unit 4 dc equivalent circuit and network theorems
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UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS 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: I2 current towards = current flowing away I1 + I2+ I3 = I4 + I5 I1 + I2 + (- I3 ) + (- I4 ) + (-I5 ) = 0 I= 0 I4 I1 I3 I5 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. R1 V R2 E = IR1 + IR2 E = I(R1 + R2 ) E + (- IR1 ) + (- IR2) = 0 Figure 2 1.1 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. MARLIANA/JKE/POLISAS/ET101-UNIT4 1 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS Example 1 Find the current flow through each resistor using mesh analysis for the circuit below. R1 R2 10Ω 20Ω V1 R3 V2 40Ω 20V 10V Figure 3 Solution: Step 1: Assign a distinct current to each closed loop of the network. R1 I1 10Ω V1 I1 I2 I13 R3 40Ω R2 20Ω I2 V2 20V 10V Figure 4 Step 2: Apply KVL around each closed loop of the network. Loop 1: Loop 2: ------------ equation 1 --------------- 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 : MARLIANA/JKE/POLISAS/ET101-UNIT4 2 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS Example 2 Find the current flow through each resistor using mesh analysis for the circuit below. R2 R1 5kΩ V1 3kΩ R3 V2 6kΩ 55V 40V Figure 5 Solution: Step 1: Assign a distinct current to each closed loop of the network. R1 I1 5kΩ V1 I1 I2 3kΩ I3 R3 R2 6kΩ I2 40V V2 55V Figure 6 Step 2: Apply KVL around each closed loop of the network. Loop 1: Loop 2: ------------ equation 1 --------------- 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 : MARLIANA/JKE/POLISAS/ET101-UNIT4 3 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS 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. R1 R2 10Ω 20Ω V1 R3 V2 40Ω 20V 10V Figure 7 Solution: 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). R1 I1 Va I2 10Ω V1 I13 R3 40Ω R2 20Ω V2 20V 10V C Figure 8 Step 3: Apply KCL at each of the common nodes in the network KCL: MARLIANA/JKE/POLISAS/ET101-UNIT4 4 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS 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. R1 R2 5kΩ 3kΩ V1 R3 V2 6kΩ 55V 40V 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). R1 I1 Va I2 5kΩ V1 3kΩ I3 R3 R2 6kΩ 40V V2 55V C Figure 10 MARLIANA/JKE/POLISAS/ET101-UNIT4 5 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS 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 MARLIANA/JKE/POLISAS/ET101-UNIT4 6 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS TUTORIAL 1 Find the current through each resistor for the networking below using Mesh Analysis and Nodal Analysis. a) R1 R2 4Ω 2Ω d) R1 4Ω V1 R3 V2 8Ω R3 6V 4V 3Ω V1 R2 10V 12Ω V2 12V b) R1 R2 20Ω 10Ω V1 R3 e) R1 V2 15Ω 15V 10V R2 5.6kΩ 3.3kΩ R3 2.2kΩ V3 V1 10V c) 30V V2 20V R1 4kΩ R3 3kΩ V1 30V R2 2kΩ V2 25V MARLIANA/JKE/POLISAS/ET101-UNIT4 7 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS 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 A Linear Network Containing Several Energy Source and Resistance IL RL RTH VTH RL B Thevenin’s Equivalent Circuit Figure 11 Thevenin’s theorem procedure: 1. Open circuit RL and find Thevenin’s voltage (VTH). 2. Find Thevenin’s resistance (RTH) when voltage source is short circuit or current source is open circuit and RL is open circuit. 