# UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS

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1 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS 1.0 Kirchoff s Law Kirchoff s Current Law (KCL) states at any junction in an electric circuit the total current flowing towards that junction is equal to the total current flowing away from the junction, i.e. I = 0 Thus, referring to figure 1: I1 I2 I3 I5 I4 current towards = current flowing away I 1 + I 2 + I 3 = I 4 + I 5 I 1 + I 2 + (- I 3 ) + (- I 4 ) + (-I 5 ) = 0 I = 0 Figure 1 Kirchoff s Voltage Law (KVL) states in any closed loop in a network, the algebraic sum Figure 2 of the voltage drops (i.e. products of current and resistance) taken around the loop is equal to the resultant e.m.f. acting in that loop. V E = I + I E = I( + ) E + (- I ) + (- I) = 0 Figure Mesh analysis Analysis using KVL to solve for the currents around each closed loop of the network and hence determine the currents through and voltages across each elements of the network. Mesh analysis procedure: 1. Assign a distinct current to each closed loop of the network. 2. Apply KVL around each closed loop of the network. 3. Solve the resulting simultaneous linear equation for the loop currents. MAIANA/JKE/POLISAS/ET101-UNIT4 1

2 Example 1 Find the current flow through each resistor using mesh analysis for the circuit below. 20V Figure 3 Solution: Step 1: Assign a distinct current to each closed loop of the network. I1 I2 I13 I1 I2 20V Figure 4 Step 2: Apply KVL around each closed loop of the network. Loop 1: Loop 2: equation equation 2 Step 3: Solve the resulting simultaneous linear equation for the loop currents. Solve equation 1 and 2 using matrix Matrix form: From KCL : MAIANA/JKE/POLISAS/ET101-UNIT4 2

3 Example 2 Find the current flow through each resistor using mesh analysis for the circuit below. 5kΩ 3kΩ 40V 6kΩ 55V Figure 5 Solution: Step 1: Assign a distinct current to each closed loop of the network. I1 I2 5kΩ I3 3kΩ 40V I1 6kΩ I2 55V Figure 6 Step 2: Apply KVL around each closed loop of the network. Loop 1: Loop 2: equation equation 2 Step 3: Solve the resulting simultaneous linear equation for the loop currents. Solve equation 1 and 2 using matrix Matrix form: From KCL : MAIANA/JKE/POLISAS/ET101-UNIT4 3

4 1.2 Nodes analysis Analysis using KCL to solve for voltages at each common node of the network and hence determines the currents through and voltages across each elements of the network. Nodal analysis procedure: 1. Determine the number of common nodes and reference node within the network. 2. Assign current and its direction to each distinct branch of the nodes in the network. 3. Apply KCL at each of the common nodes in the network 4. Solve the resulting simultaneous linear equation for the nodal voltages. 5. Determine the currents through and voltages across each the elements in the network. Example 3 Find the current flow through each resistor using mesh analysis for the circuit below. 20V Solution: Figure 7 Step 1: Determine the number of common nodes and reference node within the network (Figure 8). 1 common node (Va), reference node C Step 2: Assign current and its direction to each distinct branch of the nodes in the network (Figure 8). I1 Va I2 I13 20V C Figure 8 Step 3: Apply KCL at each of the common nodes in the network KCL: MAIANA/JKE/POLISAS/ET101-UNIT4 4

5 Step 4: Solve the resulting simultaneous linear equation for the nodal voltages. Step 5: Determine the currents through each elements Example 4 Find the current flow through each resistor using mesh analysis for the circuit below. 5kΩ 3kΩ 40V 6kΩ 55V Figure 9 Solution: Step 1: Determine the number of common nodes and reference node within the network (Figure 10). 1 common node (Va), reference node C Step 2: Assign current and its direction to each distinct branch of the nodes in the network (Figure 10). I1 Va I2 5kΩ I3 3kΩ 40V 6kΩ 55V C Figure 10 MAIANA/JKE/POLISAS/ET101-UNIT4 5

6 Step 3: Apply KCL at each of the common nodes in the network KCL: Step 4: Solve the resulting simultaneous linear equation for the nodal voltages. Step 5: Determine the currents through each elements MAIANA/JKE/POLISAS/ET101-UNIT4 6

7 TUTORIAL 1 Find the current through each resistor for the networking below using Mesh Analysis and Nodal Analysis. a) d) 4V b) 4Ω 8Ω 2Ω 6V 4Ω 3Ω 12V 12Ω e) 15Ω 15V 5.6kΩ 2.2kΩ 3.3kΩ V3 c) 20V 30V 4kΩ 3kΩ 30V 25V 2kΩ MAIANA/JKE/POLISAS/ET101-UNIT4 7

