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ConcepTest PowerPoints Chapter 16 Physics: Principles with Applications, 7 th edition Giancoli 2014 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.

ConcepTest 16.1a Electric Charge I Two charged balls are repelling each other as they hang from the ceiling. What can you say about their charges? a) one is positive, the other is negative b) both are positive c) both are negative d) both are positive or both are negative 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. Follow-up: What does the picture look like if the two balls are oppositely charged? What about if both balls are neutral?

ConcepTest 16.1b Electric Charge II From the picture, what can you conclude about the charges? a) have opposite charges b) have the same charge c) all have the same charge d) one ball must be neutral (no charge) 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 16.2a Conductors I 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: a) positive b) negative c) neutral d) positive or neutral e) negative or neutral 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. remember the ball is a conductor! Follow-up: What happens if the metal ball is replaced by a plastic ball?

ConcepTest 16.2b Conductors II Two neutral conductors are connected by a wire and a charged rod is brought near, but does not touch. The wire is taken away, and then the charged rod is removed. What are the charges on the conductors? a) b) c) d) e) 0 + + 0 + + While the conductors are connected, positive charge will flow from the blue to the green ball due to polarization. Once disconnected, the charges will remain on the separate conductors even when the rod is removed. 0 0 Follow-up: What will happen when the conductors are reconnected with a wire???

ConcepTest 16.3a Coulomb s Law I What is the magnitude of the force F 2? F 1 = 3N F 2 =? Q Q a) 1.0 N b) 1.5 N c) 2.0 N d) 3.0 N e) 6.0 N

ConcepTest 16.3a Coulomb s Law I What is the magnitude of the force F 2? F 1 = 3N F 2 =? Q Q a) 1.0 N b) 1.5 N c) 2.0 N d) 3.0 N e) 6.0 N 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!!

ConcepTest 16.3b Coulomb s Law II F 1 = 3N F 2 =? Q Q If we increase one charge to 4Q, what is the magnitude of F 1? F =? 1 4Q Q F =? 2 a) 3/4 N b) 3.0 N c) 12 N d) 16 N e) 48 N

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

ConcepTest 16.3c Coulomb s Law III 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? a) 9 F b) 3 F c) F d) 1/3 F e) 1/9 F Originally we had: F before = k(q)(q)/d 2 = F Now we have: F after = k(q)(q)/(3d) 2 = 1/9 F? Q F Q d Q F Q? 3d Follow-up: What is the force if the original distance is halved?

ConcepTest 16.4a Electric Force I Two balls with charges +Q and +4Q are fixed at a separation distance of 3R. Is it possible to place another charged ball Q 0 on the line between the two charges such that the net force on Q 0 will be zero? a) yes, but only if Q 0 is positive b) yes, but only if Q 0 is negative c) yes, independent of the sign (or value) of Q 0 d) no, the net force can never be zero A positive charge would be repelled by both charges, so a point where these two repulsive forces cancel can be found. A negative charge would be attracted by both, and the +Q +4Q 3R same argument holds. Follow-up: What happens if both charges are +Q? Where would the F = 0 point be in this case?

ConcepTest 16.4b Electric Force II Two balls with charges +Q and +4Q are separated by 3R. Where should you place another charged ball Q 0 on the line between the two charges such that the net force on Q 0 will be zero? +Q +4Q a) b) c) d) e) R 2R 3R The force on Q 0 due to +Q is: F = k(q 0 )(Q)/R 2 The force on Q 0 due to +4Q is: F = k(q 0 )(4Q)/(2R) 2 Since +4Q is 4 times bigger than +Q, then Q 0 needs to be farther from +4Q. In fact, Q 0 must be twice as far from +4Q, since the distance is squared in Coulomb s Law.

ConcepTest 16.4c Electric Force III Two balls with charges +Q and 4Q are fixed at a separation distance of 3R. Is it possible to place another charged ball Q 0 anywhere on the line such that the net force on Q 0 will be zero? a) yes, but only if Q 0 is positive b) yes, but only if Q 0 is negative c) yes, independent of the sign (or value) of Q 0 d) no, the net force can never be zero A charge (positive or negative) can be placed to the left of the +Q charge, such that the repulsive force from the +Q charge cancels the attractive +Q 4Q 3R force from 4Q. Follow-up: What happens if one charge is +Q and the other is Q?

