ECE 212Electrical Circuits II

HW 01 Assigned: Wed, Jan 11Due: Fri, Jan 13

Law 9.1

Consider the sinusoidal voltage:

v t( ) 2 120⋅ cos 377 t⋅ 30 deg⋅−( )⋅ V⋅:=

a) What is the maximum amplitude of the voltage?

Vm 2 120⋅ V⋅:= Vm 169.7 V=

b) What is the angular frequency of the voltage in radians/second?

ω1 377radsec⋅:= ω1 377.0

radsec⋅=

c) What is the frequency of the voltage in Hz?

f1ω12 π⋅

:= f1 60.0 Hz⋅=

d) What is the period of the voltage in seconds?

T11f1

:= T1 16.7 ms⋅=

e) What is the phase angle of the voltage in radians?

φ1 30− deg⋅:= φ1 0.5− rad⋅=

HW01 Solution.xmcdC:\JLAW\CLASSES\2012 Spring 212\Homework\

Page 1 of 5 January 13, 2012Solution

ECE 212Electrical Circuits II

HW 01 Assigned: Wed, Jan 11Due: Fri, Jan 13

f) What is the first time after t=0 sec. that v(t) = 80V?

0 4 8 12 16200−

100−

0

100

200Voltage versus Time

Time (msec)

Vol

tage

(vol

t)

v t 10 3−⋅( )

t

t80V

acos80 V⋅

2 120⋅ V⋅⎛⎜⎝

⎞⎟⎠

φ1−⎛⎜⎝

⎞⎟⎠

ω1:= t80V 4.3 ms⋅=

HW01 Solution.xmcdC:\JLAW\CLASSES\2012 Spring 212\Homework\

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ECE 212Electrical Circuits II

HW 01 Assigned: Wed, Jan 11Due: Fri, Jan 13

Law 9.2Consider the sinusoidal current:

i t( ) 8 cos 314.16 t⋅ 53.13 deg⋅+( )⋅ A⋅:=

Determine: a) f in hertz

ω2 314.16radsec⋅:= f2

ω22 π⋅

:= f2 50.0 Hz⋅=

b) T in milliseconds

T21f2

:= T2 20.0 ms⋅=

c) IMIm 8 A⋅:= Im 8.0 A=

d) i(t) at t =0 seconds

i 0 s⋅( ) 4.8 A=

e) φ in degrees and radians

φ2 53.13 deg⋅:= φ2 0.93 rad⋅=

f) the smallest positive value of t such that i(t) = 0 A

t0A

acos0 A⋅8 A⋅

⎛⎜⎝

⎞⎟⎠

φ2−⎛⎜⎝

⎞⎟⎠

ω2:= t0A 2.05 ms⋅=

HW01 Solution.xmcdC:\JLAW\CLASSES\2012 Spring 212\Homework\

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ECE 212Electrical Circuits II

HW 01 Assigned: Wed, Jan 11Due: Fri, Jan 13

0 5 10 3−× 10 10 3−× 0.015 0.0210−

5−

0

5

10Current versus Time

Time (sec)

Cur

rent

(A)

i t( )

t

HW01 Solution.xmcdC:\JLAW\CLASSES\2012 Spring 212\Homework\

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ECE 212Electrical Circuits II

HW 01 Assigned: Wed, Jan 11Due: Fri, Jan 13

g) the smallest positive value of t such that di t( )dt

0As

⋅=

Method I:

i t( ) Im cos ω2 t⋅ φ2+( )⋅ A⋅=

di t( )dt

ω2− Im⋅ sin ω2 t⋅ φ2+( )⋅= ω2 t⋅ φ2+ asin

di t( )dt

⎛⎜⎝

⎞⎟⎠

ω2− Im⋅

⎡⎢⎢⎣

⎤⎥⎥⎦

=

tdin

asin0−

As

⋅

ω2 Im⋅

⎛⎜⎜⎝

⎞⎟⎟⎠

φ2−

⎛⎜⎜⎝

⎞⎟⎟⎠

ω2:= tdin 2.95− ms⋅= tdi tdin

T22

+:= tdi 7.05 ms⋅=

Method II:

tx 6.5 ms⋅:= ω2− Im⋅ sin ω2 tx⋅ φ2+( )⋅ 430.8−As

=

Given

0 ω2− Im⋅ sin ω2 tx⋅ φ2+( )⋅=

tx 0 s⋅≥

tdiMethodII Find tx( ):= tdiMethodII 7.05 ms⋅=

HW01 Solution.xmcdC:\JLAW\CLASSES\2012 Spring 212\Homework\

Page 5 of 5 January 13, 2012Solution