1. SCHOOL OF ENGINEERING Introduction to Electrical and Electronic Engineering Part 2 Pr. Nazim Mir-Nasiri and Pr. Alexander Ruderman

2. Resistive Circuits Series Resistances A series combination of resistance has an equivalent resistance equal to the sum of the original resistances KVL: v=v 1 +v 2 +v 3 = R 1 i+ R 2 i+ R 3 i = (R 1 +R 2 +R 3 )i= R eq i R eq = R 1 +R 2 +R 3

3. A parallel combination of resistance has an equivalent resistance equal to the sum of the reciprocals of original resistances KCL: i=i 1 +i 2 +i 3 = v/R 1 + v/R 2 + v/R 3 = (1/R 1 +1/R 2 +1/R 3 )v= 1/R eq v 1/R eq = 1/R 1 +1/R 2 +1/R 3 Resistive Circuits Parallel Resistances

4. Network analysis is the process of determining the current, voltage, and power for each element given the circuit diagram and the element values. The steps are: 1. Begin by locating a combination of resistances that are in serious or parallel. Often the place to start is farthest from the source 2.Redraw the circuit with the equivalent resistance for the combination found in step 1 3.Repeat steps 1 and 2 until the circuit is reduced as far as possible. Often (but not always) we end up with single source and a single resistance 4.Solve for the current and voltages in the final equivalent circuit. Then, transfer results back one step and solve for additional unknown currents and voltages. Again transfer the results back one step and solve. Repeat until all of the currents and voltages 5. are known in the original circuit. Resistive Circuits Network Analysis by using Series and Parallel Equivalents

5. Resistive Circuits Network Analysis by Using and Parallel Equivalents

6. Resistive Circuits Network Analysis by Using and Parallel Equivalents

7. Resistive Circuits Network Analysis by Using and Parallel Equivalents Transfer results back v 2 =R eq1  i 1 =20   3A = 60V i 2 =v 2 /R 2 = 60V/30  = 2A i 3 =v 2 /R 3 = 60V/60  = 1A v 1 =R 1  i 1 =10   3A = 30V p s =-v s  i 1 =-(90)  3A=-270 W (opposite to the passive configuration) p 1 = R 1  i 1 2 =10  (3A) 2 =90 W p 2 = v 2 2 /R 2 =(60V) 2 /30  = 120 W p 3 = v 2 2 /R 3 =(60V) 2 /60  = 60 W p s + p 1 + p 2 + p 3 =0

8. Resistive Circuits Voltage-Divider Principle of voltage division: R eq = R 1 +R 2 +R 3 i=v total /R eq v 1 =R 1  i = R 1  (v total /R eq ) Of the total voltage, the fraction that appears across a given resistance in a serious circuit is the ratio of the given resistance to the total serious resistance v 3 =R 3  i = R 3  (v total /R eq ) v 2 =R 2  i = R 2  (v total /R eq )

9. Resistive Circuits Current-Divider Principle of current division: 1/R eq = 1/R 1 +1/R 2 R eq = (R 1  R 2 )/(R 1 + R 2 ) v=R eq  i total For two (only) resistances is parallel, the fraction of the total current flowing in a resistance is the ratio of the other resistance to the sum of two resistances. [If more than two in parallel, they must be combined only to two in the circuit.] i 2 =v/R 2 = [R 1 /(R 1 + R 2 )]  i total i 1 =v/R 1 = [R 2 /(R 1 + R 2 )]  i total

10. Resistive Circuits example

11. Resistive Circuits position transducer potentiometer K is sensitivity of the device [v/degree]

12. Resistive Circuits Node-Voltage Analysis (element voltages) It can be applied to any circuit in contrast to previous methods We first select one of the nodes as the reference node, normally one end of a voltage source and mark it as the ground symbol We label the voltages at each of the other nodes The reference polarities for the voltages at the nodes are positive and the polarity is negative at the reference node Write voltage equations around the loops using KVL Solve for the node voltages Find currents and powers for each element in the circuit

