1. Chapter 3. Circuit Analysis Techniques
3.1 Circuit Analysis by Inspection
Consider Figure Find VS that cause 2 ohm Resistance to conduct 3A flowing downward.

2. 1.Label the nodes
2. Select the reference node and ground it
3. Label the current directions
4. Start from where maximum information
Ohms Law VB-0=2X3=6 SO VB=6V
5. Consider Node B Apply KCL
i1+1A=3A i1=2A
Ohms Law VA-VB= 3 x i1
VA-6=3x2 =6+6=12V VA=12V
7. . i2 by Ohms Law. i2=VA-0/4=12/4=3A i2=3A
8. Apply KCL at node A
i3=i1+i2=2+3=5A i3=5A
9. Ohms Law at resistance 1 ohms
Vs- VA=1Xi3= Vs=1X5+12=17
Vs=17V

3. Example 2. Find value of R that cause 2A current source to release 12 watt.
1. We know for current source to release power, the current must flow in the direction of increasing voltage. This means that GND is at higher potential than VB SO P=VI, V=P/I, 0-VB=12/2, -VB=6V, OR VB=-6V
2. To find i1 using ohms law V=IR, 0-VB=i1x4, -VB=4i1, i1=-VB/4, i1=-(-6)/4=1.5A
3. For i2 Apply KCL at node VB, i2+i1=2A, i2=2A-1.5A=0.5A, i2=0.5A

4. Example continue
4. Find VA, VA-VB=20 X i2, VA= 20x0.5-VB=10+(-6), VA=4V
5. Find i3, use ohms law, 17-VA=10xi3, (17-4)/10=i3, or i3=13/10, i3=1.3
6. Find i4, Apply KCL at node VA, i3=i2+i4, i4=i3-i2= , i4=0.8A
7. Find R, Apply Ohms law, VA-0=Rx i4, R=VA/i4=4/0.8=5 Ohm, So R=5Ω

5. Example 3 Design a six stage Resistive Ladder, when driven by a 100V source, yields the following node voltages 50V, 20V, 10V, 5V, 2V, 1V as shown. All series resistors are of 10k ohms
Start from the farthest branch and move towards the source.
For Rs use ohms Law, V=IR, R=V/I and for Is use Ohms Law and KCL at your ease.
i11=2-1/10k=0.1mA , i12=i11=0.1mA, So R12=1-0/i12=1/0.1mA=10KΩ, R12=10KΩ
i9=5-2/10k=3/10k=0.3mA, i9=i10+i11, i10=i9-i11= =0.2mA, R10=2-0/i10=2/0.2 R10=20/2mA= 10K and so on We find All Rs as R2=R8=25K, R4=R10=R12=10K, and R6=20K . We also find all Is. Then we find Req = Vs/i1=100/5mA=20KΩ

6. Modify the circuit so that Req=100KΩ
. To ensure the same node voltages we must leave all resistance ratio unchanged. This can be obtained by multiplying all resistances by a common scale factor Req(new)/Req(old)=100K/20K=5
So R1=R3=R5=R7=R9=R11=10Kx5=50K and R2=R8=20Kx5=100K and R4=R10=R12=10Kx5=50K and R6=20Kx5=100K

7. DC Biasing
Diodes Transistors and ICs require a certain voltages and currents to operate properly. This is called DC Biasing. Resistances plays a fundamental roles in DC Biasing. Figure shows a Voltage and current conflicts so we use a series resistor R1 to resolve voltage conflict and R2 to resolve current conflict.
By KVL the voltage drop across R1 is 15-(5+6)=4V So R1=4/2mA=2KΩ
By KCL Current through R2 is 0.2mA-0.5mA=1.5mA So the R2=6/1.5=4KΩ X1 X2
So for proper biasing of X1 and X2 THE R1 and R2 must be 2KΩ and 4KΩ Respectively
5V 2mA R1
6V 0.5mA
15V R2

8. Nodal Analysis
In Nodal Analysis We find out the node voltages as per following steps
Choose the reference node and label the other nodes
Give direction to currents
Apply KCL at each node and get the equations and find the values of all currents using Ohms Law.
Place currents values in the equations obtained in step 3 and solve the equations using any method.

9. Example: Where i1=(12-Va)/1 ----(B) i2=(Va-0)/3-----©
Designate Nodes
Give current directions
Apply KCL at node Va
i1=4+i (A)
Where i1=(12-Va)/1 ----(B)
i2=(Va-0)/3-----©
4. Placing these values in equation A
(12-Va)/1=4+Va/ Va=(12+Va)/ Va=12+Va =4Va
Va=24/ Va=6v
5. Placing this value in equation B and C we get i1=6A and i2=2A
6. Power released by source P=V x i1= 12 x 6= P=72w
Power dissipate by resistors By 1Ω P=VI=(12-6)6=6X6=36W
By 3Ω P=VI=6X2=12W
8. Current source dissipate Power as arrow is downward so P=VI=6X4=24
9. Power is conserved….72w=24w+36w+12w

10. If current flow upward Apply KCL at node Va
i1+4=i2 =(12-Va)/1+4= Va/3 = Va+12=Va= 48=4Va, Va=48/4
Va=12V

11. Apply Nodal Analysis and Find V1 & V2
1. Apply KCL at node V1
(9-V1)/3K = V1/6K+ (V1-V2)/4K
Multiply both sides by 12 we get (9V1-V1)=2V1+3(V1-V2)
36-4V1=2V1+3V1-3V V1-3V2-36= (3V1-V2-12)= V1-V2-12= (A)

