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
𝑉𝑉 = 𝐼𝐼𝑍𝑍
= 0.2𝑚𝑚𝑚𝑚 × 18.8𝐾𝐾Ω
= 3.76 𝑉𝑉
The phase angle, 𝜃𝜃 = tan−1
𝑍𝑍 = �𝑅𝑅2 + 𝑋𝑋𝐶𝐶2
𝑋𝑋𝐶𝐶
𝑅𝑅
= �(10 𝐾𝐾)2 + (15.9 𝐾𝐾)2
= 18.8𝐾𝐾 Ω
= tan−1 �
15.9𝐾𝐾Ω
10 𝐾𝐾Ω
� = 57.8°
R
57.8
Z
Xc
1
2𝜋𝜋𝜋𝜋𝜋𝜋
1
=
Ω
2𝜋𝜋 × 1𝑘𝑘 × 0.01𝜇𝜇
= 15.9𝐾𝐾Ω
𝑋𝑋𝐶𝐶 =
7 EXAMPLE 10- 3
Determine the current in the RC circuit of Figure
𝑉𝑉
𝑍𝑍
10
=
𝐴𝐴
5.30𝐾𝐾
= 1.89𝑚𝑚𝑚𝑚
𝐼𝐼 =
𝑍𝑍 = �𝑅𝑅2 + 𝑋𝑋𝐶𝐶2
= �(2.2 𝐾𝐾)2 + (4.82 𝐾𝐾)2
= 5.30𝐾𝐾 Ω
𝑋𝑋𝐶𝐶 =
1
2𝜋𝜋𝜋𝜋𝜋𝜋
1
Ω
2𝜋𝜋 × 1.5𝑘𝑘 × 0.022𝜇𝜇
= 4.82𝐾𝐾Ω
=
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.
𝑉𝑉𝑠𝑠 = �𝑉𝑉𝑅𝑅2 + 𝑉𝑉𝐶𝐶2 = �(10 𝑉𝑉)2 + (15 𝑉𝑉)2 = 18 𝑉𝑉
The phase angle between the resistor voltage and source voltage is
𝑉𝑉
15 𝑉𝑉
𝜃𝜃 = tan−1 𝐶𝐶 = tan−1
= 56.3°
𝑉𝑉𝑅𝑅
VR
10 𝑉𝑉
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 reactance𝑋𝑋𝐶𝐶 =
1
2𝜋𝜋𝜋𝜋𝜋𝜋
=
𝜙𝜙 = 90° − tan−1
1
Ω
2𝜋𝜋(1 𝐾𝐾)(0.1𝜇𝜇)
= 1.59 𝐾𝐾Ω, so
𝜙𝜙 = 90° − tan−1
𝜙𝜙 = 23.2°
The output voltage lags the input by 23.20
𝑋𝑋𝐶𝐶
𝑅𝑅
1.59 𝐾𝐾
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
𝑋𝑋𝐶𝐶 =
The capacitive susceptance is
𝐺𝐺 =
1
1
=
= 3.03𝑚𝑚𝑚𝑚
𝑅𝑅 330
1
1
=
= 723Ω
2𝜋𝜋𝜋𝜋𝜋𝜋 2𝜋𝜋 × 1 𝐾𝐾 × 0.22𝜇𝜇
𝐵𝐵𝐶𝐶 =
1
1
=
= 1.38𝑚𝑚𝑚𝑚
𝑋𝑋𝐶𝐶 723
Total admittance, 𝑌𝑌 = �𝐺𝐺 2 + 𝐵𝐵𝐶𝐶2 = �(3.03 𝑚𝑚𝑚𝑚)2 + (1.38 𝑚𝑚𝑚𝑚)2 = 3.33 𝑚𝑚𝑚𝑚
1
1
Impedance 𝑍𝑍 = 𝑌𝑌 = 3.33 𝑚𝑚𝑚𝑚 = 300Ω
11 EXAMPLE 10-13
Convert the parallel circuit in Figure to an equivalent series form.
