Presentation on theme: "electronics fundamentals"— Presentation transcript:
1 electronics fundamentals circuits, devices, and applicationsTHOMAS L. FLOYDDAVID M. BUCHLAChapter 6 – Series and Parallel Combination Circuits
2 Identifying series-parallel relationships Most practical circuits have combinations of series and parallel components.From Chapters 4 and 5Components that are connected in series will share a common path.Components that are connected in parallel will be connected across the same two nodes.
3 Combination circuitsCircuits containing both series and parallel circuits are called COMBINATION circuitsYou 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:characteristics that are electrically the same as another circuit butis generally simpler.
4 Equivalent circuits is equivalent to There are no electrical measurements that can distinguish the boxes.
5 Equivalent circuits Another example: is equivalent to There are no electrical measurements that can distinguish the boxes.
6 Equivalent circuits is equivalent to is equivalent to There are no electrical measurements that can distinguish between the three boxes.
7 Current at all points is the IT = IR1 = IR2 What Do We KnowFor Series Circuits:Current at all points is theIT = IR1 = IR2Voltage across each resistorVT = V1 + V2For Parallel CircuitCurrentIT = IR1 + IR2Voltage at all nodes is theVT = V1 = V2samedropsdivides across each resistor in the branchsame
8 Seven Step Process for Solving a Combination Circuit 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.Calculate the total resistance (RT) of the circuit by adding all REQ’s to the other series resistances.Calculate the total current (IT) using RT in Ohm’s law.Calculate the voltage drop across any series resistances or REQ’s using Ohm’s law.Calculate the branch currents in all parallel sections of the circuit using the voltage drop across REQ and Ohm’s law.Use the branch currents and resistance values to calculate the voltage of the parallel resistances.Make a summary of the voltage drops and currents for each resistance to make sure they total correctly.
38 Loaded voltage divider The voltage-divider equation was developed for a series circuit. Recall that the output voltage is given by+AA voltage-divider with a resistive load forms a combination (parallel) circuit.The voltage divider is said to be LOADED.The loading reduces the total resistance from node A to ground.
39 Loaded voltage divider What is the voltage across R3?AForm an equivalent series circuit by combining R2 and R3; then apply the voltage-divider formula to the equivalent circuit:8.10 V
50 Loading effect of a voltmeter 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 R2.4.04 V10 V4.04 VWhat is a possible explanation of the meter not displaying 10 volts?A voltmeter has internal resistanceThis RINT can change the resistance of the circuit under test.A 1 MW internal resistance of the meter accounts for the readings.
52 Wheatstone bridgeThe Wheatstone bridge consists of:a dc voltage source andfour resistive arms forming two voltage dividers.The output is taken between the dividers.Frequently, one of the bridge resistors is adjustable. (R2).
53 When the bridge is balanced, the output voltage is Balanced Wheatstone bridgeWhen the bridge is balanced, the output voltage iszeroThe products of resistances in the opposite diagonal arms areequal.
58 Measuring a physical parameter using a transducer. Unbalanced Wheatstone BridgeUnbalance occurs when VOUT ≠ 0Used to measureMechanical StrainTemperaturePressureVOUT is converted to a digital output indicating the value of the reading.Measuring a physical parameter using a transducer.
62 Remove RL to make an open circuit between A & B Calculate R1||R3 = Wheatstone BridgeRemove RL to make an open circuit between A & BCalculate R1||R3 =Calculate R2||R4 =Calculate the voltage from A to groundCalculate the voltage from B to ground165 W179 W7.5 V6.87 V
71 FIGURE 6–54 Maximum power is transferred to the load when RL = RS.
72 Maximum power transfer Theorem The maximum power is transferred from a source to a load when: RL = RS(resistance of the voltage source)The maximum power transfer theorem assumes the source voltage and resistance are fixed.
73 Maximum power transfer Theorem What is the power delivered to the load?The voltage to the load is5.0 VThe power delivered is:
75 FIGURE 6–56 Curve showing that the load power is maximum when RL = RS.
76 Superposition theorem A way to determine currents and voltages in a linear circuit that has multiple sourcesTake one source at a time andAlgebraically summing the results.
