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electronics fundamentals circuits, devices, and applications THOMAS L. FLOYD DAVID M. BUCHLA Chapter 6 – Series and Parallel Combination Circuits Series and Parallel Circuits Identifying series-parallel relationships Most practical circuits have combinations of series and parallel components. From Chapters 4 and 5 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 Series and Parallel Circuits Combination circuits Circuits containing both series and parallel circuits are called COMBINATION circuits You can frequently simplify analysis by combining series and parallel components. Solve by forming the simplest equivalent circuit possible. An equivalent circuit is one that has: 1. characteristics that are electrically the same as another circuit but 2. is generally simpler. Series and Parallel Circuits Equivalent circuits R1 1.0 k R2 is equivalent to R1 2.0 k 1.0 k There are no electrical measurements that can distinguish the boxes. Series and Parallel Circuits Equivalent circuits Another example: is equivalent to R1 R2 1.0 k 1.0 k R1,2 500 There are no electrical measurements that can distinguish the boxes. Series and Parallel Circuits Equivalent circuits is equivalent to R1 1.0 k R3 R 1,2 R3 R2 4.7 k 3.7 k 4.7 k 2.7 k is equivalent to R1,2,3 2.07 k There are no electrical measurements that can distinguish between the three boxes. Series and Parallel Circuits What Do We Know 1. For Series Circuits: a. Current at all points is the same IT = IR1 = IR2 b. Voltage drops across each resistor VT = V 1 + V2 2. For Parallel Circuit a. Current divides across each resistor in the branch IT = IR1 + IR2 b. Voltage at all nodes is the same VT = V1 = V2 Series and Parallel Circuits Seven Step Process for Solving a Combination Circuit 1. Simplify the circuit to a series circuit by finding the effective equivalent resistance (REQ) of each parallel section in the circuit. Redraw the simplified circuit. 2. Calculate the total resistance (RT) of the circuit by adding all REQ’s to the other series resistances. 3. Calculate the total current (IT) using RT in Ohm’s law. 4. Calculate the voltage drop across any series resistances or REQ’s using Ohm’s law. 5. Calculate the branch currents in all parallel sections of the circuit using the voltage drop across REQ and Ohm’s law. 6. Use the branch currents and resistance values to calculate the voltage of the parallel resistances. 7. Make a summary of the voltage drops and currents for each resistance to make sure they total correctly. Series and Parallel Circuits A simple series-parallel resistive circuit. Series and Parallel Circuits R4 is added to the circuit in series with R1. Series and Parallel Circuits R5 is added to the circuit in series with R2. Series and Parallel Circuits R6 is added to the circuit in parallel with the series combination of R1 and R4. Series and Parallel Circuits Sketch the circuits for nodes A to B A to C B to C Series and Parallel Circuits Sketch the circuits for nodes A to B RT R1 ( R2 R3) Series and Parallel Circuits Sketch the circuits for nodes A to C RT R3 ( R1 R2) Series and Parallel Circuits Sketch the circuits for nodes B to C RT R2 ( R1 R3) Series and Parallel Circuits R ( AB ) ( R5 ( R 6 R 3)) R 2 ( R1 R 4) R 7 Series and Parallel Circuits Reduce the following circuit to REQ Series and Parallel Circuits REQ( R3 R4) 6Ω Series and Parallel Circuits REQ( R3 R4) R6 10 Ω Series and Parallel Circuits ( REQ( R3 R4) R6) R5 5 Ω Series and Parallel Circuits IT (( REQ( R3 R4) R6) R5) R1 R2 IT = 2 amps 50 Ω Series and Parallel Circuits Review of voltage relationships. Series and Parallel Circuits Kirchoff’s current law What are the readings for node A? I + I - 26.5 mA + A I VS 5.0 V + + 8.0 mA R2 470 R4 - 18.5 mA 100 R1 270 R3 330 R6 100 R5 100 Series and Parallel Circuits The voltage-divider equation was developed for a series circuit. Recall that the output voltage is given by Loaded voltage divider + R1 A R2 R3 R2 V2 VS RT 1. A voltage-divider with a resistive load forms a combination (parallel) circuit. 