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Kirchhoffs Laws Kirchhoffs circuit laws are two equalities that deal with the conservation of charge and energy in electrical circuits. There basically two Kirchhoffs law :-1. Kirchhoffs current law (KCL) – Based on principle of conservation of electric charge.2. Kirchhoffs voltage law (KVL) - Based on principle of conservation of energy.
Kirchhoffs current law (KCL)This law is also called Kirchhoffs first law, Kirchhoffs pointrule, Kirchhoffs junction rule (or nodal rule), and Kirchhoffsfirst rule.The principle of conservation of electric charge implies that: At any node (junction) in an electrical circuit, the sumof currents flowing into that node is equal to the sum ofcurrents flowing out of that node, or The algebraic sum ofcurrents in a network of conductors meeting at a point is zero.Strictly speaking KCL only applies to circuits with steadycurrents (DC).However, for AC circuits having dimensions much smallerthan a wavelength, KCL is also approximately applicable.
The current entering any junction is equal to the current leaving that junction. i1 + i4 =i2 + i3Recalling that current is a signed (positive or negative)quantity reflecting direction towards or away from a node,this principle can be stated as: I 0
Kirchhoffs voltage law (KVL)This law is also called Kirchhoffs second law, Kirchhoffs loop(or mesh) rule, and Kirchhoffs second rule.The principle of conservation of energy implies that The directed sum of the electrical potentialdifferences (voltage) around any closed circuit is zero, or More simply, the sum of the emfs in any closed loop isequivalent to the sum of the potential drops in that loopStrictly speaking KVL only applies to circuits with steadycurrents (DC).However, for AC circuits having dimensions much smaller thana wavelength, KVL is also approximately applicable.
The algebraic sum of the products of the resistances of theconductors and the currents in them in a closed loop is equalto the total emf available in that loop. Similarly to KCL, it canbe stated as: KVL: Vn 0 Vemf I R OR loop The sum of all the voltages around the loop is equal to zero. v1 + v2 + v3 - v4 = 0
Mesh AnalysisMesh analysis (or the mesh current method) is a method thatis used to solve planar circuits for the currents (and indirectlythe voltages) at any place in the circuit. Planar circuits arecircuits that can be drawn on a plane surface withno wires crossing each other.Mesh analysis works by arbitrarily assigning mesh currents inthe essential meshes. An essential mesh is a loop in the circuitthat does not contain any other loop.
Steps to Determine Mesh Currents: 1. Assign mesh currents i1, i2, .., in to the n meshes. Current direction need to be same in all meshes either clockwise or anticlockwise. 2. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents. 3. Solve the resulting n simultaneous equations to get the mesh currents
Solve for the mesh currents. R1 R3 R3 i1 V1 R3 R2 R3 i2 V2Use i for a mesh current and I for a branchcurrent. It’s evident from Fig. 3.17 that I1 i1 , I 2 i2 , I3 i1 i2
Nodal AnalysisIn electric circuits analysis, nodal analysis, node-voltageanalysis, or the branch current method is a method ofdetermining the voltage (potential difference) between"nodes" (points where elements or branches connect) inan electrical circuit in terms of the branch currents.Nodal analysis is possible when all the circuit elementsbranch constitutive relations have an admittancerepresentation. Kirchhoff’s current law is used to develop the methodreferred to as nodal analysis
STEPS FOR NODAL ANALYSIS:-• Note all connected wire segments in the circuit. These are the nodes of nodal analysis.• Select one node as the ground reference. The choice does not affect the result and is just a matter of convention. Choosing the node with most connections can simplify the analysis.• Assign a variable for each node whose voltage is unknown. If the voltage is already known, it is not necessary to assign a variable.• For each unknown voltage, form an equation based on Kirchhoffs current law. Basically, add together all currents leaving from the node and mark the sum equal to zero.
• If there are voltage sources between two unknown voltages, join the two nodes as a super node. The currents of the two nodes are combined in a single equation, and a new equation for the voltages is formed.• Solve the system of simultaneous equations for each unknown voltage.
