Network Theorem - Lecture 2 PDF

Network Theorem - Lecture 2 Network Theorem Superposition theorem Thévenin’s theorem Norton’s theorem Maximum power transfer Millman’s theorem Substitution theorem Reciprocity theorems. Introduction Superposition Theorem In general, the t...

Network Theorem - Lecture 2 Network Theorem Superposition theorem Thévenin’s theorem Norton’s theorem Maximum power transfer Millman’s theorem Substitution theorem Reciprocity theorems. Introduction Superposition Theorem In general, the theorem can be used to do the following: v Analyze networks that have two or more sources that are not in series or parallel. v The current through, or voltage across, an element in a network is equal to the algebraic sum of the currents or voltages produced independently by each source. v Since the effect of each source will be determined independently, the number of networks to be analyzed will equal the number of sources. The Superposition Theorem The superposition theorem states that: The current through, or voltage across, any element of a network is equal to the algebraic sum of the currents or voltages produced independently by each source. Only one source at a time can be used to find voltage or current. Once we have the solution for each source, we can combine the results to obtain the total solution. Please Note… When removing a voltage source from a network schematic, replace it with a direct connection (short circuit) of zero ohms. Any internal resistance associated with the source must remain in the network. When removing a current source from a network schematic, replace it by an open circuit of infinite ohms. Any internal resistance associated with the source must remain in the network. Example 1 Using superposition theorem, find I1: Solution: Open the current source E 30 I '1 = = =5A R1 6 Example 1 – Solution Cont. Replace the current source and short the voltage source: æ Rsc ö I "1 = I çç ÷÷ è Rsc + R1 ø æ 0 ö = 3ç ÷=0A è0+6ø With both sources in the circuit, the total current is therefore; I1 = I '1 + I "1 = 5+0 = 5A Example 2 Use superposition theorem to find I2: Solution Short the voltage source E2 R2 R3 R2 R3 = R2 + R3 12 ´ 4 = = 3W 12 + 4 Example 2 – Solution cont. I 's RT = R1 + R2 R3 = 24 + 3 = 27 W E1 54 I 's = = =2A RT 27 æ R3 ö æ 4 ö I '2 = I 's çç ÷÷ = 2ç ÷ = 0.5 A è R2 + R3 ø è 12 + 4 ø Replace E2 and short the voltage source E1: R1 R2 R1 R2 = R1 + R2 I "s 24 ´ 12 = =8W 24 + 12 Example 2 – Solution cont. RT = R3 + R1 R2 = 4 + 8 = 12 W E I "s = 2 RT 48 I "s = =4A 12 æ R1 ö æ 24 ö I "2 = I "s çç ÷÷ = 4ç ÷ = 2.67 A è R1 + R2 ø è 24 + 12 ø With both sources in the circuit, the total current is therefore; I 2 = I "2 -I '2 = 2.67 - 0.5 = 2.17 A Example 3 (a) Use superposition theorem to find I2 (b) Demonstrate that the superposition theorem is not applicable to power level Solution: Replace the current source with an open circuit: E I '2 = R1 + R2 36 = =2A 12 + 6 P'2 = (I '2 ) R2 = 2 2 ´ 6 = 24 W 2 Example 3 – Solution cont. Reconnect the current source and replace the voltage source with a short circuit: æ R1 ö I "2 = I çç ÷÷ R è 1 + R2 ø æ 12 ö = 9ç ÷=6A è 12 + 6 ø P"2 = (I "2 ) R2 = 6 2 ´ 6 = 216 W 2 With both sources in the circuit, the current through R2 is; I 2 = I '2 + I "2 = 6+2 =8A P2 = I 22 R2 = 82 ´ 6 = 384 W  P'2 +P"2 = 24 + 216 = 240 W which is not equal to: P2 = 384 W Hence, the superposition is not applicable to power level Example 4 Using the principle of superposition, find the current l2 through the 12 k resistor. Solution: Example 4 – Solution cont. Current divider rule: Considering the effect of the 9 V voltage source: Example 4 – Solution cont. Since have the same direction through R2, the desired current is the sum of the two: Thévenin’s Theorem In general, the theorem is used to do the following: v Analyze networks with sources that are not in series or parallel. v Reduce the number of components required to establish the same characteristics at the output terminals. v Investigate the effect of changing a particular component on the behavior of a network without having to analyze the entire network after each change. Thévenin’s theorem states the following: Any two-terminal DC network can be replaced by an equivalent circuit consisting solely of a voltage source and a series resistor Thévenin’s Theorem Procedure Preliminary: 1. Remove that portion of the network where the Thévenin equivalent circuit is found. 