13. Resistive Circuits Node-Voltage Analysis (element voltages) Example (positive polarity at the head of the arrow) If V x is unknown using KVL for the loop with this unknown we get -V 2 + V x + V 3 =0 Therefore, the voltage across any element is the V x = V 2 - V 3 difference between node voltages V y = V 2 – V 1 ; V z = V 3 – V 1 Ohm’s law can be used to define currents when voltages are known

14. Resistive Circuits Node-Voltage Analysis (KCL equations) To find the current flowing out of node n through a resistance towards node k, we subtract the voltage at node k from voltage at node n and divide the difference by the resistance between the nodes We apply KCL, adding all of the expressions for the currents leaving node n and setting the sum to zero Repeat the same for all of the nodes in the circuit to obtain the number of equations equal to the number of nodes

15. Resistive Circuits Node-Voltage Analysis (KCL equations) Example

16. Resistive Circuits Node-Voltage Analysis (KCL equations) Example

17. Resistive Circuits Solving the Network Equations For each node equation in difference of voltages the voltage of reference node comes first Once we have written the equations needed for the node voltages we put equations into standard form We group the node-voltage variables on the left-hand side of the equations and place terms that do not involve the node voltages on the right-hand sides Then we can solve for the node voltages by variety methods, such as substitution and determinants

18. Resistive Circuits Solving the Network Equations Example

19. Resistive Circuits Circuits with Voltage Sources For this circuit it is impossible to write a current equation in terms of the node voltages for any single node because a voltage source is connected to each node The circuit requires to form a supernode The supernode is done by drawing a dashed line around several nodes, including the elements (voltage sources) connected between them The modified KCL for supernodes: The net current flowing through any closed surface (enclosed by dash lines) must equal zero.

20. Resistive Circuits Circuits with Voltage Sources The modified KCL for supernodes: The net current flowing through any closed surface (enclosed by dash lines) must equal zero. Note v 3 = -15 V because node 3 connected to the negative terminal of the source Then, for the supernode enclosing the 10-V source, we sum currents leaving the supernode surface through one of the resistors

21. Resistive Circuits Circuits with Voltage Sources Note, we obtain dependent equations if we use all the nodes in writing current equations, i.e. if we would use current equations for both supernodes (they comprise all 4 nodes of the circuit) To avoid dependency we can use KVL (clockwise sense) to the loop that include the voltage source These two equation form an independent set that can be used to solve for v 1 and v 2

22. Resistive Circuits Circuits with Voltage Sources Example For independence must include KVL equation that includes the voltage source To solve for three voltages any two of the following three KCL equations can be used (one supernode and two normal nodes, including the reference node) ( super node 1-2, current source 1A into node 2) (Current source 1A out from ground node)

23. Resistive Circuits Circuits with Dependent Sources Same approach for 1, 2, 3 independent nodes Then use one additional equation for the dependent source current value i x (Currents are into the node 1) (First component of the third equation above)

24. Resistive Circuits Mesh-Current Analysis Applying KVL to the loops with normal branch currents: Applying KCL to the node: Combined equations (3 rd current excluded) :

25. Resistive Circuits Mesh-Current Analysis i 1 and i 2 are mesh currents, normally chosen to flow in clockwise When several mesh currents flow through one element, we consider the current in that element to be algebraic sum of the mesh currents The current in R 3 (referenced downwards) is i 1 - i 2 and v 3 = R 3 (i 1 - i 2 ) If we follow loops (KVL) with mesh currents i 1 and i 2 we get the same equations (i 1 - i 2 is positive in 1 st loop but negative in 2 nd )

26. Resistive Circuits Mesh-Current Analysis Mesh current analysis has more advantage for more complex networks Note how the actual current is referenced in the common circuit element for two mesh currents. For example, if current in R 2 referenced to the right then algebraic sum is i 1 – i 3, if the current in R 2 referenced to the left then the algebraic sum is i 3 -i 1. After solving for mesh currents they may take negative actual values.

