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electronics fundamentals
circuits, devices, and applications THOMAS L. FLOYD DAVID M. BUCHLA Chapter 6 – Series and Parallel Combination Circuits
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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.
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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: characteristics that are electrically the same as another circuit but is generally simpler.
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Equivalent circuits is equivalent to
There are no electrical measurements that can distinguish the boxes.
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Equivalent circuits Another example: is equivalent to
There are no electrical measurements that can distinguish the boxes.
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Equivalent circuits is equivalent to is equivalent to
There are no electrical measurements that can distinguish between the three boxes.
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Current at all points is the IT = IR1 = IR2
What Do We Know For Series Circuits: Current at all points is the IT = IR1 = IR2 Voltage across each resistor VT = V1 + V2 For Parallel Circuit Current IT = IR1 + IR2 Voltage at all nodes is the VT = V1 = V2 same drops divides across each resistor in the branch same
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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.
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A simple series-parallel resistive circuit.
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R4 is added to the circuit in series with R1.
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R5 is added to the circuit in series with R2.
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R6 is added to the circuit in parallel with the series combination of R1 and R4.
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FIGURE 6–5
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FIGURE 6–6
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FIGURE 6–7
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Sketch the circuits for nodes A to B A to C B to C
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Sketch the circuits for nodes A to B
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Sketch the circuits for nodes A to C
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Sketch the circuits for nodes B to C
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FIGURE 6–9
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FIGURE 6–10
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Reduce the following circuit to REQ
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6 Ω
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10 Ω
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5 Ω
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IT 50 Ω IT = 2 amps
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FIGURE 6–13
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FIGURE 6–14
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FIGURE 6–15
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FIGURE 6–16
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FIGURE 6–17
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FIGURE 6–18
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FIGURE 6–19
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Review of voltage relationships.
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Kirchoff’s current law
What are the readings for node A?
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FIGURE 6–21
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Loaded voltage divider
The voltage-divider equation was developed for a series circuit. Recall that the output voltage is given by + A A 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.
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Loaded voltage divider
What is the voltage across R3? A Form an equivalent series circuit by combining R2 and R3; then apply the voltage-divider formula to the equivalent circuit: 8.10 V
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FIGURE 6–22
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FIGURE 6–23
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FIGURE 6–24
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FIGURE 6–25
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FIGURE 6–26 A voltage divider with both unloaded and loaded outputs.
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FIGURE 6–28
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FIGURE 6–29
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FIGURE 6–31
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FIGURE 6–32
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The loading effect of a voltmeter.
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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 V 10 V 4.04 V What is a possible explanation of the meter not displaying 10 volts? A voltmeter has internal resistance This RINT can change the resistance of the circuit under test. A 1 MW internal resistance of the meter accounts for the readings.
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FIGURE 6–34
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Wheatstone bridge The Wheatstone bridge consists of: a dc voltage source and four resistive arms forming two voltage dividers. The output is taken between the dividers. Frequently, one of the bridge resistors is adjustable. (R2) .
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When the bridge is balanced, the output voltage is
Balanced Wheatstone bridge When the bridge is balanced, the output voltage is zero The products of resistances in the opposite diagonal arms are equal.
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Wheatstone bridge. Voltage Divider 1 Voltage Divider 2
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Wheatstone bridge. V1=V2 I1=I3 V3=V4 I2=I4
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What is the value of R2 if the bridge is balanced?
Wheatstone bridge What is the value of R2 if the bridge is balanced? 330 W 470 W 12 V 384 W 270 W
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Finding an Unknown Resistance
Scale Factor
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Measuring a physical parameter using a transducer.
Unbalanced Wheatstone Bridge Unbalance occurs when VOUT ≠ 0 Used to measure Mechanical Strain Temperature Pressure VOUT is converted to a digital output indicating the value of the reading. Measuring a physical parameter using a transducer.
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VOUT = VA-VB = 0
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Using the Thermistor chart, what is VOUT when the temperature is 50o C
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
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Example of a load cell.