8 2.0 Thevenin s Theorem Thevenins Theorem states: "Any linear circuit containing several energy source and resistances can be replaced by just a Single Voltage in series with a Single Resistor". Thevenins equivalent circuit. A Linear Network Containing Several Energy Source and Resistance A B VTH RTH Thevenin s Equivalent Circuit Figure 11 Thevenin s theorem procedure: 1. Open circuit R L and find Thevenin s voltage (V TH ). 2. Find Thevenin s resistance (R TH ) when voltage source is short circuit or current source is open circuit and R L is open circuit. 3. Draw the Thevenin s equivalent circuit such as in figure 11 with the value of V TH and R TH. Find the I L which current flow through the R L. Example 5 Find the current flow through equal to 30Ω for the circuit in Figure Ω Figure 12 MAIANA/JKE/POLISAS/ET101-UNIT4 8

9 Solution: Step 1: Open circuit R L and find Thevenin s voltage (V TH ). Using VDR find V TH VTH Figure 13 Step 2: Find Thevenin s resistance (R TH ) when voltage source is short circuit RTH Figure 14 Step 3: Draw the Thevenin s equivalent circuit with the value of V TH and R TH RTH 28Ω VTH 8V 30Ω Figure 15 MAIANA/JKE/POLISAS/ET101-UNIT4 9

10 Example 6 Find current flow through R 4. 60Ω Is 300mA R4 30Ω 90Ω 25Ω Solution : Figure 16 Step 1 : Open circuit R L and find Thevenin s voltage (V TH ). I1 60Ω I2 Using CDR, find I 2 Is 300mA 30Ω 90Ω VTH Figure 17 Step 2: Find Thevenin s resistance (R TH ) when current source,i S is open circuit. 60Ω 30Ω 90Ω RTH Figure 18 Step 3: Draw the Thevenin s equivalent circuit with the value of V TH and R TH RTH 45Ω VTH 4.5V 25Ω Figure 19 MAIANA/JKE/POLISAS/ET101-UNIT4 10

11 TUTORIAL 2 1. Refer to figure 1, find the current flow through resistor 12Ω using Thevenin s Theorem. 3Ω 4Ω R4 36V 6Ω 12Ω 5. Calculate the current flow in 30Ω resistor for the circuit in figure 5 using Thevenin s Theorem. Is R4 2A 30Ω Figure 1 2. Find the current flow through resistor 15Ω for the circuit in figure 2 using Thevenin s Theorem. 15V 3Ω 2Ω 15Ω Figure 2 4Ω 6Ω 5Ω 3. Count value stream I L by using Thevenin s Theorem. 5kΩ 4kΩ 3kΩ 2kΩ 20V Figure 5 6. Refer to figure 6, find the current flow through 50Ω using Thevenin s Theorem. Is 200mA 30Ω 50Ω Figure 6 7. Use Thevenin s Theorem, find the current flow through resistor R=. 6V 8Ω 15Ω 1kΩ Figure 7 Figure 3 4. Use Thevenin s Theorem to find the current flowing in 5Ω resistor shown in figure 4. 15V 6Ω 4Ω 5Ω 8Ω 2Ω 8. Use Thevenin s Theorem, find the current flow through resistor R=. 6V 8Ω 15Ω Figure 4 Figure 8

12 3.0 Norton s Theorem Nortons Theorem states: "Any linear circuit containing several energy sources and resistances can be replaced by a single Constant Current generator in parallel with a Single Resistor". A Linear Network Containing Several Energy Source and Resistance A B IN RN Norton Equivalent Circuit Example 7 Figure 20 Norton s theorem procedure: 1. Remove R L from the circuit. Find I N by shorting links output terminal. 2. Find R N by short-circuit voltage source or open-circuit current source. 3. Draw the Norton s equivalent circuit such as in figure 20 with the value of I N and R N. Find the I L which current flow through the R L. Find the current flow through equal to 30Ω for the circuit in Figure Ω Step 1: Figure 21 Remove R L from the circuit. Find I N by shorting links output terminal. IT I1 IN 30Ω Figure 22

13 Step 2: Find R N by short-circuit voltage source. RN Figure 23 Step3: Draw the Norton s equivalent circuit with the value of I N and R N. Find the I L which current flow through the R L. Using CDR, find I L IN RN 0.286A 28Ω 30Ω Example 6 Figure 24 Find current flow through R 4. 60Ω Is 300mA R4 30Ω 90Ω 25Ω Figure 25 MAIANA/JKE/POLISAS/ET101-UNIT4 13

14 Solution: Step 1: Remove R L from the circuit. Find I N by shorting links output terminal. Current flow at 90Ω is 0A, so. Is 300mA 60Ω IN R4 30Ω 90Ω 25Ω Step 2: Figure 26 Find R N by open-circuit current source. 30Ω 60Ω 90Ω RN Figure 27 Step3: Draw the Norton s equivalent circuit with the value of I N and R N. Find the I L which current flow through the R L. Using CDR, find I L IN RN 100mA 45Ω 25Ω Figure 28 MAIANA/JKE/POLISAS/ET101-UNIT4 14

