* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Spark-gap transmitter wikipedia, lookup

Immunity-aware programming wikipedia, lookup

Integrated circuit wikipedia, lookup

Regenerative circuit wikipedia, lookup

Negative resistance wikipedia, lookup

Transistor–transistor logic wikipedia, lookup

Integrating ADC wikipedia, lookup

Josephson voltage standard wikipedia, lookup

Valve RF amplifier wikipedia, lookup

Operational amplifier wikipedia, lookup

Electrical ballast wikipedia, lookup

Two-port network wikipedia, lookup

Power electronics wikipedia, lookup

Schmitt trigger wikipedia, lookup

Voltage regulator wikipedia, lookup

Power MOSFET wikipedia, lookup

Switched-mode power supply wikipedia, lookup

Current source wikipedia, lookup

Resistive opto-isolator wikipedia, lookup

Surge protector wikipedia, lookup

Opto-isolator wikipedia, lookup

Current mirror wikipedia, lookup

Transcript

07-Dec-11 Summary Series circuits All circuits have three common attributes. These are: Basics of electrical systems R1 1. A source of voltage. 2. A load. R2 VS + 3. A complete path. R3 Summary Summary Series circuits Series circuit rule for current: A series circuit is one that has only one current path. R1 VS Because there is only one path, the current everywhere is the same. For example, the reading on the first ammeter is 2.0 mA, What do the other meters read? R1 R2 VS R1 R2 R3 VS R3 R3 + 2.0 mA _ R2 R1 + 2.0 mA _ R2 VS _ _ 2.0 mA + Summary 2.0 mA + R1 Summary Series circuits VS 12 V Series circuit 680 Ω R2 1.5 kΩ R3 The total resistance of resistors in series is the sum of the individual resistors. For example, the resistors in a series circuit are 680 Ω, 1.5 kΩ, and 2.2 kΩ. What is the total resistance? R1 VS 12 V 680 Ω R3 2.2 kΩ R2 1.5 kΩ 4.38 kΩ 2.2 kΩ Tabulating current, resistance, voltage and power is a useful way to summarize parameters in a series circuit. Continuing with the previous example, complete the parameters listed in the Table. I1= 2.74 mA I2= 2.74 mA I3= 2.74 mA IT= 2.74 mA R1= 0.68 kΩ R2= 1.50 kΩ R3= 2.20 kΩ RT= 4.38 kΩ V1= 1.86 V V2= 4.11 V V3= 6.03 V VS= 12 V P1= P2= P3= PT= 5.1 mW 11.3 mW 16.5 mW 32.9 mW 1 07-Dec-11 Summary Summary Voltage sources in series Kirchhoff’s voltage law Voltage sources in series add algebraically. For example, the total voltage of the sources shown is 27 V Kirchhoff’s voltage law (KVL) is generally stated as: + 9V The sum of all the voltage drops around a single closed path in a circuit is equal to the total source voltage in that closed path. + 9V + What is the total voltage if one battery is accidentally reversed? 9 V KVL applies to all circuits, but you must apply it to only one closed path. In a series circuit, this is (of course) the entire circuit. + 9V n A mathematical shorthand way of writing KVL is ∑V i =1 R1 Summary Kirchhoff’s voltage law =0 Summary 680 Ω VS 12 V i R2 1.5 kΩ Voltage divider rule R3 2.2 kΩ The voltage drop across any given resistor in a series circuit is equal to the ratio of that resistor to the total resistance, multiplied by source voltage. Notice in the series example given earlier that the sum of the resistor voltages is equal to the source voltage. I1= 2.74 mA I2= 2.74 mA I3= 2.74 mA IT= 2.74 mA R1= 0.68 kΩ R2= 1.50 kΩ R3= 2.20 kΩ RT= 4.38 kΩ V1= 1.86 V V2= 4.11 V V3= 6.03 V VS= 12 V P1= P2= P3= PT= 5.1 mW 11.3 mW 16.5 mW 32.9 mW VS Assume R1 is twice the size of R2. What is the voltage across R1? 8 V Summary R1 12 V R2 Summary R1 Voltage divider Voltage divider 15 kΩ VS + 20 V R2 10 kΩ What is the voltage across R2? The total resistance is 25 kΩ. Applying the voltage divider formula: R 10 kΩ V2 = 2 VS = 20 V = 8.0 V 25 kΩ RT Notice that 40% of the source voltage is across R2, which represents 40% of the total resistance. Voltage dividers can be set up for a variable output using a potentiometer. In the circuit shown, the output voltage is variable. VS + 15 V What is the largest output voltage available? 