77 Superposition theorem Four Step ProcessLeave one voltage or current source at a time in the circuit (circuit 1) and replace the other (circuit 2) with a short.Calculate REQ/Total and then calculate the voltage or current for the resistor(s).Repeat steps 1 and 2 for the other circuit (circuit 2).Algebraically add the results for all sources.
79 Summary Summary What does the ammeter read for I2? Source 1: RT(S1)= I1= I2=Source 2: RT(S2)= I3= I2=Both sources I2=Set up a table of pertinent information and solve for each quantity listed:1.56 mA6.10 kW1.97 mA0.98 mA8.73 kW2.06 mA0.58 mA1.56 mAThe total current is the algebraic sum.
80 Thevenin’s theorem and Wheatstone Bridge 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
81 Analysis: Planning: Measurement: TroubleshootingThe effective troubleshooter must think logically about circuit operation.Understand normal circuit operation and find out the symptoms of the failure.Analysis:Decide on a logical set of steps to find the fault.Planning:Measurement:Following the steps in the plan, make measurements to isolate the problem. Modify the plan if necessary.
82 TroubleshootingThe output of the voltage-divider is 6.0 V. Describe how you would use analysis and planning in finding the fault.APlanning: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.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.Analysis:
132 Selected Key Terms Loading Load currentBleeder currentWheatstone bridgeThe effect on a circuit when an element that draws current from the circuit is connected across the output terminals.The output current supplied to a load.The current left after the load current is subtracted from the total current into the circuit.A 4-legged type of bridge circuit with which an unknown resistance can be accurately measured using the balanced state. Deviations in resistance can be measured using the unbalanced state.
133 Selected Key Terms Thevenin’s theorem Maximum power transferSuperpositionA circuit theorem that provides for reducing any two-terminal resistive circuit to a single equivalent voltage source in series with an equivalent resistance.The condition, when the load resistance equals the source resistance, under which maximum power is transferred to the load.A method for analyzing circuits with two or more sources by examining the effects of each source by itself and then combining the effects.
134 1. Two circuits that are equivalent have the same a. number of componentsb. response to an electrical stimulusc. internal power dissipationd. all of the above
135 2. If a series equivalent circuit is drawn for a complex circuit, the equivalent circuit can be analyzed witha. the voltage divider theoremb. Kirchhoff’s voltage lawc. both of the aboved. none of the above
136 3. For the circuit shown,a. R1 is in series with R2b. R1 is in parallel with R2c. R2 is in series with R3d. R2 is in parallel with R3
137 4. For the circuit shown,a. R1 is in series with R2b. R4 is in parallel with R1c. R2 is in parallel with R3d. none of the above
138 5. A signal generator has an output voltage of 2. 0 V with no load 5. A signal generator has an output voltage of 2.0 V with no load. When a 600 W load is connected to it, the output drops to 1.0 V. The Thevenin resistance of the generator isa. 300 Wb. 600 Wc. 900 Wd W.
139 6. For the circuit shown, Kirchhoff's voltage law a. applies only to the outside loopb. applies only to the A junction.c. can be applied to any closed path.d. does not apply.
140 7. The effect of changing a measured quantity due to connecting an instrument to a circuit is called a. loadingb. clippingc. distortiond. loss of precision
141 8. An unbalanced Wheatstone bridge has the voltages shown 8. An unbalanced Wheatstone bridge has the voltages shown. The voltage across R4 isa. 4.0 Vb. 5.0 Vc. 6.0 Vd. 7.0 V
142 9. Assume R2 is adjusted until the Wheatstone bridge is balanced 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 bea. 4.0 Vb. 5.0 Vc. 6.0 Vd. 7.0 V
143 10. Maximum power is transferred from a fixed source when a. the load resistor is ½ the source resistanceb. the load resistor is equal to the source resistancec. the load resistor is twice the source resistanced. none of the above