27. Resistive Circuits Mesh-Current Analysis with Current Sources Current sources forces a specified current to flow through its terminals, but the voltages across its terminals is not predetermined and it depends on the circuit to which the source is connected Distinct part - first mesh current is equal current source current KVL is applied to mesh 2

28. Resistive Circuits Mesh-Current Analysis with Current Sources Example Distinct part – Supermesh –combination of meshes 1 and 2 KVL is applied to supermesh first Then KVL is applied to mesh 3 Then for the current source : Finally we can define all of the mesh currents from the equations

29. Resistive Circuits Thevenin Equivalent for Two-Terminal Circuits Two-terminal circuit is one (that can be of any complex interconnections of resistances and sources) that has only two points that can be connected to other circuits The Thevenin equivalent to such circuits is one that consists of an only independent Thevenin voltage source in series with a Thevenin resistance

30. Resistive Circuits Thevenin Equivalent for Two-Terminal Circuits The Thevenin equivalent with open-circuited terminals has no current flowing through the circuit, therefore The Thevenin equivalent with short-circuited terminals has resistance value as ratio of open-circuit voltage of the original circuit to its short- circuit current

31. Resistive Circuits Thevenin Equivalent for Two-Terminal Circuits Example i 1 = v OC = = v t = R 2  i 1

32. Resistive Circuits Thevenin Equivalent with a Dependent Source We use node- voltage analysis for the open-circuit voltage v OC = 8.57 V Short-circuit

33. Resistive Circuits Norton Equivalent for Two-Terminal Circuits The Norton equivalent consists of an independent current source I n in parallel with the Thevenin resistance If we zero the Norton current source (disconnecting it), the Norton equivalent becomes a resistance because if we zero the voltage source in the Thevenin equivalent (by short circuiting) it becomes also a resistance that is equivalent to the internal resistance of the original circuit. Can be used as alternative for calculating instead of:

34. Resistive Circuits Norton Equivalent for Two-Terminal Circuits If we place a short circuit across the Norton equivalent, the Norton current becomes equal to the short-circuit current I n = i SC First determine the open-circuit voltage V t =v OC Next determine the short-circuit current I n = i SC Zero the current source and find the Thevenin resistance looking back into the terminals (alternative approach) or Can also be used V t =R t  I n to compute any remaining value

35. Resistive Circuits Maximum Power Transfer Question: What load resistance R L should be connected to a two- terminal circuit in order to maximize the power delivered to the load? The current through the resistor The power delivered Derivative of the power with respect to resistance Solution is Actual maximum power is

36. Resistive Circuits Wheatstone Bridge Wheatstone bridge is a circuit to measure unknown resistances The bridge is capable of responding to a very small currents (less than 1 µA) through a detector resistance R x. Resistors R 2 and R 3 are adjustable and can be tuned until the detector indicates zero current and no voltage between terminals a and b. In this conditions we say the bridge is balanced

37. Resistive Circuits Wheatstone Bridge In balanced condition (i g =0) by applying KCL at nodes a and b respectively KVL applied around the loop R 1, R 2 and detector and since v ab =0 Similarly KVL around R 3, R 4 and detector Dividing one equation by the other we obtain the resistors ratio for the balanced bridge

38. Resistive Circuits Superposition Principles Suppose we have a circuit composed resistors, linear dependent (which follow the linear function) sources and n independent sources The current flowing in each element is a response to the independent sources Consider zeroing all the independent sources (current sources become open circuits and voltage sources become short-circuited) except the first The response ( while sources are zeroed) for that source is then r 1 (could be either current or voltage) If we keep only the second source the response becomes r 2 We can repeat the process for each source in the circuit until obtain response to the n-th source r n The superposition principle states that the total response is the sum of the responses to each of the independent sources acting individually

39. Resistive Circuits Superposition Principles Example 1.Only voltage source active (apply voltage division principle 2. Only current source active (resisters in parallel) 3. Voltage across due to the current source 4. Adding the individual responses