# EEE0201 Prelab Week 2

EEE2255 - CIRCUT ANALYSIS LAB
PRELIMINARY WORK
Week 2
Theory
1. ��� ������ –�� + ��1 + ��2 + ��3 = 0 → ( − 12) + � ∗ 120 + � ∗ 4700 + � ∗ 1500 = 0
−12 + � ∗ 6320 = 0 → � = 1.9 ��.
Note that the current throught the resistors are same because they are in series.
��1 = � ∗ �1 = 1.9 �� ∗ 120 �ℎ�� = 228 �� → ��2 = � ∗ �2 = (1.9 ��) ∗ (4700 �ℎ��) = 8.93 �
��3 = � ∗ �3 = 1.9 �� ∗ 1500 �ℎ�� = 2.85 � → �� = 12 � ��� ��1 + ��2 + ��3 = 12.008 �
Total voltage is equal to the supply voltage with an unsignificant tolerance.
2. ��� ������ �ℎ�� ���� 1: − �� + ��1 = 0 ��� ���� 2: − ��1 + ��2 = 0
���� 3: − ��2 + ��3 = 0. ���� �ℎ� ���������, �� ��� �� = ��1 = ��2 = ��3 = 10 �
Note that the voltage across the resistors are same because they are in parallel.
İ�1 =
İ�2 =
İ�3 =
��1
�
= 10
�����ℎ�� = 3.3 ��
�1
3.3
��2
�
= 10
�����ℎ�� = 0.45 ��
�2
22
��3
�
= 10
�����ℎ�� = 6.67 ��
�3
1.5
3. ��� ������ �ℎ�� ���� 1: − �� + ��1 + ��2 = 0 ��� ���� 2: − ��2 + ��3 + ��4 = 0
���� 3: − �� + ��1 + ��3 + ��4 = 0
�ℎ� ������� �ℎ����ℎ �ℎ� �2 �� �2. �� ���� 1, �� ℎ��� �� = ��1 + ��2.
�� = ��1 + ��2 → �� = � ∗ �1 + �2 ∗ �2 → �� = � ∗ 120 + �2 ∗ 4700 = 15
�ℎ� ������� �ℎ����ℎ �ℎ� ��������� �3 ��� �4 ��� ���� ��� �ℎ�� �� �3.
�� ���� 2, �� ℎ��� ��2 = ��3 + ��4.
��2 = ��3 + ��4 → ��2 = �3 ∗ �3 + �3 ∗ �4 → ��2 = �3 ∗ 1500 + �3 ∗ 4700 = 6200 �3
�ℎ��� �� �ℎ� 3�� ���� ������ �ℎ� ������� ���ℎ �ℎ� �������� �� = ��1 + ��3 + ��4.
�� = ��1 + ��3 + ��4 → �� = � ∗ �1 + �3 ∗ �3 + �3 ∗ �4
�� = � ∗ 120 + �3 ∗ 1500 + �3 ∗ 4700 → �� = � ∗ 120 + �3 ∗ 6200 = 15
According the KCL, there is an equation: −� + �2 + �3 = 0 → � = �2 + �3
�� ℎ��� ���� ���������.
�� = � ∗ 120 + �2 ∗ 4700 = 15
��2 = �3 ∗ 1500 + �3 ∗ 4700 = 6200 �3
�� = � ∗ 120 + �3 ∗ 6200 = 15
� = �2 + �3
������� �ℎ� ��������� �� ���
� = 5 �� → ��1 = � ∗ �1 → 5 �� ∗ 120Ω = 0.6 �
�2 = 3 �� → ��2 = �2 ∗ �2 → ��2 = 3 �� ∗ 4.7 �Ω = 14.1 �
�3 = 2 �� → ��3 = �3 ∗ �3 → 2 �� ∗ 1.5 �Ω = 3 �
�3 = 2 �� → ��4 = �3 ∗ �4 → 2 �� ∗ 4.7 �Ω = 9.4 �
4. ���� 1: ��� �1 + �2 + �3 = 120 Ω + 4700 Ω + 1500 Ω
���� 2: ��� = 1 / (
��� = 6.32 �Ω
1
1
1
1
1
1
+
+
+
+
) = 1/(
)
�1
�2
�3
3300
22000
1500
��� = 985.07 Ω
���� 3: ��� = �1 + (�2 || (�3 + �4) ) = 120 + (4700 || (1500 + 4700) )
��� = 2793.39 Ω
5. The voltage divider rule states that
�1 = �� ∗ (
�1
120
) = 12 ∗ (
)
�1 + �2 + �3
120 + 4700 + 1500
�2 = �� ∗ (
�2
4700
) = 12 ∗ (
)
�1 + �2 + �3
120 + 4700 + 1500
�3 = �� ∗ (
�3
1500
) = 12 ∗ (
)
�1 + �2 + �3
120 + 4700 + 1500
�1 = 0.227 �
�2 = 8.924 �
�2 = 2.848 �
6. The current divider rule states that
�� = �� ∗
7. The terminal current measured from
� =
�
���
=
�1
�1 + �2
= 1 �� ∗
�� = 0.869 ��
12
(�1||(�2 + �3) + �4)
� = 0.42 ��
10000
10000 + 1500
=
12 �
28.37 �Ω
SIMULATION
1.
According to simulation, we measure Is = 19 mA, Vr1 = 228 mV, Vr2 = 8.92 V, Vr3 = 2.85 V which are the same results
as we found in KVL and KCL with an unsignificant tolerance.
2.
According to simulation, we measure Vs = 10 V, Ir1 = 3.03 mA, Ir2 = 0.455 mA, Ir3 = 6.67 mA which are the same
results as we found in KVL and KCL with an unsignificant tolerance.
3.
According to simulation, we measure I = 5.37 mA, Ir2 = 3.05 mA, Ir3-4 = 2.32 mA, Vr1 = 0.644 V, Vr2 = 14.4 V, Vr3 =
3.47 V, Vr4 = 10.9 V which are the same results as we found in KVL and KCL with an unsignificant tolerance.
4.
We found the equivalent resistance for the circuits on the theory.
5.
According to simulation, we measure V1 = 0.228 V, V2 = 8.92 V, V3 = 2.85 V which are the same results as we found
in the voltage divider rule with an unsignificant tolerance.
6.
According to simulation, we measure I0 = 0.87 mA which is the same result as we found in the current divider rule
with an unsignificant tolerance.
7.
According to simulation, we measure the terminal current as 0.423 mA which is the same result as we found in the
current divider rule with an unsignificant tolerance.