12. Example Continue 2.
Apply KCL at node V2
(V1-V2)/4K=V2/2K+5m,
Multiply eq by 4 we get V1-V2=2V V1-3V2-20=0….(B)
Solving equation A and Eq. B we get V1=2V, and V2=-6V
i1=(9-2)/3k=7/3k= 2.3mA, i2=2/6k= 0.33mA
i3=(2-(-6)/4k=8/4k=2mA i4= (0-(-6)/2k=3mA
4. Check KCL at V mA=0.33mA+2mA , 2.3mA=2.3mA
5. Apply KCL at V mA+3mA=5mA
i4=3mA IS ADDED AS IT IS OBTAINED BY -6V

13. Apply Nodal analysis to the circuit of figure a and check results
Designate nodes
Give current directions V2 V1 V3
Figure b
Figure. a

14. 1. Apply KCL at node V (V2-V1)/2=V1/1, Multiply bothsides by V1-V2=13………(A) V2=3V1-13…..(A) Apply KCL at node V =(V2-V1)/2+(V2-V3)/ Multiply by 4 both sides =2V2-2V1+V2-V3, V2-2V1-V3=12……(B) V3=3V2-2V1-12…..(B) Apply KCL at node (V2-V3)/4=V3/ Multiply both sides by V2-2V3=V V2-3V3=52……..(C) 4. Placing equation B in Equation C 2V2-3(3V2-2V1-12)= V2-9V2+6V1+36=52, 6V1-7V2=52-36, 6V1-7V2=16……(D) 5. Place Eq A in Eq D, 6V1-7(3V1-13)=16, 6V1-21V1+91=16, -15V1=-91+16, V1=5V 6. Place V1 IN Eq D, 6(5)-7V2=16, V2=16, -7V2=-30+16, -7V2=-14, V2=2V 7. Place V2 in Eq C, 2(2)-3V3=52, 4-3V3=52, -3V3=52-4, -V3=48/3, V3=-16V V2 V3 V1
Figure b

15. Check 1. i1 by Ohms Law i15/1=5A, i2=(5-2)/2 =1. 5A, i4=(2-(-16))/4=4
Check i1 by Ohms Law i15/1=5A, i2=(5-2)/2 =1.5A, i4=(2-(-16))/4=4.5A, i8=(0-(-16))/8=2A KCL at Node V A=5A+1.5A KCL AT Node V A+1.5A=4.5A KCL at Node V A=4.5A+2A Figure c shows actual current directions V2 V1 V3
Figure c

16. a)Apply Nodal Analysis and check the results
Super Node When the voltage source (Dependent or Independent) is connected between two non-reference nodes, they form a super node. In the figure below 8V source form super node.
a)Apply Nodal Analysis and check the results
b)Find magnitude and direction of current
Passing through 8V source. Is the source
Delivering or absorbing power?

17. Label all the Node voltages and currents Apply KCL at V1, i1= 0.5+i2
Surrounds the voltage source and its nodes with dotted lines and V3-V2=8, OR V3=V2+8
Label all the Node voltages and currents
Apply KCL at V1, i1= 0.5+i2
(V3-V1) /2=0.5+(V1-V2)/1.5
Multiply by 6 both sides
3V3-3V1=3+4V1-4V2, 4V1+3V1-4V2-3V3+3=0
7V1-4V2-3V3+3=0…….(A)
Place V3 in equation A
7V1-4V2-3(V2+8)+V3=0, 7V1-4V2-3V2-24+3
7V1-7V2-21=0, 7(V1-V2-3)=0,
V1-V2=3…….(B) V2 V3 V1
4. Apply KCL at super node, i2=i1+i3+i4, (V1-V2)/1.5=(V3-V1)/2+V2/3+V3/10
Multiply both 30 both sides, 20V1-20V2=10V2+3V3+15V3-15V1
35V1-30V2-18V3=0, 35V1-30V2-18(V2+8)=0, 35V1-30V2-18V2-144=0
35V1-48V2=144………© Solving Eq B & C we get V1=0V, V2=-3V and V3=5V

18. Check i2=(V1-V2)/1.5=0-(-3)/1.5=2A i1=(V3-V1)/2=5-0/2=5/2=2.5A
i3=(0-V3)/3=-(-3)/3=1A(upward)
i4=(V3-0)/10=5/10=0.5A
KCL at node V1 2.5=0.5+2
KCL at super node
I2+i3=i1+i4, 2A+1A=2.5A+0.5A
Check results shows that obtained results are correct
B) The current through 8V source(The currents coming out from the positive side of super node of the source) are
i1+i4=2.5A+0.5A=3A,
This conforms the active sign conversion So the 8V Source is delivering power. V2 V1 V3

19. Apply Nodal analysis to the circuit
4V source is forming a super node.
We label the node voltages and all current directions as shown in the next slide

20. Apply KCL at node V1 i2=i1+i3 (12-V1)/18=V1/3+(V1-(V2+4))….(A)
2. Apply KCL at the super node
I3+i4=i5
{V1-(V2+4)}/8+{12-(V2+4)}/5V2/6..(B)
Multiply Eq A and B by 72 and 120 respectively and we get the equations
37V1-9V2=84
15V1-59V2=-132
Solving these equations we get
V1=3V and V2=3V
And the voltage at the positive side of 4V Source is V2+4=3+4=7V
Check
1.i1=v1/3=3/3=1A, i2=(12-3)/18=0.5A, i3=(7-3)/8=0.5A(Anticlockwise)
i4=(12-7)/5=5/5=1A, i5=V2/6=3/6=0.5A
2. Apply KCL at node V1, 0.5A+0.5A=1A
3. Apply KCL at super node= 1A=0.5A+0.5A
Important :1O Ω resistor does not effect our results, it only absorb Power from 10V source