5.0 V R1 20 kΩ R2 10 kΩ VOUT 2 07-Dec-11 Summary Summary A Power in Series Circuits Voltage measurements R1 470 Ω Use the voltage divider rule to find V1 and V2. Then find the power in R1 and R2 and PT. Applying the voltage divider rule: 470 Ω V1 = 20 V = 11.75 V 800 Ω 330 Ω V2 = 20 V = 8.25 V 800 Ω R2 330 Ω VS + 20 V The power dissipated by each resistor is: P1 = (11.75 V ) P2 = 2 470 Ω 2 (8.25 V ) 330 Ω = 0.29 W } P0.5=W Voltage is relative and is measured with respect to another point in the circuit. B R2 10 kΩ C What are VA, VB, and VAB for the circuit shown? VA = 12 V VB = 8 V VAB = 4 V Summary Summary Voltage measurements Ground reference is not always at the lowest point in a circuit. Assume the ground is moved to B as shown. VS + 12 V Voltages that are given with respect to ground are shown with a single subscript. For example, VA means the voltage at point A with respect to ground (called reference ground). VB means the voltage at point B with respect to ground. VAB means the voltage between points A and B. T = 0.21 W R1 5.0 kΩ A VS + 12 V Voltage measurements R1 5.0 kΩ B R2 10 kΩ C What are VA, VB, and VC for the circuit? VA = 4 V VB = 0 V VC = −8 V Has VAB changed from the previous circuit? No, it is still 4 V Assume that R2 is open. For this case, what are VA, VB, and VC for the circuit? A VS + 12 V R1 5.0 kΩ B R2 10 kΩ C If R2 is open, there is no current. Notice that VB = 0 V because it is ground and VA = 0 V because it has the same potential as VB. VC = −12 V because the source voltage is across the open. Selected Key Terms Selected Key Terms Series In an electric circuit, a relationship of components in which the components are connected such that they provide a single path between two points. Reference ground The metal chassis that houses the assembly or a large conductive area on a printed circuit board is used as a common or reference point; also called common. Kirchhoff’s A law stating that (1) the sum of the voltage voltage law drops around a closed loop equals the source voltage in that loop or (2) the algebraic sum of all of the voltages (drops and source) is zero. Open A circuit condition in which the current path is broken. Short A circuit condition in which there is zero or an abnormally low resistance between two points; usually an inadvertent condition. Voltage divider A circuit consisting of series resistors across which one or more output voltages are taken. 3 07-Dec-11 Quiz 1. In a series circuit with more than one resistor, the current is Quiz 2. In a series circuit with more than one resistor, the voltage is a. larger in larger resistors a. larger across larger resistors b. smaller in larger resistors b. smaller across larger resistors c. always the same in all resistors c. always the same across all resistors d. there is not enough information to say d. there is not enough information to say Quiz 3. If three equal resistors are in series, the total resistance is a. one third the value of one resistor b. the same as one resistor c. three times the value of one resistor Quiz 4. A series circuit cannot have a. more than two resistors b. more than one voltage source c. more than one path d. all of the above d. there is not enough information to say Quiz 5. In a closed loop, the algebraic sum of all voltages (both sources and drops) a. is zero Quiz 6. The current in the 10 kΩ resistor is a. 0.5 mA b. 2.0 mA b. is equal to the smallest voltage in the loop c. 2.4 mA c. is equal to the largest voltage