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Unformatted text preview: 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 C H A P T E R 1 8 Problem Solutions 18.1 From ( 29 V I R r = + , the internal resistance is 9.00 V 72.0 0.117 A V r R I =- =- = 4 .92 18.2 (a) 1 2 3 4.0 8.0 12 eq R R R R = + + = + + + = 24 (b) The same current exists in all resistors in a series combination. 24 V 24 eq V I R = = = 1.0 A (c) If the three resistors were connected in parallel, 1 1 1 2 3 1 1 1 1 1 1 4.0 8.0 12 eq R R R R-- = + + = + + = 2 .18 Resistors in parallel have the same potential difference across them, so 4 4 24 V 4.0 V I R = = = 6.0 A , 8 24 V 8.0 I = = 3.0 A , and 12 24 V 12 I = = 2 .0 A 18.3 For the bulb in use as intended, ( 29 ( 29 2 2 120 V 192 75.0 W bulb V R = = = Now, presuming the bulb resistance is unchanged, the current in the circuit shown is 120 V 0.620 A 0.800 192 0.800 eq V I R = = = + + , and the actual power dissipated in the bulb is ( 29 ( 29 2 2 0.620 A 192 bulb I R = = =...
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This homework help was uploaded on 04/21/2008 for the course PHY 213 taught by Professor Cao during the Spring '08 term at Kentucky.
- Spring '08