2. The voltage divider is said to be LOADED. 3. The loading reduces the total resistance from node A to ground. Series and Parallel Circuits Loaded voltage divider What is the voltage across R3? VS = +15 V Form an equivalent series circuit by combining R2 and R3; then apply the voltagedivider formula to the equivalent circuit: R1 330 R2 470 A R3 2.2 k R2,3 R2 R3 470 2.2 k = 387 V3 V2,3 R2,3 387 VS 15 V 8.10 V R R 330 387 2,3 1 Series and Parallel Circuits The loading effect of a voltmeter. Series and Parallel Circuits Given: VS = 10 V and R1 and R2 are not defective but the meter reads only 4.04 V when it is across either R1 or R2. What is a possible explanation of the meter not displaying 10 volts? Loading effect of a voltmeter VS + 10 V R1 470 k + R2 470 k + 4.04 10 VV 4.04 V 1. A voltmeter has internal resistance 2. This RINT can change the resistance of the circuit under test. 3. A 1 M internal resistance of the meter accounts for the readings. . VS R3 R1 Output - • The Wheatstone bridge consists of: 1. a dc voltage source and 2. four resistive arms forming two voltage dividers. • The output is taken between the dividers. • Frequently, one of the bridge resistors is adjustable. (R2) Wheatstone bridge + Series and Parallel Circuits R2 R4 Series and Parallel Circuits Balanced Wheatstone bridge When the R bridge is V Output balanced, the R output voltage is zero The products of resistances in the opposite diagonal arms are equal. 1 + - S R3 2 R4 Series and Parallel Circuits Wheatstone bridge. Voltage Divider 1 Voltage Divider 2 Series and Parallel Circuits R2 R1 R 3 R4 Wheatstone bridge. V1=V2 I1=I3 V3=V4 I2=I4 Series and Parallel Circuits 470 330 Output - R2 R1 R 3 R4 12 V VS R3 R1 + What is the value of R2 if the bridge is balanced? 384 Wheatstone bridge R2 R4 270 Series and Parallel Circuits Finding an Unknown Resistance R2 RX RV R4 Scale Factor Series and Parallel Circuits Unbalanced Wheatstone Bridge • Unbalance occurs when VOUT ≠ 0 • Used to measure 1. Mechanical Strain 2. Temperature 3. Pressure • VOUT is converted to a digital output indicating the value of the reading. Measuring a physical parameter using a transducer. Series and Parallel Circuits VOUT = VA-VB = 0 RX VX VS RT Series and Parallel Circuits 10 10 10 10 10 Using the Thermistor chart, what is VOUT when the temperature is 50o C VA=8.8v VB=6.0v VA-B=2.8v Series and Parallel Circuits Example of a load cell. Series and Parallel Circuits Wheatstone Bridge 1. Remove RL to make an open circuit between A & B 2. Calculate R1||R3 = 165 3. Calculate R2||R4 = 179 4. Calculate the voltage from A to ground 7.5 V 5. Calculate the voltage from B to ground 6.87 V VS +15 V + - R1 R2 330 390 RL A B 150 R3 R4 330 330 Series and Parallel Circuits Maximum power transfer Theorem The maximum power is transferred from a source to a load when: RL = RS(resistance of the voltage source) RS VS + RL The maximum power transfer theorem assumes the source voltage and resistance are fixed. Series and Parallel Circuits Maximum power transfer Theorem What is the power delivered to the load? RS VS + 10 V The voltage to the load is 5.0 V The power delivered is: 5.0 V V PL = 0.5 W RL 50 2 2 50 RL 50 Series and Parallel Circuits Superposition theorem A way to determine currents and voltages in a linear circuit that has multiple sources 1. Take one source at a time and 2. Algebraically summing the results. R3 R1 + - VS2 18 V R2 6.8 k - + - VS1 12 V 6.8 k I2 + 2.7 k Series and Parallel Circuits Superposition theorem Four Step Process 1. Leave one voltage or current source at a time in the circuit (circuit 1) and replace the other (circuit 2) with a short. 