ConcepTest 16.5a Proton and Electron I A proton and an electron are held apart a distance of 1 m and then released. As they approach each other, what happens to the force between them? a) it gets bigger b) it gets smaller c) it stays the same By Coulomb s Law, the force between the two charges is inversely proportional to the distance squared. So, the closer they get to each other, the bigger the electric force between them gets! p e Follow-up: Which particle feels the larger force at any one moment?

ConcepTest 16.5b Proton and Electron II A proton and an electron are held apart a distance of 1 m and then released. Which particle has the larger acceleration at any one moment? a) proton b) electron c) both the same p e The two particles feel the same force. Since F = ma, the particle with the smaller mass will have the larger acceleration. This would be the electron.

ConcepTest 16.5c Proton and Electron III A proton and an electron are held apart a distance of 1 m and then let go. Where would they meet? a) in the middle b) closer to the electron s side c) closer to the proton s side By Newton s 3rd Law, the electron and proton feel the same force. But, since F = ma, and since the proton s mass is much greater, the proton s acceleration will be much smaller! Thus, they will meet closer to the proton s original position. p e Follow-up: Which particle will be moving faster when they meet?

ConcepTest 16.6 Forces in 2D a) b) c) Which of the arrows best represents the direction of the net force on charge +2Q d +Q d) e) +Q due to the other two charges? d +4Q

ConcepTest 16.6 Forces in 2D a) b) c) Which of the arrows best represents the direction of the net force on charge +2Q d +Q d) e) +Q due to the other two charges? d +4Q The charge +2Q repels +Q towards the right. The charge +4Q repels +Q upwards, but with a stronger force. Therefore, the net force is up and to +2Q the right, but mostly up. Follow-up: What happens if the yellow charge would be +3Q? +4Q

ConcepTest 16.7 Electric Field You are sitting a certain distance from a point charge, and you measure an electric field of E 0. If the charge is doubled and your distance from the charge is also doubled, what is the electric field strength now? a) 4 E 0 b) 2 E 0 c) E 0 d) 1/2 E 0 e) 1/4 E 0 Remember that the electric field is: E = kq/r 2. Doubling the charge puts a factor of 2 in the numerator, but doubling the distance puts a factor of 4 in the denominator, because it is distance squared!! Overall, that gives us a factor of 1/2. Follow-up: If your distance is doubled, what must you do to the charge to maintain the same E field at your new position?

ConcepTest 16.8a Field and Force I Between the red and the blue charge, which of them experiences the greater electric field due to the green charge? a) b) +1 +2 c) the same for both +1 +2 +1 +1 d d

ConcepTest 16.8a Field and Force I Between the red and the blue charge, which of them experiences the greater electric field due to the green charge? a) b) +1 +2 c) the same for both +1 +2 +1 +1 d d Both charges feel the same electric field due to the green charge because they are at the same point in space! E = k Q r 2

ConcepTest 16.8b Field and Force II Between the red and the blue charge, which of them experiences the greater electric force due to the green charge? a) b) +1 +2 c) the same for both +1 +2 +1 +1 d d

ConcepTest 16.8b Field and Force II Between the red and the blue charge, which of them experiences the greater electric force due to the green charge? a) b) +1 +2 c) the same for both +1 +2 +1 +1 d d The electric field is the same for both charges, but the force on a given charge also depends on the magnitude of that specific charge. F = qe

ConcepTest 16.9a Superposition I What is the electric field at the center of the square? -2 C b) a) c) d) e) E = 0-2 C

ConcepTest 16.9a Superposition I What is the electric field at the center of the square? -2 C b) a) c) d) e) E = 0-2 C For the upper charge, the E field vector at the center of the square points towards that charge. For the lower charge, the same thing is true. Then the vector sum of these two E field vectors points to the left. Follow-up: What if the lower charge was +2 C? What if both charges were +2 C?

ConcepTest 16.9b Superposition II What is the electric field at the center of the square? -2 C -2 C b) a) c) d) e) E = 0-2 C -2 C

ConcepTest 16.9b Superposition II What is the electric field at the center of the square? -2 C -2 C b) a) c) d) e) E = 0-2 C -2 C The four E field vectors all point outwards from the center of the square toward their respective charges. Because they are all equal, the net E field is zero at the center!! Follow-up: What if the upper two charges were +2 C? What if the right-hand charges were +2 C?

