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.

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

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

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=1.3-0.5, i4=0.8A 7. Find R, Apply Ohms law, VA-0=Rx i4, R=VA/i4=4/0.8=5 Ohm, So R=5Ω

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=0.3-0.1=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

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

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

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.

Example: Where i1=(12-Va)/1 ----(B) i2=(Va-0)/3-----© Designate Nodes Give current directions Apply KCL at node Va i1=4+i2-------(A) Where i1=(12-Va)/1 ----(B) i2=(Va-0)/3-----© 4. Placing these values in equation A (12-Va)/1=4+Va/3 12-Va=(12+Va)/3 36-3Va=12+Va 24=4Va Va=24/4 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= 72 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

If current flow upward Apply KCL at node Va i1+4=i2 =(12-Va)/1+4= Va/3 = 36-3Va+12=Va= 48=4Va, Va=48/4 Va=12V

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 4(9V1-V1)=2V1+3(V1-V2) 36-4V1=2V1+3V1-3V2 9V1-3V2-36=0 3(3V1-V2-12)=0 3V1-V2-12=0---------(A)

Example Continue 2. Apply KCL at node V2 (V1-V2)/4K=V2/2K+5m, Multiply eq by 4 we get V1-V2=2V2+20. 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 V1 2.3mA=0.33mA+2mA , 2.3mA=2.3mA 5. Apply KCL at V2 2mA+3mA=5mA i4=3mA IS ADDED AS IT IS OBTAINED BY -6V

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

1. Apply KCL at node V1 6.5+(V2-V1)/2=V1/1, Multiply bothsides by 2 3V1-V2=13………(A) V2=3V1-13…..(A) 2. Apply KCL at node V2 3=(V2-V1)/2+(V2-V3)/4 Multiply by 4 both sides 12=2V2-2V1+V2-V3, 3V2-2V1-V3=12……(B) V3=3V2-2V1-12…..(B) 3. Apply KCL at node 3 (V2-V3)/4=V3/8+6.5 Multiply both sides by 8 2V2-2V3=V3+52 2V2-3V3=52……..(C) 4. Placing equation B in Equation C 2V2-3(3V2-2V1-12)=52 2V2-9V2+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, 30-7V2=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

Check 1. i1 by Ohms Law i15/1=5A, i2=(5-2)/2 =1. 5A, i4=(2-(-16))/4=4 Check 1. i1 by Ohms Law i15/1=5A, i2=(5-2)/2 =1.5A, i4=(2-(-16))/4=4.5A, i8=(0-(-16))/8=2A 2. KCL at Node V1 6.5A=5A+1.5A 3. KCL AT Node V2 3A+1.5A=4.5A 4. KCL at Node V3 6.5A=4.5A+2A Figure c shows actual current directions V2 V1 V3 Figure c

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?

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

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

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

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