1. Reference Node 500 500 +I1 V 1k 500 I2 500 –The reference node is called the ground node where V = 0
Example 500 500 V1 V2 V3 1 2 3I1 1k 500 I2 500V1, V2, and V3 are unknowns for which we solve using KCL
Steps of Nodal Analysis1. Choose a reference (ground) node.2. Assign node voltages to the other nodes.3. Apply KCL to each node other than the reference node; express currents in terms of node voltages.4. Solve the resulting system of linear equations for the nodal voltages.
Superposition Theorem• It is used to find the solution to networks with two or more sources that are not in series or parallel• The current through, or voltage across, an element in a linear bilateral network is equal to the algebraic sum of the currents or voltages produced independently by each source.• For a two-source network, if the current produced by one source is in one direction, while that produced by the other is in the opposite direction through the same resistor, the resulting current is the difference of the two and has the direction of the larger• If the individual currents are in the same direction, the resulting current is the sum of two in the direction of either current
Superposition Theorem• The total power delivered to a resistive element must be determined using the total current through or the total voltage across the element and cannot be determined by a simple sum of the power levels established by each source For applying Superposition theorem:-• Replace all other independent voltage sources with a short circuit (thereby eliminating difference of potential. i.e. V=0, internal impedance of ideal voltage source is ZERO (short circuit)).• Replace all other independent current sources with an open circuit (thereby eliminating current. i.e. I=0, internal impedance of ideal current source is infinite (open circuit).
Example:- Determine the branches current using Superposition theorem. 6 2 i1 i3 120 V i2 3 i4 4 12 A Figure 1 Solution:• The application of the superposition theorem is shown in Figure 1, where it is used to calculate the branch current. We begin by calculating the branch current caused by the voltage source of 120 V. By substituting the ideal current with open circuit, we deactivate the current source, as shown in Figure 2.
6 v1 2 i1 i3 120 V i2 i4 4 3 Figure 2• To calculate the branch current, the node voltage across the 3Ω resistor must be known. Therefore v 1 120 v1 v1 =0 6 3 2 4 where v1 = 30 VThe equations for the current in each branch,
120 30 i1 = = 15 A 6 30 i2 = = 10 A 3 = i = 30 i3 4 =5A 6 6 2 i 1" i 3" i 2" 3 i4" 4 12 AIn order to calculate the current cause by the current source, wedeactivate the ideal voltage source with a short circuit, as shown
To determine the branch current, solve the nodevoltages across the 3Ω dan 4Ω resistors as shown inFigure 4 6 2 + + v3 3 v4 4 12 A - - v3 v3 v3 v4 =0 3 6 2 v4 v3 v4 12 = 0 2 4The two node voltages are
• By solving these equations, we obtain v3 = -12 V v4 = -24 V Now we can find the branches current,
To find the actual current of the circuit, add the currents due to both the current and voltage source,
Thevenins theoremThevenins theorem for linear electrical networks states thatany combination of voltage sources, current sources,and resistors with two terminals is electrically equivalent to asingle voltage source V and a single series resistor R.Any two-terminal, linear bilateral dc network can be replacedby an equivalent circuit consisting of a voltage source and aseries resistor
Thévenin’s Theorem The Thévenin equivalent circuit provides an equivalence at the terminals only – the internal construction and characteristics of the original network and the Thévenin equivalent are usually quite different• This theorem achieves two important objectives: – Provides a way to find any particular voltage or current in a linear network with one, two, or any other number of sources – We can concentration on a specific portion of a network by replacing the remaining network with an equivalent circuit
Calculating the Thévenin equivalent• Sequence to proper value of RTh and ETh• Preliminary – 1. Remove that portion of the network across which the Thévenin equation circuit is to be found. In the figure below, this requires that the load resistor RL be temporarily removed from the network.