2. Mark the terminals of the remaining two-terminal network. RTh : 3. Calculate RTh by first setting all sources to zero (voltage sources are replaced by short circuits, and current sources by open circuits) and then finding the resultant resistance between the two marked terminals. (If the internal resistance of the voltage and/or current sources is included in the original network, it must remain when the sources are set to zero. ) ETh: 4. Calculate ETh by first returning all sources to their original position and finding the open-circuit voltage between the marked terminals. Conclusion: 5. Draw the Thévenin equivalent circuit with the portion of the circuit previously removed replaced between the terminals of the equivalent circuit. Example 5 Find the Thevenin equivalent circuit for the network in the shaded area: Solution: Remove the component external to the relevant network i.e. R3. Example 5 – solution cont. Replace the current source with an open circuit and calculate RTh: RTh = R1 + R2 = 2+4 = 6W Reconnect the current source and calculate ETh: ETh = IR1 = 12 ´ 4 = 48 V Example 5 – solution cont. Draw the Thevenin equivalent circuit: The original circuit The corresponding Thevenin equivalent circuit Example 6 Find the Thevenin equivalent circuit for the network in the shaded area: Solution: Remove the component external to the relevant network i.e. R3. Example 6 – Solution cont. Replace the voltage source with a short circuit: Redraw the circuit and calculate RTh: R1 R2 RTh = R1 R2 = R1 + R2 6´ 4 = = 2.4 W 6+4 Example 6 – Solution cont. Reconnect the voltage source: Redraw the circuit: Redraw the circuit further and calculate ETh: æ R1 ö ÷÷ = 8æç 6 ö RTh = E çç ÷ = 4.8 V R è 1 + R2 ø è 6 + 3 ø Example 6 – Solution cont. Draw the corresponding Thevenin equivalent circuit: Reconnect R3 which had been removed previously Example 6 – Solution cont. Find the Thevenin circuit for network within the shaded area. Solution: Redraw the circuit: RL Example 6 – Solution cont. Disconnect the component(s) external to the network: Replace the voltage sources with short circuits and find RTh: R2 R3 Ra = R2 + R3 4´ 6 = 4+6 = 2.4 kW Example 6 – Solution cont. Replace R2 and R3 with Ra: R1 Ra Rb = R1 + Ra Ra 2.4 kW 0.8 ´ 2.4 = = 0.6 kW 0.8 + 2.4 Replace R1 and Ra with Rb: RTh = Rb + R4 = 0.6 + 1.4 Rb 0.6 kW = 2 kW Example 6 – Solution cont. Use superposition theorem to find ETh: E Th Replace E2 with a short circuit and find E’Th: E 'Th = V3 æ R2 R3 ö = E1 çç ÷ ÷ è R1 + R2 R3 ø æ 46 ö = 6çç ÷ = 4.5 V ÷ è 0.8 + 4 6 ø Example 6 – Solution cont. Reconnect E2 and replace E1 with a short circuit and find E”Th: E"Th = V3 æ R1 R3 ö = E2 çç ÷ ÷ è R2 + R1 R3 ø æ 0.8 6 ö = 10çç ÷ = 1.5 V ÷ è 4 + 0.8 6 ø Since E’Th and E”Th are of opposite polarities: ETh = E 'Th - E"Th = 4.5 - 1.5 = 3 V Draw the corresponding Thevenin equivalent circuit and reconnect RL. Norton’s Theorem The theorem states the following: Any two-terminal linear bilateral dc network can be replaced by an equivalent circuit consisting of a current source and a parallel resistor Norton’s Theorem Procedure Preliminary: 1. Remove that portion of the network across which the Norton equivalent circuit is found. 2. Mark the terminals of the remaining two-terminal network. Norton’s Theorem Procedure RN: 3. Calculate RN by first setting all sources to zero (voltage sources are replaced with short circuits and current sources with open circuits) and then finding the resultant resistance between the two marked terminals. (If the internal resistance of the voltage and/or current sources is included in the original network, it must remain when the sources are set to zero.) Since RN = RTh , the procedure and value obtained using the approach described for Thévenin’s theorem will determine the proper value of RN. Norton’s Theorem Procedure IN: 4. Calculate IN by first returning all sources to their original position and then finding the short-circuit current between the marked terminals. It is the same current that would be measured by an ammeter placed between the marked terminals. 