The equivalent series values are
𝑅𝑅𝑒𝑒𝑒𝑒 = 𝑍𝑍 cos 𝜃𝜃
𝑋𝑋𝐶𝐶(𝑒𝑒𝑒𝑒) = 𝑍𝑍 sin 𝜃𝜃
Now,
𝑍𝑍 =
1
𝑎𝑎𝑎𝑎𝑎𝑎 𝑌𝑌 = �𝐺𝐺 2 + 𝐵𝐵𝐶𝐶2
𝑌𝑌
1
1
=
= 5.56 𝑚𝑚𝑚𝑚
𝑅𝑅 180
1
1
𝑎𝑎𝑎𝑎𝑎𝑎, 𝐵𝐵𝐶𝐶 =
=
= 3.70 𝑚𝑚𝑚𝑚
𝑋𝑋𝐶𝐶 270
𝐺𝐺 =
𝑠𝑠𝑠𝑠, 𝑌𝑌 = �(5.56 𝑚𝑚𝑚𝑚)2 + (3.70 𝑚𝑚𝑚𝑚)2 = 6.68 𝑚𝑚𝑚𝑚
Equivalent series values
𝑎𝑎𝑎𝑎𝑎𝑎 𝑍𝑍 =
1
1
=
= 150Ω
𝑌𝑌 6.68 𝑚𝑚𝑚𝑚
𝑅𝑅𝑒𝑒𝑒𝑒 = 150 × cos 33.6° = 125Ω
𝑋𝑋𝐶𝐶(𝑒𝑒𝑒𝑒) = 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 𝐻𝐻/𝑚𝑚.
Solution:
Inductance, 𝐿𝐿 =
𝑁𝑁 2 𝜇𝜇𝜇𝜇
𝑙𝑙
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
𝑑𝑑 2
0.005 2
𝐴𝐴 = 𝜋𝜋𝑟𝑟 = 𝜋𝜋 � � = 𝜋𝜋 �
� = 1.96 × 10−5 𝑚𝑚2
2
2
2
So, 𝐿𝐿 =
(350)2 �0.25×
10−3 𝐻𝐻
��1.96×10−5 𝑚𝑚2 �
𝑚𝑚
0.015𝑚𝑚
= 40𝑚𝑚𝑚𝑚
Example 11-2: Determine the total inductance for each of the series
connections
Solution
a. 𝐿𝐿 𝑇𝑇 = 1 𝐻𝐻 + 2 𝐻𝐻 + 1.5 𝐻𝐻 + 5 𝐻𝐻 = 9.5 𝐻𝐻
b. 𝐿𝐿 𝑇𝑇 = 5 𝑚𝑚𝑚𝑚 + 2 𝑚𝑚𝑚𝑚 + 10 𝑚𝑚𝑚𝑚 + 1 𝑚𝑚𝑚𝑚 = 18 𝑚𝑚𝐻𝐻
Example 11-3: Determine LT
Solution
𝐿𝐿 𝑇𝑇 =
1
1
1
=
=
1
1
1
1
1
1
0.8𝑚𝑚𝑚𝑚
𝐿𝐿1 + 𝐿𝐿2 + 𝑙𝑙3 10𝑚𝑚𝑚𝑚 + 5𝑚𝑚𝑚𝑚 + 2𝑚𝑚𝑚𝑚
= 1.25𝑚𝑚𝑚𝑚
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
𝜏𝜏 =
The final current is
𝐿𝐿 10𝑚𝑚𝑚𝑚
=
= 8.33 𝜇𝜇𝜇𝜇
𝑅𝑅 1.2𝑘𝑘Ω
𝐼𝐼𝐹𝐹 =
𝑉𝑉𝑆𝑆
12𝑉𝑉
=
= 10𝑚𝑚𝑚𝑚
𝑅𝑅 1.2𝑘𝑘Ω
Using the time-constant percent value
At 1𝜏𝜏: 𝑖𝑖 = 0.63(10𝑚𝑚𝑚𝑚) = 6.3𝑚𝑚𝑚𝑚; 𝑡𝑡 = 8.33 𝜇𝜇𝜇𝜇
At 2𝜏𝜏: 𝑖𝑖 = 0.86(10𝑚𝑚𝑚𝑚) = 8.6𝑚𝑚𝑚𝑚; 𝑡𝑡 = 16.7 𝜇𝜇𝜇𝜇
At 3𝜏𝜏: 𝑖𝑖 = 0.95(10𝑚𝑚𝑚𝑚) = 9.5𝑚𝑚𝑚𝑚; 𝑡𝑡 = 25.0 𝜇𝜇𝜇𝜇
At 4𝜏𝜏: 𝑖𝑖 = 0.98(10𝑚𝑚𝑚𝑚) = 9.8𝑚𝑚𝑚𝑚; 𝑡𝑡 = 33.3 𝜇𝜇𝜇𝜇
At 5𝜏𝜏: 𝑖𝑖 = 0.99(10𝑚𝑚𝑚𝑚) = 9.9𝑚𝑚𝑚𝑚 ≅ 10𝑚𝑚𝑚𝑚; 𝑡𝑡 = 41.7 𝜇𝜇𝜇𝜇
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
𝐻𝐻𝐻𝐻𝐻𝐻ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝑓𝑓 =
1
10𝜏𝜏
𝜏𝜏 =
𝑓𝑓 =
𝐿𝐿 15 𝑚𝑚𝑚𝑚
=
= 0.454 𝜇𝜇𝜇𝜇
𝑅𝑅 33 𝑘𝑘Ω
1
= 220 𝑘𝑘𝑘𝑘𝑘𝑘
10 × 0.454𝜇𝜇𝜇𝜇
b. voltage across resistor will have same shape as the current wave form.