in the loop d. 10 mA VS + 24 V R1 10 kΩ R2 2.0 kΩ d. depends on the source voltage 4 07-Dec-11 Quiz Quiz 8. The smallest output voltage available from the voltage divider is 7. The output voltage from the voltage divider is a. 2.0 V a. 0 V b. 4.0 V VS + 24 V c. 12 V R1 10 kΩ R2 2.0 kΩ d. 20 V b. 1.5 V VOUT R1 10 k Ω VS + 15 V c. 5.0 V R2 10 kΩ d. 7.5 V Quiz 9. The total power dissipated in a series circuit is equal to the a. power in the largest resistor b. power in the smallest resistor c. average of the power in all resistors VOUT Quiz 10. The meaning of the voltage VAB is the voltage at a. Point A with respect to ground b. Point B with respect to ground c. The average voltage between points A and B. d. The voltage difference between points A and B. d. sum of the power in all resistors Quiz Summary Resistors in parallel Answers: 1. c 6. b 2. a 7. b 3. c 8. a 4. c 9. d 5. a 10. d Resistors that are connected to the same two points are said to be in parallel. A R1 R2 R3 R4 B 5 07-Dec-11 Summary Summary Parallel circuits Parallel circuit rule for voltage Because all components are connected across the same voltage source, the voltage across each is the same. A parallel circuit is identified by the fact that it has more than one current path (branch) connected to a common voltage source. VS For example, the source voltage is 5.0 V. What will a volt- meter read if it is placed across each of the resistors? +5.0 V + + R1 R2 R3 +5.0 V + VS +5.0 V Summary Parallel circuit rule for resistance R1 680 Ω VS + R2 1.5 kΩ R3 2.2 kΩ R2 1.5 kΩ R3 2.2 kΩ R1 R2 The resistance of two parallel resistors can be found by either: RT = 1 1 1 + R1 R2 or RT = R1 R2 R1 + R2 What is the total resistance if R1 = 27 kΩ and R2 = 56 kΩ? 18.2 kΩ Summary Parallel circuit R1 680 Ω Special case for resistance of two parallel resistors For example, the resistors in a parallel circuit are 680 Ω, 1.5 kΩ, and 2.2 kΩ. What is the total resistance? 386 Ω + +5.0 V + Summary The total resistance of resistors in parallel is the reciprocal of the sum of the reciprocals of the individual resistors. VS +5.0 V + R4 Summary R1 680 Ω R2 1.5 kΩ Tabulating current, resistance, voltage and power is a useful way to summarize parameters in a parallel circuit. Continuing with the previous example, complete the parameters listed in the Table. R3 2.2 kΩ Kirchhoff’s current law Kirchhoff’s current law (KCL) is generally stated as: The sum of the currents entering a node is equal to the sum of the currents leaving the node. Notice in the previous example that the current from the source is equal to the sum of the branch currents. I1= 7.4 mA R1= 0.68 kΩ V1= 5.0 V P1= 36.8 mW I1= 7.4 mA R1= 0.68 kΩ V1= 5.0 V P1= 36.8 mW I2= 3.3 mA R2= 1.50 kΩ V2= 5.0 V P2= 16.7 mW I2= 3.3 mA R2= 1.50 kΩ V2= 5.0 V P2= 16.7 mW I3= 2.3 mA R3= 2.20 kΩ V3= 5.0 V P3= 11.4 mW I3= 2.3 mA R3= 2.20 kΩ V3= 5.0 V P3= 11.4 mW IT= 13.0 mA RT= 386 Ω VS= 5.0 V PT= 64.8 mW IT= 13.0 mA RT= 386 Ω VS= 5.0 V PT= 64.8 mW 6 07-Dec-11 Summary Current divider Current divider When current enters a node (junction) it divides into currents with values that are inversely proportional to the resistance values. The most widely used formula for the current divider is the tworesistor equation. For resistors R1 and R2, R2 I1 = I T and R1 + R2 R1 I2 = IT R1 + R2 Notice the subscripts. The resistor in the numerator is not the same as the one for which current is found. Summary Power in parallel circuits Power in each resistor can be calculated with any of the standard power formulas. Most of the time, the voltage is known, so the equation Summary P= V 2 is most convenient. R As in the series case, the total power is the sum of the powers dissipated in each resistor. What