (102t + 45) A

16. 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

17. 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

18. 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

19. 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

20. 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

21. 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

22. 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

23. Converting it to rectangular form
4∠ 0 = 4 + j0 This is the final answer