– 2. Mark the terminals of the remaining two- terminal network. (The importance of this step will become obvious as we progress through some complex networks)– 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. (This step is invariably the one that will lead to the most confusion and errors. In all cases, keep in mind that it is the open-circuit potential between the two terminals marked in step 2)
• Conclusion: – 5. Draw the Thévenin equivalent circuit with Insert Figure 9.26(b) the portion of the circuit previously removed replaced between the terminals of the equivalent circuit. This step is indicated by the placement of the resistor RL between the terminals of the Thévenin equivalent circuit
Another way of Calculating the Thévenin equivalent• Measuring VOC and ISC – The Thévenin voltage is again determined by measuring the open-circuit voltage across the terminals of interest; that is, ETh = VOC. To determine RTh, a short-circuit condition is established across the terminals of interest and the current through the short circuit Isc is measured with an ammeter – Using Ohm’s law: RTh = Voc / Isc
Example:- find the Thevenin equivalent circuit. 5 4 a + + 25 V 3A v1 vab 20 - - b Solution• In order to find the Thevenin equivalent circuit for the circuit shown in Figure1 , calculate the open circuit voltage, Vab. Note that when the a, b terminals are open, there is no current flow to 4Ω resistor. Therefore, the voltage vab is the same as the voltage across the 3A current source, labeled v1.• To find the voltage v1, solve the equations for the singular node voltage. By choosing the bottom right node as the reference node,
v 1 25 v1 3 0 5 20• By solving the equation, v1 = 32 V. Therefore, the Thevenin voltage Vth for the circuit is 32 V.• The next step is to short circuit the terminals and find the short circuit current for the circuit shown in Figure 2. Note that the current is in the same direction as the falling voltage at the terminal. 5 4 a + + 25 V 3A v2 vab isc 20 - - b Figure 2
Current isc can be found if v2 is known. By using the bottomright node as the reference node, the equationfor v2 becomesBy solving the above equation, v2 = 16 V. Therefore, the shortcircuitcurrent isc is v2 25 v2 v2 3 0 5 20 4The Thevenin resistance RTh isFigure 3 shows the Thevenin equivalent circuit for the Figure 1.
Norton theoremNortons theorem for linear electrical networks states thatany collection of voltage sources, current sources,and resistors with two terminals is electrically equivalent to anideal current source, I, in parallel with a single resistor.Any two linear bilateral dc network can be replaced by anequivalent circuit consisting of a current and a parallelresistor.
Calculating the Norton equivalent• The steps leading to the proper values of IN and RN• 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
• 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• IN : – 4. Calculate IN by first returning all the 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. – Conclusion: – 5. Draw the Norton equivalent circuit with the portion of the circuit previously removed replaced between the terminals of the equivalent circuit
ExampleDerive the Norton equivalent circuitSolutionStep 1: Source transformation (The 25V voltage source is converted to a 5 A current source.) 5 4 a 25 V 3A 20 b 4 a 5A 5 20 3A b
Step 2: Combination of parallel source and parallel resistance 4 a 8A 4 bStep 3: Source transformation (combined serial resistance to produce the Thevenin equivalent circuit.) 8 a 32 V b
• Step 4: Source transformation (To produce the Norton equivalent circuit. The current source is 4A (I = V/R = 32 V/8 )) a 4A 8Ω b Norton equivalent circuit.
Maximum power transfer theoremThe maximum power transfer theorem states that, toobtain maximum external power from a source with a finiteinternal resistance, the resistance of the load must be equalto the resistance of the source as viewed from the outputterminals. A load will receive maximum power from a linear bilateral dc network when its total resistive value is exactly equal to the Thévenin resistance of the network as “seen” by the load RL = RTh
Resistance network which contains dependent and independent sources• Maximum power transfer happens when the load resistance RL is equal to the Thevenin equivalent resistance, RTh. To find the maximum power delivered to RL , 2 2 VTh RL VTh pmax = 2 = 2RL 4R L
Application of Network Theorems• Network theorems are useful in simplifying analysis of some circuits. But the more useful aspect of network theorems is the insight it provides into the properties and behaviour of circuits• Network theorem also help in visualizing the response of complex network.• The Superposition Theorem finds use in the study of alternating current (AC) circuits, and semiconductor (amplifier) circuits, where sometimes AC is often mixed (superimposed) with DC