2. Calculate REQ/Total and then calculate the voltage or current for the resistor(s). 3. Repeat steps 1 and 2 for the other circuit (circuit 2). 4. Algebraically add the results for all sources. Series and Parallel Circuits Series and Parallel Circuits Summary - VS1 12 V 6.8 kk k 6.8 VS2S2 -++1.56 mA 18 V RR222 6.8 kk k 6.8 II222 - Source 1: Source 2: 2.7 2.7 kkk + Set up a table of pertinent information and solve for each quantity listed: RR333 RR111 ++ What does the ammeter read for I2? RT(S1)= 6.10 k I1= 1.97 mA I2= 0.98 mA RT(S2)= 8.73 k I3= 2.06 mA I2= 0.58 mA Both sources The total current is the algebraic sum. I2= 1.56 mA Series and Parallel Circuits Thevenin’s theorem and Wheatstone Bridge RTH A RL VTH 7.5 V 165 150 B RTH' 179 VTH' 6.87 V 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 .0152-.0191 Series and Parallel Circuits Troubleshooting 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. Series and Parallel Circuits The output of the voltagedivider is 6.0 V. Describe how you would use analysis and planning in finding the fault. Troubleshooting VS = +15 V R1 330 R2 470 A R3 2.2 k Decide From an onearlier a logical calculation, set of steps V3 to should locateequal the fault. 8.10 V. A You low could voltage decide is most to 1) likely check caused the source by a low voltage, source2) disconnect voltage or incorrect the load and resistors check(possibly the outputR1voltage, and R2 and ifreversed). it is correct, If the 3) circuit check the is new, loadincorrect resistance. components If R3 is correct, are possible. check other resistors. Series and Parallel Circuits FIGURE 6–97 Series and Parallel Circuits Selected Key Terms 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. Series and Parallel Circuits 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. Series and Parallel Circuits 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 Series and Parallel Circuits 2. If a series equivalent circuit is drawn for a complex circuit, the equivalent circuit can be analyzed with a. the voltage divider theorem b. Kirchhoff’s voltage law c. both of the above d. none of the above Series and Parallel Circuits 3. For the circuit shown, a. R1 is in series with R2 d. R2 is in parallel with R3 - c. R2 is in series with R3 VS + b. R1 is in parallel with R2 R1 R2 R3 Series and Parallel Circuits 4. For the circuit shown, R4 a. R1 is in series with R2 R1 b. R4 is in parallel with R1 - d. none of the above + c. R2 is in parallel with R3 VS R2 R3 Series and Parallel Circuits 5. A signal generator has an output voltage of 2.0 V with no load. When a 600 load is connected to it, the output drops to 1.0 V. The Thevenin resistance of the generator is a. 300 b. 600 c. 900 d. 1200 . Series and Parallel Circuits 6. For the circuit shown, Kirchhoff's voltage law a. applies only to the outside loop b. applies only to the A junction. c. can be applied to any closed path. d. does not apply. VS + 10 V R1 270 A R2 330 R3 470 Series and Parallel Circuits 7. The effect of changing a measured quantity due to connecting an instrument to a circuit is called a. loading b. clipping c. distortion d. loss of precision Series and Parallel Circuits 8. An unbalanced Wheatstone bridge has the voltages shown. The voltage across R4 is a. 4.0 V d. 7.0 V - c. 6.0 V VS 12 V + b. 5.0 V R1 7.0 V R2 R3 + RL 1.0 V R4 Series and Parallel Circuits 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 a. 4.0 V d. 7.0 V + RL - - c. 6.0 V VS 12 V + b. 5.0 V R3 R1 R2 R4 5.0 V Series and Parallel Circuits 10. Maximum power is transferred from a fixed source when a. the load resistor is ½ the source resistance b. the load resistor is equal to the source resistance c. the load resistor is twice the source resistance d. none of the above Series and Parallel Circuits Quiz Answers: 1. b 6. c 2. c 7. a 3. d 8. a 4. d 9. d 5. b 10. b