1. ConcepTest 16.1aElectric Charge I ConcepTest 16.1a Electric Charge I 1) one is positive, the other is negative 2) both are positive 3) both are negative 4) both are positive or both are negative Two charged balls are repelling each other as they hang from the ceiling. What can you say about their charges?

2. ConcepTest 16.1aElectric Charge I ConcepTest 16.1a Electric Charge I same charge The fact that the balls repel each other only can tell you that they have the same charge, but you do not know the sign. So they can be either both positive or both negative. 1) one is positive, the other is negative 2) both are positive 3) both are negative 4) both are positive or both are negative Two charged balls are repelling each other as they hang from the ceiling. What can you say about their charges? Follow-up: What does the picture look like if the two balls are oppositely charged? What about if both balls are neutral?

3. 1) have opposite charges 2) have the same charge 3) all have the same charge 4) one ball must be neutral (no charge) From the picture, what can you conclude about the charges? ConcepTest 16.1bElectric Charge II ConcepTest 16.1b Electric Charge II

4. 1) have opposite charges 2) have the same charge 3) all have the same charge 4) one ball must be neutral (no charge) From the picture, what can you conclude about the charges? The GREEN and PINK balls must have the same charge, since they repel each other. The YELLOW ball also repels the GREEN, so it must also have the same charge as the GREEN (and the PINK). ConcepTest 16.1bElectric Charge II ConcepTest 16.1b Electric Charge II

5. ConcepTest 16.2aConductors I ConcepTest 16.2a Conductors I 1) positive 2) negative 3) neutral 4) positive or neutral 5) negative or neutral A metal ball hangs from the ceiling by an insulating thread. The ball is attracted to a positive-charged rod held near the ball. The charge of the ball must be:

6. negative neutral induction Clearly, the ball will be attracted if its charge is negative. However, even if the ball is neutral, the charges in the ball can be separated by induction (polarization), leading to a net attraction. 1) positive 2) negative 3) neutral 4) positive or neutral 5) negative or neutral A metal ball hangs from the ceiling by an insulating thread. The ball is attracted to a positive-charged rod held near the ball. The charge of the ball must be: remember the ball is a conductor! ConcepTest 16.2aConductors I ConcepTest 16.2a Conductors I Follow-up: What happens if the metal ball is replaced by a plastic ball?

7. Q Q F 1 = 3N F 2 = ? 1) 1.0 N 2) 1.5 N 3) 2.0 N 4) 3.0 N 5) 6.0 N What is the magnitude of the force F 2 ? ConcepTest 16.3aCoulomb’s Law I ConcepTest 16.3a Coulomb’s Law I

8. same magnitude force of one on the other of a pair is the same as the reverse Note that this sounds suspiciously like Newton’s 3rd Law!! The force F 2 must have the same magnitude as F 1. This is due to the fact that the form of Coulomb’s Law is totally symmetric with respect to the two charges involved. The force of one on the other of a pair is the same as the reverse. Note that this sounds suspiciously like Newton’s 3rd Law!! Q Q F 1 = 3N F 2 = ? 1) 1.0 N 2) 1.5 N 3) 2.0 N 4) 3.0 N 5) 6.0 N What is the magnitude of the force F 2 ? ConcepTest 16.3aCoulomb’s Law I ConcepTest 16.3a Coulomb’s Law I

9. ConcepTest 16.3bCoulomb’s Law II ConcepTest 16.3b Coulomb’s Law II 1) 3/4 N 2) 3.0 N 3) 12 N 4) 16 N 5) 48 N If we increase one charge to 4Q, what is the magnitude of F 1 ? 4Q Q F 1 = ? F 2 = ? Q Q F 1 = 3N F 2 = ?

10. ConcepTest 16.3bCoulomb’s Law II ConcepTest 16.3b Coulomb’s Law II Originally we had: F 1 = k(Q)(Q)/r 2 = 3 N Now we have: F 1 = k(4Q)(Q)/r 2 4 times bigger which is 4 times bigger than before. 1) 3/4 N 2) 3.0 N 3) 12 N 4) 16 N 5) 48 N If we increase one charge to 4Q, what is the magnitude of F 1 ? 4Q4Q4Q4Q Q F 1 = ? F 2 = ? Q Q F 1 = 3N F 2 = ? Follow-up: Now what is the magnitude of F 2 ?

