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CHAPTER 20 |
ELECTRIC CIRCUITS |
CONCEPTUAL QUESTIONS
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1. REASONING AND SOLUTION
a. When S
1 is set to position A, current can flow from the generator only along the path that requires S2 to be set to position A. Hence, the light will be on when S2 is in position A.b. When S
1 is set to position B, current can flow from the generator only along the path that requires S2 to be set to position B. Hence, the light will be on when S2 is in position B._
____________________________________________________________________________________________2. REASONING AND SOLUTION When an incandescent light bulb is turned on, the tungsten filament becomes white hot. Since the voltage is constant, the power delivered to the light bulb is given by Equation 20.6c: 
. From Equation 20.5, 
, where 
is the temperature coefficient of resistivity and is a positive number. Thus, as the filament temperature increases, the resistance of the wire increases, and as the filament heats up, the power delivered to the bulb decreases.
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3. REASONING AND SOLUTION Two materials have different resistivities. Two wires of the same length are made, one from each of the materials. The resistance of each wire is given by Equation 20.3: 
, where
is the same for each.
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4. REASONING AND SOLUTION The resistance of a wire is given by Equation 20.3: 
, where
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5. REASONING AND SOLUTION One electrical appliance operates with a voltage of 120 V, while another operates with a voltage of 240 V. The power used by either appliance is given by Equation 20.6c: 
. Without knowing the resistance R of each appliance, no conclusion can be reached as to which appliance, if either, uses more power.
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6. REASONING AND SOLUTION Two light bulbs are designed for use at 120 V and are rated at 75 W and 150 W. The power used by either bulb is given by Equation 20.6c: . We see from Equation 20.6c that, at constant voltage, the power used by a bulb is inversely proportional to the resistance of the filament. Therefore, the filament resistance is greater for the 75-W bulb.
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7. REASONING AND SOLUTION Rather than state how many watts of power an appliance uses, appliance instructions often give statements such as "10 A, 120 V" instead. This statement gives the current used by the appliance for a specific voltage. It provides all the information necessary to determine the power usage of the appliance from Equation 20.6a: P = IV. Thus, the "current-voltage" rating given in the instructions is equivalent to a statement of the power consumption of the appliance.
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8. REASONING AND SOLUTION When the switch is initially closed, a current appears in the circuit, because charges flow through the heater wire, the bimetallic strip, the contact point, and the light bulb. The bulb glows in response. As charges flow through the heater wire, it becomes hot thereby heating the bimetallic strip. As the strip is heated, the brass, having a larger coefficient of thermal expansion, expands more than the steel. Thus, the bimetallic strip bends into an arc away from the contact point, and the electric current drops to zero because the charges no longer have a continuous path along which to flow. The bulb ceases to glow. Since there is no longer a current, the resistance wire, and hence, the bimetallic strip begin to cool down to room temperature. As the bimetallic strip cools, it bends back to its initial position, as shown in the drawing in the text. When it reaches room temperature it will again touch the contact point, and a current will begin to flow again. The wire and the bimetallic strip will become hot and the bimetallic strip will bend away from the contact point and the current will again drop to zero. This cycle will continue as long as the switch remains closed. Therefore, as long as the switch remains closed, the bulb will flash on and off.
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9. REASONING AND SOLUTION The power rating of a 1000-W heater specifies the power consumed when the heater is connected to an ac voltage of 120 V. The power consumed by the heater is given by Equation 20.6c: 
. When two of these heaters are connected in series, the equivalent resistance of the combination is R + R = 2R. The power consumed by two of the heaters connected in series is, therefore, 
.
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10. REASONING AND SOLUTION A number of light bulbs are to be connected to a single electrical outlet. The bulbs will provide more brightness when they are connected in such a way that their power output is greatest. Since the voltage at the outlet is constant, the power delivered by the bulbs is given by Equation 20.6c: , where R is the equivalent resistance of the combination of light bulbs. When the light bulbs are connected in series, their equivalent resistance is given by Equation 20.16: 
When the light bulbs are connected in parallel, their equivalent resistance is given by Equation 20.17: 
Clearly, the equivalent resistance will be less when the bulbs are connected in parallel. From Equation 20.6c, we can conclude that the power output will be greatest, and the light bulbs will provide more brightness, when they are connected in parallel.