is the total power if 10 V is applied to the parallel combination of R1 = 270 Ω and R2 = 150 Ω? 1.04 W Key Terms Assume that R1is a 2.2 kΩ resistor that is in parallel with R2, which is 4.7 kΩ. If the total current into the resistors is 8.0 mA, what is the current in each resistor? R2 4.7 kΩ 5.45 mA I1 = IT = 8.0 mA = 6.9 kΩ R1 + R2 R1 2.2 kΩ 2.55 mA I2 = IT = 8.0 mA = 6.9 kΩ R1 + R2 Notice that the larger resistor has the smaller current. Summary Assume there are 8 resistive wires that form a rear window defroster for an automobile. (a) If the defroster dissipates 90 W when connected to a 12.6 V source, what power is dissipated by each resistive wire? (b) What is the total resistance of the defroster? (a) Each of the 8 wires will dissipate 1/8 of the total power or 90 W = 11.25 W 8 wires 2 (12.6 V ) = 1.76 Ω V2 (b) The total resistance is R = = P 90 W What is the resistance of each wire? 1.76 Ω x 8 = 14.1 Ω Quiz Parallel The relationship in electric circuits in which two or more current paths are connected between two separate points (nodes). Branch One current path in a parallel circuit. Kirchhoff’s A law stating the total current into a node equals the current law total current out of the node. Node A point or junction in a circuit at which two or more 1. The total resistance of parallel resistors is equal to a. the sum of the resistances b. the sum of the reciprocals of the resistances c. the sum of the conductances d. none of the above components are connected. Current divider A parallel circuit in which the currents divide inversely proportional to the parallel branch resistances. 7 07-Dec-11 Quiz 2. The number of nodes in a parallel circuit is Quiz 3. The total resistance of the parallel resistors is a. one a. 2.52 kΩ b. two b. 3.35 kΩ c. three c. 5.1 kΩ d. any number d. 25.1 kΩ Quiz 4. If three equal resistors are in parallel, the total resistance is R1 10 kΩ R2 10 kΩ Quiz 5. In any circuit the total current entering a node is a. one third the value of one resistor. a. less than the total current leaving the node. b. the same as one resistor. b. equal to the total current leaving the node. c. three times the value of one resistor. c. greater than the total current leaving the node. d. the product of the three resistors d. can be any of the above, depending on the circuit. Quiz 6. The current divider formula to find I1 for the special case of two resistors is Quiz 7. The total current leaving the source is R1 IT RT a. 1.0 mA R b. I1 = 2 I T RT c. 6.0 mA a. I1 = R3 5.1 kΩ b. 1.2 mA VS + 12 V R1 10 kΩ R2 2.0 kΩ d. 7.2 mA c. I1 = R2 IT R1 + R2 R d. I1 = 1 I T R1 + R2 8 07-Dec-11 Quiz Quiz 9. The voltage across R2 is 8. The current in R1 is a. 0 V a. 6.7 mA b. 0.67 V b. 13.3 mA c. 20 mA I = 20 mA R1 100 Ω R2 200 Ω c. 1.33 V I = 20 mA R1 100 Ω R2 200 Ω d. 4.0 V d. 26.7 mA Quiz Quiz 10. The total power dissipated in a parallel circuit is equal to the Answers: a. power in the largest resistor. 1. d 6. c b. power in the smallest resistor. 2. b 7. d c. average of the power in all resistors. 3. a 8. b d. sum of the power in all resistors. 4. a 9. c 5. b 10. d Summary Summary Identifying series-parallel relationships Combination circuits Most practical circuits have combinations of series and parallel components. Most practical circuits have various combinations of series and parallel components. You can frequently simplify analysis by combining series and parallel components. An important analysis method is to form an equivalent circuit. An equivalent circuit is one that has characteristics that are electrically the same as another circuit but is generally simpler. Components that are connected in series will share a common path. Components that are connected in parallel will be connected across the same two nodes. 