AC System Analysis (Week4 Studio Solutions)

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SEJ102_Week 4
(Studio Solutions)
6 EXAMPLE 10-2
The current in figure is 0.2mA. Determine the source voltage and the phase angle. Draw the
impedance triangle
 = 
= 0.2 × 18.8Ω
= 3.76 
The phase angle,  = tan−1
 = �2 + 2


= �(10 )2 + (15.9 )2
= 18.8 Ω
= tan−1 �
15.9Ω
10 Ω
� = 57.8°
R
57.8
Z
Xc
1
2
1
=
Ω
2 × 1 × 0.01
= 15.9Ω
 =
7 EXAMPLE 10- 3
Determine the current in the RC circuit of Figure


10
=

5.30
= 1.89
 =
 = �2 + 2
= �(2.2 )2 + (4.82 )2
= 5.30 Ω
 =
1
2
1
Ω
2 × 1.5 × 0.022
= 4.82Ω
=
8 EXAMPLE 10-4
Determine the source voltage and the phase angle in Figure. Draw the voltage phasor diagram.
Since VR and VC are 90° out of phase, you cannot add them directly. The source voltage is the phasor sum of VR
and VC.
 = �2 + 2 = �(10 )2 + (15 )2 = 18 
The phase angle between the resistor voltage and source voltage is

15 
 = tan−1  = tan−1
= 56.3°

VR
10 
56.3
VC
VS
9 EXAMPLE 10-6
Determine the amount of phase lag from the input to output in the lag circuit in Figure.
The phase lab between the output voltage and input voltage is
Capacitive reactance =
1
2
=
 = 90° − tan−1
1
Ω
2(1 )(0.1)
= 1.59 Ω, so
 = 90° − tan−1
 = 23.2°
The output voltage lags the input by 23.20


1.59 
680
10 EXAMPLE 10-10
Determine the total admittance in Figure, and then convert it to impedance.
To determine admittance, Y, first calculate the values for G and BC.
The capacitive reactance is
 =
The capacitive susceptance is
 =
1
1
=
= 3.03
 330
1
1
=
= 723Ω
2 2 × 1  × 0.22
 =
1
1
=
= 1.38
 723
Total admittance,  = � 2 + 2 = �(3.03 )2 + (1.38 )2 = 3.33 
1
1
Impedance  =  = 3.33  = 300Ω
11 EXAMPLE 10-13
Convert the parallel circuit in Figure to an equivalent series form.
The equivalent series values are
 =  cos 
() =  sin 
Now,
 =
1
  = � 2 + 2

1
1
=
= 5.56 
 180
1
1
,  =
=
= 3.70 
 270
 =
,  = �(5.56 )2 + (3.70 )2 = 6.68 
Equivalent series values
  =
1
1
=
= 150Ω
 6.68 
 = 150 × cos 33.6° = 125Ω
() = 150 × sin 33.6° = 83Ω
Example 11-1: Determine the inductance of the coil in Figure. The
permeability of the core is 0.25 × 10−3 /.
Solution:
Inductance,  =
 2 

where N is number of turns, A is
area of the core and l is length.
We have, l=1.5cm=0.015m, d=0.5cm=0.005m
 2
0.005 2
 =  =  � � =  �
� = 1.96 × 10−5 2
2
2
2
So,  =
(350)2 �0.25×
10−3 
��1.96×10−5 2 �

0.015
= 40
Example 11-2: Determine the total inductance for each of the series
connections
Solution
a.   = 1  + 2  + 1.5  + 5  = 9.5 
b.   = 5  + 2  + 10  + 1  = 18 
Example 11-3: Determine LT
Solution
  =
1
1
1
=
=
1
1
1
1
1
1
0.8
1 + 2 + 3 10 + 5 + 2
= 1.25
Example 11-5: Calculate the RL time constant for the bellow circuit.
Then determine current and the time at each timeconstant interval, measure from the instant the switch is
closed.
Solution
The RL time constant is
 =
The final current is
 10
=
= 8.33 
 1.2Ω
 =

12
=
= 10
 1.2Ω
Using the time-constant percent value
At 1:  = 0.63(10) = 6.3;  = 8.33 
At 2:  = 0.86(10) = 8.6;  = 16.7 
At 3:  = 0.95(10) = 9.5;  = 25.0 
At 4:  = 0.98(10) = 9.8;  = 33.3 
At 5:  = 0.99(10) = 9.9 ≅ 10;  = 41.7 
Example 11-7: a. the circuit below has a square wave input. What is the
highest frequency that can be used and still observe the
complete waveform across the inductor?
b. assume the generator is set to the frequency
determined in (a). Describe the voltage waveform across
the resistor?
Solution:
a. the period needs to be 10 times the time-constant to
observe the entire wave. So
ℎ ,  =
1
10
 =
 =
 15 
=
= 0.454 
 33 Ω
1
= 220 
10 × 0.454
b. voltage across resistor will have same shape as the current wave form.
Example 11-8: determine the inductor current 30 after the switch is
closed
Solution:

 =  (1 −  − ) Where  is final current and  is time
constant
The final current  =


=
The RL time constant  =


12 
2.2 Ω
=
= 5.45  and
100 
2.2 
So  30 = (5.45  × �1 − 
−
= 45.5 
30
45.5
� = 2.63 
Example 11-10: a sinusoidal voltage is applied to the circuit. The
frequency is 10 kHz. Determine the inductive reactance.
Solution:
Inductive reactance,  = 2 = 2 × 10  × 5  =
2(10 × 103 × 5 × 10−3 ) = 314 Ω
Example 11-12: determine the rms current in below circuit
Solution:
 =


 = 2 = 2(10 × 103 × 100 × 10−3 ) = 6283 Ω
So,  =
5 
6283 Ω
= 796 
Example 12-2: the current in the below circuit is 200. Determine the
source voltage.
Solution:
Applying ohm’s law,  = 
The impedance  = � 2 + 2 and
Inductive reactance  = 2 = 2(10 × 103 × 100 ×
10−3 ) = 6.28Ω
So,  = �(10 Ω)2 + (6.28 Ω)2 = 11.8 Ω
Now,  = 200  × 11.8 Ω = 2.36
Example 12-3: determine the source voltage and the phase angel in
below circuit. Draw the voltage phasor diagram.
Solution:
Since    are 90° out of phase, we cannot add them
directly. The source voltage is the phasor sum of    .
 = �2 + 2 = �502 + 352 = 61 
The phase angle between the resistor voltage and source
voltage is
35 
 = tan−1(
) = 35°
50 
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