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# Nodal analysis for KCL and KVL

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Nodal analysis using KCL and KVL for a simple circuit

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### Nodal analysis for KCL and KVL

1. 1. NODAL AND LOOP ANALYSIS TECHNIQUES LEARNING GOALS NODAL ANALYSIS LOOP ANALYSIS Develop systematic techniques to determine all the voltages and currents in a circuit
2. 2. NODE ANALYSIS • One of the systematic ways to determine every voltage and current in a circuit The variables used to describe the circuit will be “Node Voltages” -- The voltages of each node with respect to a pre-selected reference node
3. 3. LEARNING EXAMPLE V1 V1 − V2 @ V1 : − 4mA + + =0 6k 12k @ V2 : 2mA + V2 V2 − V1 + =0 6k 12k BY “INSPECTION” 1  1  1 + V1 − V2 = 4mA  ÷ 12k  6k 12k  1 1   1 − V1 +  + ÷V2 = − 2mA 12k  6k 12k  USING KCL
4. 4. LEARNING EXAMPLE SUPERNODE V1 = 6V V4 = −4V SOURCES CONNECTED TO THE REFERENCE CONSTRAINT EQUATION KCL @ SUPERNODE V3 − V2 = 12V V2 − 6 V2 V3 V3 − ( −4) + + + = 0 * / 2k 2k 1k 2k 2k V2 IS NOT NEEDED FOR I O 3V2 + 2V3 = 2V − V2 + V3 = 12V * / 3 and add 5V3 = 38V V OHM'S LAW I O = 3 = 3.8mA 2k
5. 5. CIRCUITS WITH DEPENDENT SOURCES PRESENT NO SIGNIFICANT ADDITIONAL COMPLEXITY. THE DEPENDENT SOURCES ARE TREATED AS REGULAR SOURCES WE MUST ADD ONE EQUATION FOR EACH CONTROLLING VARIABLE
6. 6. LEARNING EXAMPLE FIND I O VOLTAGE SOURCE CONNECTED TO REFERENCE V1 = 3V V − V1 V2 KCL@V2 : 2 + − 2I x = 0 3k 6k CONTROLLING VARIABLE IN TERMS OF NODE VOLTAGES Ix = V2 − V1 V2 V + − 2 2 = 0 * / 6k 3k 6k 6k V2 − 2V1 = 0 ⇒ V2 = 6V IO = V1 − V2 = −1mA 3k REPLACE V2 6k
7. 7. SUPER NODE WITH DEPENDENT SOURCE VOLTAGE SOURCE CONNECTED TO REFERENCE V3 = 6V SUPERNODE CONSTRAINT V1 − V2 = 2Vx CONTROLLING VARIABLE IN TERMS OF NODES KCL AT SUPERNODE Vx = V2 ⇒ V1 = 3V2 * / 12k 2(V1 − 6) + V1 + 2V2 + V2 − 6 = 0 3V1 + 3V2 = 18 ⇒ 4V1 = 18
8. 8. LEARNING EXAMPLE FIND THE VOLTAGE Vo @V4 : V4 = 4V AT SUPER NODE V1 − V2 = 2VX V V − V3 V1 − V3 V1 − 4V −2mA + 2 + 2 + + =0 1k × 1k 1k 1k 1k 1k × @V : − 2mA + V − V + V − V = 0 3 3 2 1k CONTROLLING VARIABLE 3 1 1k VX = V2 SOLVE EQUATIONS NOW V1 = 3VX 2V1 + 2VX − 2V3 = 6V −V1 − VX + 2V3 = 2V VARIABLE OF INTEREST VO = V1 − V3
9. 9. LEARNING EXAMPLE Find the current Io @ V2 : V2 = 12V @ V3 : V3 = 2VX @ super node: V4 − V1 = 6V (constraint eq.) FIND NODES – AND SUPER NODES V − V3 V4 − V5 V4 V1 − V2 V1 − V3 + + 2I X + 4 + + 1k 1k 1k 1k 1k =0 V5 − V4 V5 @ V5 : − 2 I X + + =0 1k 1k CONTROLLING VARIABLES VX = V1 − V2 IX = V4 1k 7 eqs in 7 variables VARIABLE OF INTEREST IO = V5 1k
10. 10. LEARNING EXAMPLE Find the current Io @ V2 : V2 = 12V @ V3 : V3 = 2VX @ super node: V4 − V1 = 6V (constraint eq.) FIND NODES – AND SUPER NODES V − V3 V4 − V5 V4 V1 − V2 V1 − V3 + + 2I X + 4 + + 1k 1k 1k 1k 1k =0 V5 − V4 V5 @ V5 : − 2 I X + + =0 1k 1k CONTROLLING VARIABLES VX = V1 − V2 IX = V4 1k 7 eqs in 7 variables VARIABLE OF INTEREST IO = V5 1k