11. 1) 9 F 2) 3 F 3) F 4) 1/3 F 5) 1/9 F The force between two charges separated by a distance d is F. If the charges are pulled apart to a distance 3d, what is the force on each charge? QF QFd Q ? Q ? 3d3d3d3d ConcepTest 16.3cCoulomb’s Law III ConcepTest 16.3c Coulomb’s Law III

12. Originally we had: F before = k(Q)(Q)/d 2 = F Now we have: F after = k(Q)(Q)/(3d) 2 = 1/9 F 1) 9 F 2) 3 F 3) F 4) 1/3 F 5) 1/9 F The force between two charges separated by a distance d is F. If the charges are pulled apart to a distance 3d, what is the force on each charge? QF QFd Q?Q? 3d3d3d3d ConcepTest 16.3cCoulomb’s Law III ConcepTest 16.3c Coulomb’s Law III Follow-up: What is the force if the original distance is halved?

13. ConcepTest 18.1Connect the Battery Which is the correct way to light the lightbulb with the battery? 4) all are correct 5) none are correct 1) 3) 2)

14. continuous connection Current can only flow if there is a continuous connection from the negative terminal through the bulb to the positive terminal. This is only the case for Fig. (3). ConcepTest 18.1Connect the Battery Which is the correct way to light the lightbulb with the battery? 4) all are correct 5) none are correct 1) 3) 2)

15. Ohm’s law is obeyed since the current still increases when V increases 1) Ohm’s law is obeyed since the current still increases when V increases Ohm’s law is not obeyed 2) Ohm’s law is not obeyed This has nothing to do with Ohm’s law 3) This has nothing to do with Ohm’s law ConcepTest 18.2Ohm’s Law You double the voltage across a certain conductor and you observe the current increases three times. What can you conclude?

16. Ohm’s law is obeyed since the current still increases when V increases 1) Ohm’s law is obeyed since the current still increases when V increases Ohm’s law is not obeyed 2) Ohm’s law is not obeyed This has nothing to do with Ohm’s law 3) This has nothing to do with Ohm’s law V = I R linear Ohm’s law, V = I R, states that the relationship between voltage and current is linear. Thus for a conductor that obeys Ohm’s Law, the current must double when you double the voltage. ConcepTest 18.2Ohm’s Law You double the voltage across a certain conductor and you observe the current increases three times. What can you conclude? Follow-up: Where could this situation occur?

17. ConcepTest 18.4Dimmer When you rotate the knob of a light dimmer, what is being changed in the electric circuit? 1) the power 2) the current 3) the voltage 4) both (1) and (2) 5) both (2) and (3)

18. increases the resistance decreases the current The voltage is provided at 120 V from the outside. The light dimmer increases the resistance and therefore decreases the current that flows through the lightbulb. ConcepTest 18.4 Dimmer When you rotate the knob of a light dimmer, what is being changed in the electric circuit? 1) the power 2) the current 3) the voltage 4) both (1) and (2) 5) both (2) and (3) Follow-up: Why does the voltage not change?

19. ConcepTest 18.5aLightbulbs Two lightbulbs operate at 120 V, but one has a power rating of 25 W while the other has a power rating of 100 W. Which one has the greater resistance? 1) the 25 W bulb 2) the 100 W bulb 3) both have the same 4) this has nothing to do with resistance

20. P = V 2 / R lower power ratinghigher resistance Since P = V 2 / R the bulb with the lower power rating has to have the higher resistance. ConcepTest 18.5a Lightbulbs Two lightbulbs operate at 120 V, but one has a power rating of 25 W while the other has a power rating of 100 W. Which one has the greater resistance? 1) the 25 W bulb 2) the 100 W bulb 3) both have the same 4) this has nothing to do with resistance Follow-up: Which one carries the greater current?

21. ConcepTest 18.5bSpace Heaters I Two space heaters in your living room are operated at 120 V. Heater 1 has twice the resistance of heater 2. Which one will give off more heat? 1) heater 1 2) heater 2 3) both equally

22. P = V 2 / R, smaller resistance larger power Using P = V 2 / R, the heater with the smaller resistance will have the larger power output. Thus, heater 2 will give off more heat. ConcepTest 18.5bSpace Heaters I Two space heaters in your living room are operated at 120 V. Heater 1 has twice the resistance of heater 2. Which one will give off more heat? 1) heater 1 2) heater 2 3) both equally Follow-up: Which one carries the greater current?

