Date CONCEPTUAL FAUSIGS PRACTICE PAGE Chapter 23 Electric Current Circuit Happenings-continued ohm's lows VEIk 6. Consider two identical bulbs, Figure 2, connected in series. The circuit has twice as much resistance. The voltage is the same because the battery is the same, and in accord with Ohm's Law, twice the resistance for the same voltage means the current is [half as much]) [twice as much]. 7. So the current in each bulb is [less than 3 amps] [3 amps] [more than 3 amps], and in every part of the circuit. 8. In which bulb does the charge first flow? [The bulb on the left] [The bulb on the right] [Both at once] Figure 2 9. Current occurs in all parts of the circuit [step by step] [instantaneously]. 10. The 6 volts across the circuit divides among the two bulbs. Since bulb resistance is the same, the voltage impressed across each bulb in this series circuit is [less than 3 volts] [3 volts] [more than 3 volts]. 11. This checks with Ohm's Law: (3 V/1 0) = _ amps. In the overall circuit, (6 V/2 0) = amps. So compared with the single bulb of Figure 1 the bulbs are [brighter] [dimmer]. 12. When the two identical bulbs are connected in parallel, Figure 3, voltage among them [divides] [does not divide]. 13. Each bulb still connects across the 6-V battery, so compared with the brightness of the lone bulb in Figure 1, each bulb glows [dimmer] [the same] [brighter] 14. That's because each bulb is energized with a full 6 volts, and the current in each bulb is [3 amps] [4 amps] [6 amps]. 15. A little thought will show that the current in the battery must be Figure 3 [6 amps] [8 amps] [12 amps]. 16. In accord with Ohm's Law, the battery supplies twice the current to the circuit because the equivalent resistance of the circuit is half that of Figure 1. Similar to increased checkout lanes in a supermarket, more branches in a parallel circuit reduce resistance and allow for a flow that is [less] [greater].CONCEPTUAL Physics PRACTICE PAGE Chapter 23 Electric Current Circuit Happenings-continued 17. Adding a third identical bulb in parallel, Figure 4, further reduces the overall circuit resistance, resulting in a total current that is [less] [the same] [more]. 18. Each bulb draws 6 amps, so the current supplied by the battery O (and the current in the battery) is [6 amps] [12 amps] [18 amps]. This is consistent with Ohm's Law; for the same voltage, one-third the resistance results in current increased by [three] [six] [twelve]. 19. Consider points a, b, and c, in Figure 4. A little thought will show that current through point a is [6 amps] [12 amps] [18 amps]. Figure 4 Current through point b is [6 amps] [12 amps] [18 amps], and current through point c is [6 amps] [12 amps] [18 amps]. It's like buses leaving a terminal that branch into three streets. If 18 buses leave the terminal and branch equally along three streets, then six buses occupy a street. How many return to the terminal? [6] [12] [All 18] Likewise with electric current-6 amps in three bulbs means current in the battery is [6 amps] [12 amps] [18 amps]. Adding bulbs in parallel can't continue indefinitely because current in the battery increases with each addition, eventually producing an internal heating problem. Then the internal resistance of the battery is no longer negligible. 20. Notice that in the three circuits in Figure 5, each bulb is connected across the full 6 volts of the battery. The battery "senses" these three circuits as [entirely different] [the same]. Figure 5 Hewitt IAINT