3. Draw the Thevenin’s equivalent circuit such as in figure 11 with the value of VTH and RTH. Find the IL which current flow through the RL. Example 5 Find the current flow through RL equal to 30Ω for the circuit in Figure 12. V1 R1 R2 10Ω 20Ω R3 40Ω 10V RL 30Ω Figure 12 MARLIANA/JKE/POLISAS/ET101-UNIT4 8 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS Solution: Step 1: Open circuit RL and find Thevenin’s voltage (VTH). R1 R2 10Ω 20Ω Using VDR find VTH V1 R3 VTH 40Ω 10V Figure 13 Step 2: Find Thevenin’s resistance (RTH) when voltage source is short circuit R1 R2 10Ω 20Ω R3 RTH 40Ω Figure 14 Step 3: Draw the Thevenin’s equivalent circuit with the value of VTH and RTH RTH IL 28Ω VTH RL 30Ω 8V Figure 15 MARLIANA/JKE/POLISAS/ET101-UNIT4 9 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS Example 6 Find current flow through R4. R2 60Ω R1 30Ω Is 300mA R3 90Ω R4 25Ω Figure 16 Solution : Step 1 : Open circuit RL and find Thevenin’s voltage (VTH). R2 I1 R1 30Ω Is 300mA I2 60Ω R3 90Ω Using CDR, find I2 VTH Figure 17 Step 2: Find Thevenin’s resistance (RTH) when current source,IS is open circuit. R2 60Ω R1 30Ω R3 90Ω RTH Figure 18 Step 3: Draw the Thevenin’s equivalent circuit with the value of VTH and RTH RTH 45Ω VTH IL RL 25Ω 4.5V Figure 19 MARLIANA/JKE/POLISAS/ET101-UNIT4 10 TUTORIAL 2 1. Refer to figure 1, find the current flow through resistor 12Ω using Thevenin’s Theorem. 5. Calculate the current flow in 30Ω resistor for the circuit in figure 5 using Thevenin’s Theorem. R1 R3 R2 3Ω 4Ω 20Ω 36V R2 R4 6Ω 12Ω Is 2A R1 10Ω Figure 1 6. Refer to figure 6, find the current flow through 50Ω using Thevenin’s Theorem. 10Ω 4Ω 15Ω Is 200mA 5Ω 15V 2Ω R4 40Ω Figure 5 2. Find the current flow through resistor 15Ω for the circuit in figure 2 using Thevenin’s Theorem. 3Ω R3 30Ω 30Ω 50Ω 40Ω 6Ω Figure 6 Figure 2 7. Use Thevenin’s Theorem, find the current flow through resistor R=10Ω. 3. Count value stream IL by using Thevenin’s Theorem. IL 4kΩ 2kΩ R2 8Ω 15Ω V1 3kΩ 5kΩ R1 R3 Figure 7 Figure 3 8. Use Thevenin’s Theorem, find the current flow through resistor R=10Ω. 4. Use Thevenin’s Theorem to find the current flowing in 5Ω resistor shown in figure 4. R1 R2 8Ω 15Ω 8Ω 4Ω V1 15V 10V 6V 20V 1kΩ 6Ω V2 10Ω 5Ω Figure 4 2Ω R3 10Ω 6V Figure 8 V2 10V 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 A Linear Network Containing Several Energy Source and Resistance IL RL IN RN RL B Norton Equivalent Circuit Figure 20 Norton’s theorem procedure: 1. Remove RL from the circuit. Find IN by shorting links output terminal. 2. Find RN 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 IN and RN. Find the IL which current flow through the RL. Example 7 Find the current flow through RL equal to 30Ω for the circuit in Figure 21. R1 R2 10Ω 20Ω V1 R3 40Ω RL 10V Step 1: 30Ω Figure 21 Remove RL from the circuit. Find IN by shorting links output terminal. R2 R1 IT V1 10Ω I1 R3 40Ω 10V Figure 22 20Ω IN RL 30Ω UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS Step 2: Find RN by short-circuit voltage source. R1 R2 10Ω 20Ω R3 RN 40Ω Figure 23 Step3: Draw the Norton’s equivalent circuit with the value of IN and RN. Find the IL which current flow through the RL. IL RN 28Ω IN 0.286A Using CDR, find IL RL 30Ω Figure 24 Example 6 Find current flow through R4. R2 60Ω Is 300mA R1 30Ω R3 90Ω R4 25Ω Figure 25 MARLIANA/JKE/POLISAS/ET101-UNIT4 13 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS Solution: Step 1: Remove RL from the circuit. Find IN by shorting links output terminal. Current flow at 90Ω is 0A, so R2 . 60Ω Is R1 30Ω R3 90Ω IN R4 25Ω 300mA Figure 26 Find RN by open-circuit current source. Step 2: R2 60Ω R1 30Ω R3 90Ω RN Figure 27 Step3: Draw the Norton’s equivalent circuit with the value of IN and RN. Find the IL which current flow through the RL. IL IN 100mA RN 45Ω Using CDR, find IL RL 25Ω Figure 28 MARLIANA/JKE/POLISAS/ET101-UNIT4 14 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS TUTORIAL 3 1. Refer to figure 1, find the current flow through resistor 12Ω using Norton’s Theorem. 