1 2 9 07-Dec-11 Summary Summary Equivalent circuits Equivalent circuits For example: Another example: is equivalent to R1 1.0 kΩ R1 is equivalent to 2.0 kΩ R2 1.0 kΩ R1 R2 R1,2 500 Ω 1.0 kΩ 1.0 kΩ There are no electrical measurements that can distinguish the boxes. There are no electrical measurements that can distinguish the boxes. Summary Summary is equivalent to R1 1.0 kΩ R3 R 1,2 R3 R2 4.7 kΩ 3.7 kΩ 4.7 kΩ 2.7 kΩ is equivalent to Kirchhoff’s voltage law and Kirchhoff’s current law can be applied to any circuit, including combination circuits. For example, applying KVL, the path shown will have a sum of 0 V. 2.07 kΩ Start/Finish Summary I I − 26.5 mA + A I VS 5.0 V + + 8.0 mA − − R1 270 Ω 100 Ω R3 330 Ω R6 100 Ω R4 100 Ω 100 Ω Start/Finish Combination circuits R5 100 Ω VS + 10 V R1 270 Ω R2 330 Ω R3 470 Ω Tabulating current, resistance, voltage and power is a useful way to summarize parameters. Solve for the unknown quantities in the circuit shown. R2 470 Ω R4 18.5 mA R11 270 Ω Summary Kirchoff’s current law can also be applied to the same circuit. What are the readings for node A? + VS 5.0 V So will this path! R3 330 Ω R6 There are no electrical measurements that can distinguish between the three boxes. R 1,2,3 R2 470 Ω R5 100 Ω I1= 21.6 mA I2= 12.7 mA I3= 8.9 mA IT= 21.6 mA R1= 270 Ω R2= 330 Ω R3= 470 Ω RT= 464 Ω V1= 5.82 V P1= 126 mW V2= 4.18 V P2= 53.1 mW V3= 4.18 V P3= 37.2 mW VS= 10 V PT= 216 mW 10 07-Dec-11 Summary Loaded voltage divider R2 330 Ω R3 470 Ω Notice that the current in R1 is equal to the sum of the branch currents in R2 and R3. The sum of the voltages around the outside loop is zero. R1= 270 Ω R2= 330 Ω R3= 470 Ω RT= 464 Ω V1= 5.82 V P1= 126 mW V2= 4.18 V P2= 53.1 mW V3= 4.18 V P3= 37.2 mW VS= 10 V PT= 216 mW VS = +15 V R1 330 Ω Stiff voltage divider A R3 2.2 kΩ Form an equivalent series circuit by combining R2 and R3; then apply the voltage-divider formula to the equivalent circuit: R2,3 = R2 R3 = 470 Ω 2.2 kΩ = 387 Ω R2,3 387 Ω V3 = V2,3 = V = 15 V = 8.10 V R + R S 330 Ω + 387 Ω 2,3 1 Summary Loading effect of a voltmeter VS + 10 V R3 Summary R2 470 Ω What is the voltage across R3? A R2 A voltage-divider with a resistive load is a combinational circuit and the voltage divider is said to be loaded. The loading reduces the total resistance from node A to ground. Summary Loaded voltage divider R1 + The voltage-divider equation was developed for a series circuit. Recall that the output voltage is given by R V2 = 2 VS RT R1 VS A stiff voltage-divider is one in R2 RL which the loaded voltage nearly the same as the no-load voltage. To accomplish this, the load current must be small compared to the bleeder current (or RL is large compared to the divider resistors). If R1 = R2 = 1.0 kΩ, what value of RL will make the divider a stiff voltage divider? What fraction of the unloaded voltage is the loaded voltage? RL > 10 R2; RL should be 10 kΩ or greater. For a 10 kΩ load, R2 || RL 0.91 kΩ This is 95% of the VL = VS = VS = ( 0.476 ) VS 1.0 kΩ + 0.91 kΩ unloaded voltage. R1 + R2 || RL Summary R1 470 kΩ + Wheatstone bridge 4.04 10 VV R2 Assume VS = 10 V, but the + 4.04 V 470 kΩ meter reads only 4.04 V when it is across either R1 or R2. Can you explain what is happening? All measurements affect the quantity being measured. A voltmeter has internal resistance, which can change the resistance of the circuit under test. In this case, a 1 MΩ internal resistance of the meter accounts for the readings. The Wheatstone bridge consists R3 R1 of a dc voltage source and four VS Output resistive arms forming two voltage dividers. The output is R2 R4 taken between the dividers. Frequently, one of the bridge resistors is adjustable. When the bridge is balanced, the output voltage is zero, and the products of resistances in the opposite diagonal arms are equal. - I1= 21.6 mA I2= 12.7 mA I3= 8.9 mA IT= 21.6 mA Summary + Kirchhoff’s laws can be applied as a check on the answer. R1 270 Ω VS + 10 V 11 07-Dec-11 Summary Summary Wheatstone bridge VS 12 V + R1 470 Ω R3 330 Ω Output - Example: What is the value of R2 if the bridge is balanced? 