5. Draw the Norton equivalent circuit with the portion of the circuit previously removed replaced between the terminals of the equivalent circuit. The Norton and Thévenin equivalent circuits can also be found from each other by using the source transformation Example 7 Find the Norton equivalent circuit for the network in the shaded area Solution Step 1 & 2 Step 3 Example 7 – Solution cont. Step 4: The short-circuit connection between terminals a and b is in parallel with R2 and eliminates its effect. IN is therefore the same as through R1, and the full battery voltage appears across R1 Step 5: Example 7 – Solution cont. Converting the Norton equivalent circuit to a Thévenin equivalent circuit. Example 8 (Two sources) Find the Norton equivalent circuit for the portion of the network to the left of a-b Solution Step 1 & 2 Step 3 Example 8 – solution cont. Step 4: (Using superposition) For the 7 V battery Step 5: Maximum Power Transfer Theorem A load will receive maximum power from a network when its resistance is exactly equal to the Thévenin resistance of the network applied to the load. That is, For the Thévenin equivalent circuit, when the load is set equal to the Thévenin resistance, the load will receive maximum power from the network. The maximum power delivered to the load can be determined by first finding the current Then substitute into the power equation: And: MAXIMUM POWER TRANSFER occurs when the load voltage and current are one-half of their maximum possible values. The current through the load is determined by The voltage is determined by: The Power is determined by: Millman’s Theorem Any number of parallel voltage sources can be reduced to one. The three voltage sources can be reduced to one. This permits finding the current through or voltage across RL without having to apply a method such as mesh analysis, nodal analysis, superposition. Procedures for Millman’s Theorem Step 1: Convert all voltage sources to current sources. Step 2: Combine parallel current sources. Step 3: Convert the resulting current source to a voltage source, and the desired single-source network is obtained. In general, Millman’s theorem states that for any number of parallel voltage sources, The plus-and-minus signs include those cases where the sources may not be supplying energy in the same direction. The equivalent resistance is In terms of the resistance values: The dual of Millman’s theorem:  Example 9 Using Millman’s theorem, find the current through and voltage across the resistor RL. Substitution Theorem If the voltage across and the current through any branch of a dc bilateral network are known, this branch can be replaced by any combination of elements that will maintain the same voltage across and current through the chosen branch. The theorem states that for branch equivalence, the terminal voltage and current must be the same A known potential difference and current in a network can be replaced by an ideal voltage source and current source, respectively. This theorem cannot be used to solve networks with two or more sources that are not in series or parallel. For it to be applied, a potential difference or current value must be known.  Reciprocity Theorem The current I in any branch of a network, due to a single voltage source E anywhere else in the network, will equal the current through the branch in which the source was originally located if the source is placed in the branch in which the current I was originally measured. The current I due to the voltage source E was determined. If the position of each is interchanged, the current I will be the same value as indicated. The total resistance is