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Remove RL to make an open circuit between A & B Calculate R1||R3 =
Wheatstone Bridge Remove RL to make an open circuit between A & B Calculate R1||R3 = Calculate R2||R4 = Calculate the voltage from A to ground Calculate the voltage from B to ground 165 W 179 W 7.5 V 6.87 V
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FIGURE 6–43
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FIGURE 6–44
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FIGURE 6–47
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FIGURE 6–48
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FIGURE 6–49 A Wheatstone bridge with a load resistor connected between the output terminals is not a straightforward series-parallel circuit.
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FIGURE 6–51
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FIGURE 6–52
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FIGURE 6–53
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FIGURE 6–54 Maximum power is transferred to the load when RL = RS.
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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.
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Maximum power transfer Theorem
What is the power delivered to the load? The voltage to the load is 5.0 V The power delivered is:
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FIGURE 6–55
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FIGURE 6–56 Curve showing that the load power is maximum when RL = RS.
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Superposition theorem
A way to determine currents and voltages in a linear circuit that has multiple sources Take one source at a time and Algebraically summing the results.
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Superposition theorem
Four Step Process Leave 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.
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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 mA 6.10 kW 1.97 mA 0.98 mA 8.73 kW 2.06 mA 0.58 mA 1.56 mA The total current is the algebraic sum.
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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
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Analysis: Planning: Measurement:
Troubleshooting The 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.
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Troubleshooting The output of the voltage-divider is 6.0 V. Describe how you would use analysis and planning in finding the fault. A Planning: 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:
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FIGURE 6–58
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FIGURE 6–59
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FIGURE 6–60
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FIGURE 6–61
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FIGURE 6–62
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FIGURE 6–63
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FIGURE 6–64
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FIGURE 6–65
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FIGURE 6–66
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FIGURE 6–67
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FIGURE 6–68
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FIGURE 6–69
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FIGURE 6–73
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FIGURE 6–74 The meters indicate the correct readings for this circuit.
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FIGURE 6–75
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FIGURE 6–76
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FIGURE 6–77
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FIGURE 6–78
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FIGURE 6–79
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FIGURE 6–80
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FIGURE 6–81
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FIGURE 6–82
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FIGURE 6–83
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FIGURE 6–84
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FIGURE 6–85
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FIGURE 6–86
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FIGURE 6–87
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FIGURE 6–88
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FIGURE 6–89
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FIGURE 6–90
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FIGURE 6–91
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FIGURE 6–92
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FIGURE 6–93
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FIGURE 6–94
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FIGURE 6–95
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FIGURE 6–96
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FIGURE 6–97
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FIGURE 6–98
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FIGURE 6–99
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FIGURE 6–100
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FIGURE 6–101
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FIGURE 6–103
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FIGURE 6–104
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FIGURE 6–105
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FIGURE 6–106
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FIGURE 6–107
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FIGURE 6–108
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FIGURE 6–109
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FIGURE 6–110
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Selected Key Terms Loading
Load current Bleeder current Wheatstone bridge The 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.
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Selected Key Terms Thevenin’s theorem
Maximum power transfer Superposition A 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.
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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
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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
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3. For the circuit shown, a. R1 is in series with R2 b. R1 is in parallel with R2 c. R2 is in series with R3 d. R2 is in parallel with R3
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4. For the circuit shown, a. R1 is in series with R2 b. R4 is in parallel with R1 c. R2 is in parallel with R3 d. none of the above
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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 is a. 300 W b. 600 W c. 900 W d W.
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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.
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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
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8. An unbalanced Wheatstone bridge has the voltages shown
8. An unbalanced Wheatstone bridge has the voltages shown. The voltage across R4 is a. 4.0 V b. 5.0 V c. 6.0 V d. 7.0 V
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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 be a. 4.0 V b. 5.0 V c. 6.0 V d. 7.0 V
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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
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Quiz Answers: 1. b 2. c 3. d 4. d 5. b 6. c 7. a 8. a 9. d 10. b
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