Chapter 18 - CHAPTER 18 Quick Quizzes 1. Bulb R1 becomes...

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C HAPTER 18 Quick Quizzes 1. Bulb R 1 becomes brighter. Connecting a wire from b to c provides a nearly zero resistance path from b to c and decreases the total resistance of the circuit from R 1 + R 2 to just R 1 . Ignoring internal resistance, the potential difference maintained by the battery is unchanged while the resistance of the circuit has decreased. The current passing through bulb R 1 increases, causing this bulb to glow brighter. Bulb R 2 goes out because essentially all of the current now passes through the wire connecting b and c and bypasses the filament of Bulb R 2 . 2. (b). When the switch is opened, resistors R 1 and R 2 are in series, so that the total circuit resistance is larger than when the switch was closed. As a result, the current decreases. 3. (a). When the switch is closed, resistors R 1 and R 2 are in parallel, so that the total circuit resistance is smaller than when the switch was open. As a result, the total current increases. 4. (a) decrease; (b) decrease; (c) increase; (d) decrease; (e) decrease. As you add more lightbulbs in parallel , the overall resistance of the circuit decreases. The current leaving the battery increases with the addition of each bulb. This increase in current results in an increase in power transferred from the battery. As a result, the battery lifetime decreases. As the current rises, the terminal voltage across the battery drops further below the battery emf. Because of the drop in terminal voltage, which is applied to each lightbulb, the current in each lightbulb is reduced, leading to a reduction in brightness. 89
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CHAPTER 18 Problem Solutions 18.1 From ( ) VI Rr ∆= + , the internal resistance is 9.00 V 72.0 0.117 A V rR I =− = − Ω = 4.92 18.2 (a) 123 4.0 8.0 12 eq RRRR =++= +Ω + Ω = 24 (b) The same current exists in all resistors in a series combination. 24 V V I R == = 1.0 A (c) If the three resistors were connected in parallel, 1 1 111 1 1 1 R RRR ⎛⎞ =++ = + + = 2.18 ⎜⎟ ⎝⎠ ΩΩΩ Resistors in parallel have the same potential difference across them, so 4 4 V I R = 6.0 A , 8 I = = 3.0 A , and 12 I = = 2.0 A 18.3 For the bulb in use as intended, () 22 120 V 192 75.0 W bulb V R = Now, presuming the bulb resistance is unchanged, the current in the circuit shown is 0.620 A 0.800 V I R = Ω+ , and the actual power dissipated in the bulb is ( ) ( ) 2 2 IR ℘= = Ω = 73.8 W 0.800 192 0.800 120 V 90
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CHAPTER 18 18.4 (a) The current through this series combination is ( ) 12 V 2.0 A 6.0 bc V I R === Therefore, the terminal potential difference of the power supply is ( ) () 2.0 A9 . 0 eq VI R ∆= = + Ω= 30 V ab V 9.0 6.0 c (b) When connected in parallel, the potential difference across either resistor is the voltage setting of the power supply. Thus, = 2.3 V 99 0.25 . R = 18.5 (a) The equivalent resistance of the two parallel resistors is 1 11 4.12 7.00 10.0 p R ⎛⎞ =+ = ⎜⎟ ⎝⎠ ΩΩ Thus, ( ) 49 4.00 4.12 9.00 p RRRR =++= + + = 17.1 7.00 4.00 10.0 9.00 (b) ( ) 34.0 V 1.99 A V I R == , so = II = = Also, ( )( ) 1.99 A4 .
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Chapter 18 - CHAPTER 18 Quick Quizzes 1. Bulb R1 becomes...

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