5. Calculate the current flow in 30Ω resistor for the circuit in figure 5 using Norton Theorem. R1 R3 R2 3Ω 4Ω 20Ω 36V R2 R4 6Ω 12Ω Is 2A R1 10Ω Figure 1 R3 30Ω R4 40Ω Figure 5 2. Find the current flow through resistor 15Ω for the circuit in figure 2 using Norton Theorem. 3Ω 6. Refer to figure 6, find the current flow through 50Ω using Norton Theorem. 10Ω 4Ω 15Ω Is 200mA 5Ω 15V 2Ω 30Ω 50Ω 40Ω 6Ω Figure 6 Figure 2 7. Use Norton Theorem, find the current flow through resistor R=10Ω. 3. Count value stream IL by using Norton Theorem. IL 4kΩ 2kΩ R2 8Ω 15Ω V1 3kΩ 5kΩ R1 R3 Figure 7 Figure 3 8. Use Norton Theorem, find the current flow through resistor R=10Ω. 4. Use Norton Theorem to find the current flowing in 5Ω resistor shown in figure 4. R1 R2 8Ω 15Ω 8Ω 4Ω V1 15V 10V 6V 20V 1kΩ 6Ω V2 10Ω 5Ω Figure 4 MARLIANA/JKE/POLISAS/ET101-UNIT4 2Ω R3 V2 10Ω 10V 6V Figure 8 15 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 IL IL RN IN RL RL Norton Equivalent Circuit Thevenin 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. 2. 3. 4. Value of RL equal to RTH (RL=RTH). Value of current is half of the current when RL is short circuited. Value of load voltage is half the Thevenin’s voltage (VL = ½VTH). 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 IL RL 10V Figure 30 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS Solution: a) d) b) e) c) RTH 75Ω 75Ω 75Ω 75Ω 75Ω 75Ω RL 0 25Ω 50Ω 75Ω 100Ω 125Ω I 0.133A 0.1A 0.08A 0.067A 0.057A 0.05A VTH 10V 10V 10V 10V 10V 10V VL=IRL 0V 2.5V 4V 5.0V 5.7V 6.5V % 0% 25% 40% 50% 57% 65% PL 0W 0.25W 0.32W 0.336W 0.325W 0.312W Load Power (W) Load Power (PL) vs Load Resistance(RL) 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 50, 0.32 75, 0.336 100, 0.325 125, 0.312 25, 0.25 0, 0 0 20 40 60 80 Load Resistance (Ω) 100 120 140 Figure 31 MARLIANA/JKE/POLISAS/ET101-UNIT4 17 UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS 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. R1 R3 6Ω 8Ω R2 R 3Ω 12V Figure 32 Solution: Open circuit R and find Thevenin’s voltage (VTH). 6Ω 8Ω VTH 3Ω Using VDR find VTH 12V Figure 33 Find Thevenin’s resistance (RTH) when voltage source is short circuit 6Ω 8Ω RTH 3Ω Figure 34 Draw the Thevenin’s equivalent circuit with the value of VTH and RTH RTH IL 10Ω VTH Maximum power transfer occur when R=RTH. So, the value of R is 10Ω. R 4V Figure 35 MARLIANA/JKE/POLISAS/ET101-UNIT4 Maximum power transfer, 18 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 Open circuit Short circuit Voltage source Current source Example 9 Determine the current through resistor R2=5Ω for the network in figure 36 using superposition theorem. R1 10Ω V 15V R2 5Ω Figure 36 Solution: Step 1: V active , I inactive. So current source is open circuit. R1 10Ω V 15V Ia R2 5Ω Figure 37 I 9A UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS Step 2: V inactive, I active. So voltage source is short circuit. R1 Using CDR Ib 10Ω I 9A 5Ω R2 Figure 38 Step 3: Total current through R2=5Ω. Ia 1A Ib 6A Example 10 Find the current flow through each resistor for the network in figure 39. R1 R2 10Ω 20Ω V1 R3 40Ω 10V V2 20V Figure 39 Solution: Step 1: V1 active, V2 inactive R1 I1' 10Ω V1 I2' I3' R3 R2 20Ω 40Ω 10V Figure 40 MARLIANA/JKE/POLISAS/ET101-UNIT4 20 Step 2: V1 inactive, V2 active R1 I1'’ I2'’ I3'’ 10Ω R3 R2 20Ω V2 40Ω 20V Figure 41 Step 3: Total current flow through each resistor IR1 I1’=0.429A So I1’’=0.571A IR2 I2’=0.286A So I2’’=0.714A IR3 I3’=0.143A I3’’=0.143A So UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS TUTORIAL 4 Find the current through each resistor for the networking below using Superposition Theorem. b) R1 R2 4Ω 2Ω d) R1 4Ω V1 R3 V2 8Ω R3 6V 4V 3Ω V1 R2 10V 12Ω V2 12V b) R1 R2 20Ω 10Ω V1 R3 e) R1 V2 15Ω 15V 10V R2 5.6kΩ 3.3kΩ R3 2.2kΩ V3 V1 10V c) 30V V2 20V R1 4kΩ R3 3kΩ V1 30V R2 2kΩ V2 25V MARLIANA/JKE/POLISAS/ET101-UNIT4 22