384 Ω Thevenin’s theorem R2 R4 270 Ω Thevenin’s theorem states that any two-terminal, resistive circuit can be replaced with a simple equivalent circuit when viewed from two output terminals. The equivalent circuit is: RTH VTH Summary Summary Thevenin’s theorem Thevenin’s theorem VTH is defined as the open circuit voltage between the two output terminals of a circuit. RTH is defined as the total resistance appearing between the two output terminals when all sources have been replaced by their internal resistances. What is the Thevenin voltage for the circuit? 8.76 V RTH What is the Thevenin resistance for the circuit? 7.30 kΩ Output terminals R1 VS 12 V R2 27 kΩ VTH Summary Thevenin’s theorem Thevenin’s theorem is useful for solving the Wheatstone bridge. One way to Thevenize the bridge is to create two Thevenin circuits − from A to ground and from B to ground. The resistance between point R1 R2 A and ground is R1||R3 and the VS + RL A B resistance from B to ground is R2||R4. The voltage on each R3 R4 side of the bridge is found using the voltage divider rule. Remember, the load resistor has no affect on the Thevenin parameters. 10 kΩ RL 68 kΩ Summary Thevenin’s theorem For the bridge shown, R1||R3 = 165 Ω and R2||R4 = 179 Ω. The voltage from A to ground (with no load) is 7.5 V and from B to ground (with no load) is 6.87 V . VS +15 V + R1 330 Ω A R3 330 Ω RL 150 Ω R2 390 Ω B R4 330 Ω The Thevenin circuits for each of the bridge are shown on the following slide. 12 07-Dec-11 Summary Summary Thevenin’s theorem Maximum power transfer VTH 7.5 V 165 Ω The maximum power is transferred from a source to a load when the load resistance is equal to the internal source resistance. B RTH' RTH A RL 150 Ω VTH' 6.87 V 179 Ω RS VS + RL Putting the load on the Thevenin circuits and applying the superposition theorem allows you to calculate the load current. The load current is: 1.27 mA Summary Summary Maximum power transfer Superposition theorem Summary + - VS1 12 V 6.8 kkΩ kΩ Ω 6.8 VS2S2 -++1.56 mA 18 V RR222 6.8 kkΩ Ω 6.8 Ω II22 - Set up a table of pertinent information and solve for each quantity listed: VS1 12 V 6.8 kΩ I2 + - VS2 18 V R2 6.8 kΩ Troubleshooting RR333 RR111 2.7 kkΩ Ω 2.7 Ω 2.7 kΩ Summary ++ What does the ammeter read for I2? What does the ammeter read for I2? (See next slide for the method and the answer). - V 2 ( 5.0 V ) = = 0.5 W RL 50 Ω 2 PL = R3 R1 RL 50 Ω + VS + 10 V - RS 50 Ω The superposition theorem is a way to determine currents and voltages in a linear circuit that has multiple sources by taking one source at a time and algebraically summing the results. + What is the power delivered to the matching load? The voltage to the load is 5.0 V. The power delivered is The maximum power transfer theorem assumes the source voltage and resistance are fixed. Source 1: RT(S1)= 6.10 kΩ I1= 1.97 mA I2= 0.98 mA Source 2: RT(S2)= 8.73 kΩ I3= 2.06 mA I2= 0.58 mA Both sources I2= 1.56 mA The total current is the algebraic sum. The effective troubleshooter must think logically about circuit operation. Understand normal circuit operation and find out the symptoms of the failure. Decide on a logical set of steps to find the fault. Following the steps in the plan, make measurements to isolate the problem. Modify the plan if necessary. 