15 TUTORIAL 3 1. Refer to figure 1, find the current flow through resistor 12Ω using Norton s Theorem. 3Ω 4Ω R4 36V 6Ω 12Ω 5. Calculate the current flow in 30Ω resistor for the circuit in figure 5 using Norton Theorem. Is R4 2A 30Ω Figure 1 2. Find the current flow through resistor 15Ω for the circuit in figure 2 using Norton Theorem. Figure 5 6. Refer to figure 6, find the current flow through 50Ω using Norton Theorem. 15V 3Ω 2Ω 15Ω Figure 2 4Ω 6Ω 5Ω 3. Count value stream I L by using Norton Theorem. 5kΩ 4kΩ 3kΩ 2kΩ 20V Is 200mA 30Ω 50Ω Figure 6 7. Use Norton Theorem, find the current flow through resistor R=. 6V 8Ω 15Ω 1kΩ Figure 7 Figure 3 4. Use Norton Theorem to find the current flowing in 5Ω resistor shown in figure 4. 15V 6Ω 4Ω 5Ω 8Ω 2Ω 8. Use Norton Theorem, find the current flow through resistor R=. 6V 8Ω 15Ω Figure 4 Figure 8 MAIANA/JKE/POLISAS/ET101-UNIT4 15

16 4.0 Maximum Power Transfer theorem The maximum power transfer theorem states: A load will receive maximum power from a linear bilateral dc network when its total resistive value equal to the Thevenin s or Norton resistance of the network as seen by the load. RTH VTH IN RN Thevenin Equivalent Circuit Norton Equivalent Circuit Figure 29 For the Thevenin equivalent circuit above, maximum power will be delivered to the load when: For the Norton equivalent circuit above, maximum power will be delivered to the load when: There are four conditions occur when maximum power transfer took place in a circuit: 1. Value of equal to R TH (R L =R TH ). 2. Value of current is half of the current when is short circuited. 3. Value of load voltage is half the Thevenin s voltage (V L = ½V TH ). 4. Percentage of efficiency, % = 50%. Where: Example 7 Refer to figure 30, determine the load power for each of the following value of the variable load resistance and sketch the graph load power versus load resistance. a) 25Ω b) 50Ω c) 75Ω d) 100Ω e) 125Ω RTH 75Ω VTH Figure 30

17 Load Power (W) UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS Solution: a) d) b) e) c) R TH R L I V TH VL=IR L % P L 75Ω A 0V 0% 0W 75Ω 25Ω 0.1A 2.5V 25% 0.25W 75Ω 50Ω 0.08A 4V 40% 0.32W 75Ω 75Ω 0.067A 5.0V 50% 0.336W 75Ω 100Ω 0.057A 5.7V 57% 0.325W 75Ω 125Ω 0.05A 6.5V 65% 0.312W Load Power (PL) vs Load Resistance() 0, 0 25, , , , , Load Resistance (Ω) Figure 31 MAIANA/JKE/POLISAS/ET101-UNIT4 17

18 Example 8 For the network in figure 32, determine the value of R for maximum power transfer to R and hence calculate the maximum power using Thevenin s equivalent circuit. 6Ω 8Ω 12V 3Ω R Figure 32 Solution: Open circuit R and find Thevenin s voltage (V TH ). 6Ω 8Ω 12V 3Ω VTH Using VDR find V TH Figure 33 Find Thevenin s resistance (R TH ) when voltage source is short circuit 6Ω 8Ω 3Ω RTH Figure 34 Draw the Thevenin s equivalent circuit with the value of V TH and R TH RTH VTH 4V R Maximum power transfer occur when R=R TH. So, the value of R is. Figure 35 Maximum power transfer, MAIANA/JKE/POLISAS/ET101-UNIT4 18

19 5.0 Superposition Theorem The superposition theorem states: In any network made up of linear resistances and containing more than one source of e.m.f, the resultant current flowing in any branch is the algebraic sum of the currents that would flow in that branch if each source was considered separately, all other sources being replaced at that time by their respective internal resistances. Removing the effect of voltage and current source Short circuit Open circuit Voltage source Current source Example 9 Determine the current through resistor =5Ω for the network in figure 36 using superposition theorem. V 15V 5Ω I 9A Figure 36 Solution: Step 1: V active, I inactive. So current source is open circuit. Ia V 15V 5Ω Figure 37

20 Step 2: V inactive, I active. So voltage source is short circuit. Ib Using CDR 5Ω I 9A Figure 38 Step 3: Total current through =5Ω. Ia 1A Ib 6A Example 10 Find the current flow through each resistor for the network in figure V Figure 39 Solution: Step 1: active, inactive I1' I2' I3' Figure 40 MAIANA/JKE/POLISAS/ET101-UNIT4 20

21 Step 2: inactive, active I1' I2' I3' 20V Figure 41 Step 3: Total current flow through each resistor I I 1 =0.429A So I 1 =0.571A I I 2 =0.286A So I 2 =0.714A I I 3 =0.143A I 3 =0.143A So

22 TUTORIAL 4 Find the current through each resistor for the networking below using Superposition Theorem. b) d) 4V b) 4Ω 8Ω 2Ω 6V 4Ω 3Ω 12V 12Ω e) 15Ω 15V 5.6kΩ 2.2kΩ 3.3kΩ V3 c) 20V 30V 4kΩ 3kΩ 30V 25V 2kΩ MAIANA/JKE/POLISAS/ET101-UNIT4 22

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