# ConcepTest Clicker Questions. Chapter 26 Physics: for Scientists & Engineers with Modern Physics, 4th edition Giancoli

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1 ConcepTest Clicker Questions Chapter 26 Physics: for Scientists & Engineers with Modern Physics, 4th edition Giancoli 2008 Pearson Education, Inc. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

2 ConcepTest 26.1a Series Resistors I 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 9 V

3 ConcepTest 26.1a Series Resistors I 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? 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. 1) 12 V 2) zero 3) 3 V 4) 4 V 5) you need to know the actual value of R 9 V Follow-up: What would be the potential difference if R = 1 W, 2 W, 3 W?

4 ConcepTest 26.1b Series Resistors II 1) 12 V In the circuit below, what is the voltage across R 1? 2) zero 3) 6 V 4) 8 V 5) 4 V R 1 = 4 W R 2 = 2 W 12 V

5 ConcepTest 26.1b Series Resistors II 1) 12 V In the circuit below, what is the voltage across R 1? 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. 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 W) = 2 A, and then use R 1 = 4 W R 2 = 2 W 12 V Ohm s law to get voltages. Follow-up: What happens if the voltage is doubled?

6 ConcepTest 26.2a Parallel Resistors I 1) 10 A In the circuit below, what is the current through R 1? 2) zero 3) 5 A 4) 2 A 5) 7 A R 2 = 2 W R 1 = 5 W 10 V

7 ConcepTest 26.2a Parallel Resistors I 1) 10 A In the circuit below, what is the current through R 1? 2) zero 3) 5 A 4) 2 A 5) 7 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. R 2 = 2 W R 1 = 5 W Follow-up: What is the total current through the battery? 10 V

8 ConcepTest 26.2b 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? Parallel Resistors II 1) increases 2) remains the same 3) decreases 4) drops to zero

9 ConcepTest 26.2b 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? Parallel Resistors II 1) increases 2) remains the same 3) decreases 4) drops to zero 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. Follow-up: What happens to the current through each resistor?

10 ConcepTest 26.3a Short Circuit I Current flows through a lightbulb. If a wire is now connected across the bulb, what happens? 1) all the current continues to flow through the bulb 2) half the current flows through the wire, the other half continues through the bulb 3) all the current flows through the wire 4) none of the above

11 ConcepTest 26.3a Short Circuit I Current flows through a lightbulb. If a wire is now connected across the bulb, what happens? 1) all the current continues to flow through the bulb 2) half the current flows through the wire, the other half continues through the bulb 3) all the current flows through the wire 4) none of the above 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. Follow-up: Doesn t the wire have SOME resistance?

12 ConcepTest 26.3b Short 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: 1) glow brighter than before 2) glow just the same as before 3) glow dimmer than before 4) go out completely 5) explode

13 ConcepTest 26.3b Short 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: 1) glow brighter than before 2) glow just the same as before 3) glow dimmer than before 4) go out completely 5) explode Since bulb B is bypassed by the wire, the total resistance of the circuit decreases. This means that the current through bulb A increases. Follow-up: What happens to bulb B?

14 ConcepTest 26.4a The lightbulbs in the circuits below are identical with the same resistance R. Which circuit produces more light? (brightness power) Circuits I 1) circuit I 2) circuit II 3) both the same 4) it depends on R

15 ConcepTest 26.4a The lightbulbs in the circuits below are identical with the same resistance R. Which circuit produces more light? (brightness power) Circuits I 1) circuit I 2) circuit II 3) both the same 4) it depends on R In circuit I, the bulbs are in parallel, lowering the total resistance of the circuit. Thus, circuit I will draw a higher current, which leads to more light, because P = IV.

16 ConcepTest 26.4b The three lightbulbs in the circuit all have the same resistance of 1 W. By how much is the brightness of bulb B greater or smaller than the brightness of bulb A? (brightness power) Circuits II 1) twice as much 2) the same 3) 1/2 as much 4) 1/4 as much 5) 4 times as much 10 V

17 ConcepTest 26.4b The three lightbulbs in the circuit all have the same resistance of 1 W. By how much is the brightness of bulb B greater or smaller than the brightness of bulb A? (brightness power) Circuits II 1) twice as much 2) the same 3) 1/2 as much 4) 1/4 as much 5) 4 times as much We can use P = V 2 /R to compare the power: P A = (V A ) 2 /R A = (10 V) 2 /1 W = 100 W P B = (V B ) 2 /R B = (5 V) 2 /1 W = 25 W 10 V Follow-up: What is the total current in the circuit?

18 ConcepTest 26.5a More Circuits I What happens to the voltage across the resistor R 1 when the switch is closed? The voltage will: 1) increase 2) decrease 3) stay the same R 1 V S R 2 R 3

19 ConcepTest 26.5a More Circuits I What happens to the voltage across the resistor R 1 when the switch is closed? The voltage will: 1) increase 2) decrease 3) stay the same With the switch closed, the addition of R 1 R 2 to R 3 decreases the equivalent resistance, so the current from the battery increases. This will cause an increase in the voltage across R 1. V S R 2 R 3 Follow-up: What happens to the current through R 3?