23. ConcepTest 19.1aSeries Resistors I 9 V Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? 1) 12 V 2) zero 3) 3 V 4) 4 V 5) you need to know the actual value of R

24. equal evenly 3 V drop Since the resistors are all equal, the voltage will drop evenly across the 3 resistors, with 1/3 of 9 V across each one. So we get a 3 V drop across each. ConcepTest 19.1aSeries Resistors I 9 V Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? 1) 12 V 2) zero 3) 3 V 4) 4 V 5) you need to know the actual value of R Follow-up: What would be the potential difference if R=  Follow-up: What would be the potential difference if R= 1 

25. ConcepTest 19.1bSeries Resistors II 12 V R 1 = 4  R 2 = 2  In the circuit below, what is the voltage across ? In the circuit below, what is the voltage across R 1 ? 1) 12 V 2) zero 3) 6 V 4) 8 V 5) 4 V

26. ConcepTest 19.1bSeries Resistors II 12 V R 1 = 4  R 2 = 2  In the circuit below, what is the voltage across ? In the circuit below, what is the voltage across R 1 ? 1) 12 V 2) zero 3) 6 V 4) 8 V 5) 4 V The voltage drop across R 1 has to be twice as big as the drop across R 2.V 1 = 8 V The voltage drop across R 1 has to be twice as big as the drop across R 2. This means that V 1 = 8 V and V 2 = 4 V. Or else you could find the current I = V/R = (12 V)/(6  = 2 A, then use Ohm’s Law to get voltages. Follow-up: What happens if the voltage is doubled?

27. ConcepTest 19.2aParallel Resistors I In the circuit below, what is the current through ? In the circuit below, what is the current through R 1 ? 10 V R 1 = 5  R 2 = 2  1) 10 A 2) zero 3) 5 A 4) 2 A 5) 7 A

28. voltagesame V 1 = I 1 R 1 I 1 = 2 A The voltage is the same (10 V) across each resistor because they are in parallel. Thus, we can use Ohm’s Law, V 1 = I 1 R 1 to find the current I 1 = 2 A. ConcepTest 19.2aParallel Resistors I In the circuit below, what is the current through ? In the circuit below, what is the current through R 1 ? 10 V R 1 = 5  R 2 = 2  1) 10 A 2) zero 3) 5 A 4) 2 A 5) 7 A Follow-up: What is the total current through the battery?

29. ConcepTest 19.2bParallel Resistors II 1) increases 2) remains the same 3) decreases 4) drops to zero Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit?

30. ConcepTest 19.2bParallel Resistors II 1) increases 2) remains the same 3) decreases 4) drops to zero resistance of the circuit drops resistance decreasescurrent must increase As we add parallel resistors, the overall resistance of the circuit drops. Since V = IR, and V is held constant by the battery, when resistance decreases, the current must increase. Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit? Follow-up: What happens to the current through each resistor?

31. ConcepTest 19.3aShort Circuit Current flows through a lightbulb. If a wire is now connected across the bulb, what happens? all the current continues to flow through the bulb 1) all the current continues to flow through the bulb half the current flows through the wire, the other half continues through the bulb 2) half the current flows through the wire, the other half continues through the bulb all the current flows through the wire 3) all the current flows through the wire none of the above 4) none of the above

32. zeroALL The current divides based on the ratio of the resistances. If one of the resistances is zero, then ALL of the current will flow through that path. ConcepTest 19.3aShort Circuit Current flows through a lightbulb. If a wire is now connected across the bulb, what happens? all the current continues to flow through the bulb 1) all the current continues to flow through the bulb half the current flows through the wire, the other half continues through the bulb 2) half the current flows through the wire, the other half continues through the bulb all the current flows through the wire 3) all the current flows through the wire none of the above 4) none of the above Follow-up: Doesn’t the wire have SOME resistance?

33. ConcepTest 19.3bShort Circuit II Two lightbulbs A and B are connected in series to a constant voltage source. When a wire is connected across B, bulb A will: glow brighter than before 1) glow brighter than before glow just the same as before 2) glow just the same as before glow dimmer than before 3) glow dimmer than before 4) go out completely 5) explode

34. total resistance of the circuit decreasescurrent through bulb A increases Since bulb B is bypassed by the wire, the total resistance of the circuit decreases. This means that the current through bulb A increases. ConcepTest 19.3bShort Circuit II Two lightbulbs A and B are connected in series to a constant voltage source. When a wire is connected across B, bulb A will: glow brighter than before 1) glow brighter than before glow just the same as before 2) glow just the same as before glow dimmer than before 3) glow dimmer than before 4) go out completely 5) explode Follow-up: What happens to bulb B?

35. ConcepTest 19.4aCircuits I circuit 1 1) circuit 1 circuit 2 2) circuit 2 both the same 3) both the same it depends on R 4) it depends on R The lightbulbs in the circuit below are identical with the same resistance R. Which circuit produces more light? (brightness  power)

36. ConcepTest 19.4aCircuits I circuit 1 1) circuit 1 circuit 2 2) circuit 2 both the same 3) both the same it depends on R 4) it depends on R The lightbulbs in the circuit below are identical with the same resistance R. Which circuit produces more light? (brightness  power) parallel lowering the total resistance #1 willdraw a higher current P = I V In #1, the bulbs are in parallel, lowering the total resistance of the circuit. Thus, circuit #1 will draw a higher current, which leads to more light, because P = I V.