13 07-Dec-11 Troubleshooting Selected Key Terms R1 330 Ω A R2 470 Ω R3 The output of the voltage2.2 kΩ divider is 6.0 V. Describe how you would use analysis and planning in finding the fault. From an earlier calculation, V3 should equal 8.10 V. A low voltage is most likely caused by a low source voltage or incorrect resistors (possibly R1 and R2 reversed). If the circuit is new, incorrect components are possible. Decide on a logical set of steps to locate the fault. You could decide to 1) check the source voltage, 2) disconnect the load and check the output voltage, and if it is correct, 3) check the load resistance. If R3 is correct, check other resistors. Loading The effect on a circuit when an element that draws current from the circuit is connected across the output terminals. Load current The output current supplied to a load. Bleeder The current left after the load current is current subtracted from the total current into the circuit. Wheatstone A 4-legged type of bridge circuit with which an bridge unknown resistance can be accurately measured using the balanced state. Deviations in resistance can be measured using the unbalanced state. Selected Key Terms Thevenin’s A circuit theorem that provides for reducing theorem any two-terminal resistive circuit to a single equivalent voltage source in series with an equivalent resistance. Maximum power The condition, when the load resistance transfer equals the source resistance, under which maximum power is transferred to the load. Superposition A method for analyzing circuits with two or more sources by examining the effects of each source by itself and then combining the effects. Quiz 1. Two circuits that are equivalent have the same a. number of components b. response to an electrical stimulus c. internal power dissipation d. all of the above Quiz 2. If a series equivalent circuit is drawn for a complex circuit, the equivalent circuit can be analyzed with Quiz 3. For the circuit shown, a. R1 is in series with R2 a. the voltage divider theorem b. R1 is in parallel with R2 b. Kirchhoff’s voltage law c. R2 is in series with R3 c. both of the above d. R2 is in parallel with R3 R1 VS R2 R3 - VS = +15 V + Summary d. none of the above 14 07-Dec-11 Quiz Quiz 4. For the circuit shown, 5. A signal generator has an output voltage of 2.0 V with no load. When a 600 Ω load is connected to it, the output drops to 1.0 V. The Thevenin resistance of the generator is R4 a. R1 is in series with R2 VS + c. R2 is in parallel with R3 R2 R1 b. R4 is in parallel with R1 - d. none of the above a. 300 Ω R3 b. 600 Ω c. 900 Ω d. 1200 Ω. Quiz Quiz 6. For the circuit shown, Kirchhoff's voltage law 7. The effect of changing a measured quantity due to connecting an instrument to a circuit is called a. applies only to the outside loop b. applies only to the A junction. a. loading c. can be applied to any closed path. b. clipping d. does not apply. VS + 10 V A R1 270 Ω c. distortion d. loss of precision R2 330 Ω R3 470 Ω Quiz Quiz 8. An unbalanced Wheatstone bridge has the voltages shown. The voltage across R4 is a. 4.0 V + - d. 7.0 V R1 7.0 V R2 R3 + RL 1.0 V b. 5.0 V c. 6.0 V R4 d. 7.0 V VS 12 V R3 R1 + RL - - c. 6.0 V VS 12 V + a. 4.0 V b. 5.0 V 9. Assume R2 is adjusted until the Wheatstone bridge is balanced. At this point, the voltage across R4 is measured and found to be 5.0 V. The voltage across R1 will be R2 R4 5.0 V 15 07-Dec-11 Quiz 10. Maximum power is transferred from a fixed source when Quiz Answers: 1. b 6. c a. the load resistor is ½ the source resistance 2. c 7. a b. the load resistor is equal to the source resistance 3. d 8. a c. the load resistor is twice the source resistance 4. d 9. d d. none of the above 5. b 10. b 16