20 ConcepTest 26.5b More Circuits II What happens to the voltage across the resistor R 4 when the switch is closed? 1) increases 2) decreases 3) stays the same R 1 V S R 2 R 3 R 4

21 ConcepTest 26.5b More Circuits II What happens to the voltage across the resistor R 4 when the switch is closed? 1) increases 2) decreases 3) stays the same We just saw that closing the switch causes an increase in the voltage across R 1 (which is V AB ). The A R 1 S B voltage of the battery is constant, V R 3 R 4 so if V AB increases, then V BC must R 2 decrease! C Follow-up: What happens to the current through R 4?

22 ConcepTest 26.6 Which resistor has the greatest current going through it? Assume that all the resistors are equal. Even More Circuits 1) R 1 2) both R 1 and R 2 equally 3) R 3 and R 4 4) R 5 5) all the same V

23 ConcepTest 26.6 Which resistor has the greatest current going through it? Assume that all the resistors are equal. Even More Circuits 1) R 1 2) both R 1 and R 2 equally 3) R 3 and R 4 4) R 5 5) all the same The same current must flow through the left and right combinations of resistors. On the LEFT, the current splits equally, so I 1 = I 2. On the RIGHT, more current will V go through R 5 than R 3 + R 4, since the branch containing R 5 has less resistance. Follow-up: Which one has the smallest voltage drop?

24 ConcepTest 26.7 When you rotate the knob of a light dimmer, what is being changed in the electric circuit? Dimmer 1) the power 2) the current 3) the voltage 4) both (1) and (2) 5) both (2) and (3)

25 ConcepTest 26.7 When you rotate the knob of a light dimmer, what is being changed in the electric circuit? Dimmer 1) the power 2) the current 3) the voltage 4) both (1) and (2) 5) both (2) and (3) 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. Follow-up: Why does the voltage not change?

26 ConcepTest 26.8a 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? Lightbulbs 1) the 25 W bulb 2) the 100 W bulb 3) both have the same 4) this has nothing to do with resistance

27 ConcepTest 26.8a 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? Lightbulbs 1) the 25 W bulb 2) the 100 W bulb 3) both have the same 4) this has nothing to do with resistance Since P = V 2 / R, the bulb with the lower power rating has to have the higher resistance. Follow-up: Which one carries the greater current?

28 ConcepTest 26.8b 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? Space Heaters I 1) heater 1 2) heater 2 3) both equally

29 ConcepTest 26.8b 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? Space Heaters I 1) heater 1 2) heater 2 3) both equally 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. Follow-up: Which one carries the greater current?

30 ConcepTest 26.9 What is the current in branch P? Junction Rule 1) 2 A 2) 3 A 3) 5 A 4) 6 A 5) 10 A 5 A 8 A P 2 A

31 ConcepTest 26.9 What is the current in branch P? Junction Rule 1) 2 A 2) 3 A 3) 5 A 4) 6 A 5) 10 A The current entering the junction in red is 8 A, so the current leaving must also be 8 A. One exiting branch has 2 A, so the other branch (at P) must have 6 A. 5 A S 8 A Junction 2 A P 6 A

32 ConcepTest Kirchhoff s Rules The lightbulbs in the 1) both bulbs go out circuit are identical. When 2) intensity of both bulbs increases the switch is closed, what 3) intensity of both bulbs decreases happens? 4) A gets brighter and B gets dimmer 5) nothing changes

33 ConcepTest Kirchhoff s Rules The lightbulbs in the circuit are identical. When the switch is closed, what happens? 1) both bulbs go out 2) intensity of both bulbs increases 3) intensity of both bulbs decreases 4) A gets brighter and B gets dimmer 5) nothing changes When the switch is open, the point between the bulbs is at 12 V. But so is the point between the batteries. If there is no potential difference, then no current will flow once the switch is closed!! Thus, nothing changes. Follow-up: What happens if the bottom battery is replaced by a 24 V battery? 24 V

34 ConcepTest Wheatstone Bridge An ammeter A is connected 1) l between points a and b in the 2) l/2 circuit below, in which the four 3) l/3 resistors are identical. The current 4) l/4 through the ammeter is: 5) zero I V

35 ConcepTest Wheatstone Bridge An ammeter A is connected 1) l between points a and b in the 2) l/2 circuit below, in which the four 3) l/3 resistors are identical. The current 4) l/4 through the ammeter is: 5) zero Since all resistors are identical, the voltage drops are the same across the upper branch and the lower branch. Thus, the potentials at points a and b are also the same. Therefore, no current flows. V I

36 ConcepTest Which of the equations is valid for the circuit below? More Kirchhoff s Rules 1) 2 I 1 2I 2 = 0 2) 2 2I 1 2I 2 4I 3 = 0 3) 2 I 1 4 2I 2 = 0 4) I 3 4 2I = 0 5) 2 I 1 3I 3 6 = 0 1 W I 2 2 W 2 V 6 V 4 V I I W 3 W

37 ConcepTest Which of the equations is valid for the circuit below? More Kirchhoff s Rules 1) 2 I 1 2I 2 = 0 2) 2 2I 1 2I 2 4I 3 = 0 3) 2 I 1 4 2I 2 = 0 4) I 3 4 2I = 0 5) 2 I 1 3I 3 6 = 0 Eq. 3 is valid for the left loop: The left battery gives +2 V, then there is a drop through a 1 W resistor with current I 1 flowing. Then we go through the middle battery (but from + to!), which gives 4 V. Finally, there is a drop through a 2 W resistor with current I 2. I 2 1 W 2 W 2 V 6 V 4 V I I W 3 W

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