Example 11-8: determine the inductor current 30𝜇𝜇𝜇𝜇 after the switch is
closed
Solution:
𝑡𝑡
𝑖𝑖𝐿𝐿 = 𝐼𝐼𝐹𝐹 (1 − 𝑒𝑒 −𝜏𝜏 ) Where 𝐼𝐼𝐹𝐹 is final current and 𝜏𝜏 is time
constant
The final current 𝐼𝐼𝐹𝐹 =
𝑣𝑣𝑆𝑆
𝑅𝑅
=
The RL time constant 𝜏𝜏 =
𝐿𝐿
𝑅𝑅
12 𝑉𝑉
2.2 𝑘𝑘Ω
=
= 5.45 𝑚𝑚𝑚𝑚 and
100 𝑚𝑚𝑚𝑚
2.2 𝐾𝐾
So 𝑖𝑖𝑎𝑎𝑎𝑎 30𝜇𝜇𝜇𝜇 = (5.45 𝑚𝑚𝑚𝑚 × �1 − 𝑒𝑒
−
= 45.5 𝜇𝜇𝜇𝜇
30𝜇𝜇𝜇𝜇
45.5𝜇𝜇𝜇𝜇
� = 2.63 𝑚𝑚𝑚𝑚
Example 11-10: a sinusoidal voltage is applied to the circuit. The
frequency is 10 kHz. Determine the inductive reactance.
Solution:
Inductive reactance, 𝑋𝑋𝐿𝐿 = 2𝜋𝜋𝜋𝜋𝜋𝜋 = 2𝜋𝜋 × 10 𝑘𝑘𝑘𝑘𝑘𝑘 × 5 𝑚𝑚𝑚𝑚 =
2𝜋𝜋(10 × 103 × 5 × 10−3 ) = 314 Ω
Example 11-12: determine the rms current in below circuit
Solution:
𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 =
𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟
𝑋𝑋𝐿𝐿
𝑋𝑋𝐿𝐿 = 2𝜋𝜋𝜋𝜋𝜋𝜋 = 2𝜋𝜋(10 × 103 × 100 × 10−3 ) = 6283 Ω
So, 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 =
5 𝑉𝑉
6283 Ω
= 796 𝜇𝜇𝜇𝜇
Example 12-2: the current in the below circuit is 200𝜇𝜇𝜇𝜇. Determine the
source voltage.
Solution:
Applying ohm’s law, 𝑉𝑉𝑠𝑠 = 𝐼𝐼𝐼𝐼
The impedance 𝑍𝑍 = �𝑅𝑅 2 + 𝑋𝑋𝐿𝐿2 and
Inductive reactance 𝑋𝑋𝐿𝐿 = 2𝜋𝜋𝜋𝜋𝜋𝜋 = 2𝜋𝜋(10 × 103 × 100 ×
10−3 ) = 6.28𝐾𝐾Ω
So, 𝑍𝑍 = �(10 𝑘𝑘Ω)2 + (6.28 𝑘𝑘Ω)2 = 11.8 𝑘𝑘Ω
Now, 𝑉𝑉𝑠𝑠 = 200 𝜇𝜇𝜇𝜇 × 11.8 𝑘𝑘Ω = 2.36𝑉𝑉
Example 12-3: determine the source voltage and the phase angel in
below circuit. Draw the voltage phasor diagram.
Solution:
Since 𝑉𝑉𝑅𝑅 𝑎𝑎𝑎𝑎𝑎𝑎 𝑉𝑉𝐿𝐿 are 90° out of phase, we cannot add them
directly. The source voltage is the phasor sum of 𝑉𝑉𝑅𝑅 𝑎𝑎𝑎𝑎𝑎𝑎 𝑉𝑉𝐿𝐿 .
𝑉𝑉𝑆𝑆 = �𝑉𝑉𝑅𝑅2 + 𝑉𝑉𝐿𝐿2 = �502 + 352 = 61 𝑉𝑉
The phase angle between the resistor voltage and source
voltage is
35 𝑉𝑉
𝜃𝜃 = tan−1(
) = 35°
50 𝑉𝑉