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11. REASONING AND SOLUTION A car has two headlights. The filament of one burns out. However, the other headlight stays on. We can immediately conclude that the bulbs are not connected in series. The figure below shows the series arrangement of two such bulbs.
When the bulbs are connected in series, charges must flow through the filaments of both lights in order to have a complete circuit. Since the filament of the second bulb is burned out, charges will not be able to flow around the circuit, and neither headlight will stay on.
On the other hand, if the bulbs are connected in parallel, as shown at the right, the current will split at the junction J. Charges will be able to flow through the branch of the circuit that contains the good bulb, and that headlight will stay on. Notice that the order of the bulbs does not matter in either case. The results are the same.
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12. REASONING AND SOLUTION When two or more circuit elements are connected in series, they are connected such that the same electric current flows through each element. When two or more circuit elements are connected in parallel, they are connected such that the same voltage is applied across each element.
The circuit in Figure (a) can be shown to be a combination of series and parallel arrangements of resistors. The circuit can be redrawn as shown below.
We can see in the redrawn figure that the current through resistors 2 and 3 is the same; therefore, resistors 2 and 3 are in series and can be represented by an equivalent resistance 23 as shown in the following drawing.
The voltage across resistance 23 and resistor 4 is the same, so these two resistances are in parallel; they can be represented by an equivalent resistance 234. The current through resistance 234 is the same as that through resistor 7, so resistance 234 is in series with resistor 7; they can be represented by an equivalent resistance 2347 as shown in the figure below.
The voltage across 2347 is the same as that across resistor 5; therefore, resistance 2347 is in parallel with resistor 5. They can be represented by an equivalent resistance 23475. Similarly, resistance 23475 is in series with resistor 8, giving an equivalent resistance 234758. Resistance 234758 is in parallel with resistor 6, giving an equivalent resistance 2347586.
Finally, the current through resistor 1 and resistance 2347586 is the same, so they are in series as shown at the right. |
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The circuit in Figure (b) can also be shown to be a combination of series and parallel arrangements of resistors. Since both ends of resistors 2 and 3 are connected, the voltage across resistors 2 and 3 is the same. The same statement can be made for resistors 4 and 5, and resistors 6 and 7. Therefore, resistor 2 is in parallel with the resistor 3 to give an equivalent resistance labeled 23. Resistor 4 is in parallel with resistor 5 to give an equivalent resistance 45, and resistor 6 is in parallel with resistor 7 to give an equivalent resistance 67. From the right-hand portion of the drawing below, it is clear that the resistances 23, 45, and 67 are in series with resistor 1.
The drawing at the right shows the circuit in Figure (c). No such simplifying arguments can be made for this circuit. No two resistors carry the same current; thus, no two of the resistors are in series. Furthermore, no two resistors have the same voltage applied across them; thus, no two of the resistors are in parallel. Circuit (c) contains resistors that are neither in series nor in parallel. |
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13. REASONING AND SOLUTION One way is to make two combinations that consist of two resistors in parallel. The resistance of each combination is These two combinations can then be placed in series, as shown at the right. The equivalent resistance of the series combination is |
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A second way is to make two combinations that consist of two resistors in series. The resistance of each combination is
These two combinations can then be placed in parallel, as shown at the right. The equivalent resistance of the parallel combination is
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14. REASONING AND SOLUTION An ammeter is used to measure the current in a particular branch of a circuit, while a voltmeter is used to measure the voltage across two points in a circuit. An ideal ammeter must have a low resistance so that its presence in the circuit does not affect the current measurement. Conversely, an ideal voltmeter must have a large resistance so that it does not draw current and change the voltage between the two points in question.
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15. REASONING AND SOLUTION A voltmeter is inadvertently mistaken for an ammeter and placed in a circuit. A voltmeter is a high resistance instrument. Since an ammeter is placed in series with the circuit, we would be placing a large resistance in series with the circuit, and the resistance of the circuit would greatly increase. From Ohm's law, V = IR, the current in the circuit would drop markedly.
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16. REASONING AND SOLUTION
From Ohm's law, V = IR, 1 ohm = 1 volt/ampere = 1 volt/(coulomb/second). From the definition of capacitance, q = CV, 1 farad = 1 coulomb/volt. Then
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