ConcepTest 16.9c Superposition III -Q +Q What is the direction of the electric field at the position of the X? +Q b) c) a) e) d)

ConcepTest 16.9c Superposition III -Q +Q What is the direction of the electric field at the position of the X? +Q b) c) a) e) d) The two +Q charges give a resultant E field that is down and to the right. The Q charge has an E field up and to the left, but smaller in magnitude. Therefore, the total electric field is down and to the right. Follow-up: What if all three charges reversed their signs?

ConcepTest 16.10 Find the Charges Two charges are fixed along the x-axis. They produce an electric field E directed along the negative y-axis at the indicated point. Which of the following is true? a) charges are equal and positive b) charges are equal and negative c) charges are equal and opposite d) charges are equal, but sign is undetermined e) charges cannot be equal The way to get the resultant PINK vector is to use the GREEN and BLUE vectors. These E vectors correspond to equal charges (because the lengths are equal) that are both negative (because their y E directions are toward the charges). Q 1 Q 2 x Follow-up: How would you get the E field to point toward the right?

ConcepTest 16.11 Uniform Electric Field In a uniform electric field in empty space, a 4 C charge is placed and it feels an electrical force of 12 N. If this charge is removed and a 6 C charge is placed at that point instead, what force will it feel? a) 12 N b) 8 N c) 24 N d) no force e) 18 N Since the 4 C charge feels a force, there must be an electric field present, with magnitude: E = F / q = 12 N / 4 C = 3 N/C Once the 4 C charge is replaced with a 6 C charge, this new charge will feel a force of: F = q E = (6 C)(3 N/C) = 18 N Q Follow-up: What if the charge is placed at a different position in the field?

ConcepTest 16.12a Electric Field Lines I a) What are the signs of the charges whose electric fields are shown at right? b) c) d) e) no way to tell

ConcepTest 16.12a Electric Field Lines I a) What are the signs of the charges whose electric fields are shown at right? b) c) d) e) no way to tell Electric field lines originate on positive charges and terminate on negative charges.

ConcepTest 16.12b Electric Field Lines II Which of the charges has the greater magnitude? a) b) c) Both the same

ConcepTest 16.12b Electric Field Lines II Which of the charges has the greater magnitude? a) b) c) Both the same The field lines are denser around the red charge, so the red one has the greater magnitude. Follow-up: What is the red/green ratio of magnitudes for the two charges?

ConcepTest PowerPoints Chapter 18 Physics: Principles with Applications, 7 th edition Giancoli 2014 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.

ConcepTest 18.1 Which is the correct way to light the lightbulb with the battery? Connect the Battery d) all are correct e) none are correct

ConcepTest 18.1 Which is the correct way to light the lightbulb with the battery? Connect the Battery d) all are correct e) none are correct 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 (c).

ConcepTest 18.2 Ohm s Law You double the voltage across a certain conductor and you observe the current increases three times. What can you conclude? a) Ohm s law is obeyed since the current still increases when V increases b) Ohm s law is not obeyed c) This has nothing to do with Ohm s law 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. Follow-up: Where could this situation occur?

ConcepTest 18.3a Two wires, A and B, are made of the same metal and have equal length, but the resistance of wire A is four times the resistance of wire B. How do their diameters compare? Wires I a) d A = 4 d B b) d A = 2 d B c) d A = d B d) d A = 1/2 d B e) d A = 1/4 d B The resistance of wire A is greater because its area is less than wire B. Since area is related to radius (or diameter) squared, the diameter of A must be two times less than B. R L A

ConcepTest 18.3b Wires II A wire of resistance R is stretched uniformly (keeping its volume constant) until it is twice its original length. What happens to the resistance? a) it decreases by a factor 4 b) it decreases by a factor 2 c) it stays the same d) it increases by a factor 2 e) it increases by a factor 4 Keeping the volume (= area x length) constant means that if the length is doubled, the area is halved. L Since R, this increases the resistance by four. A

ConcepTest 18.4 When you rotate the knob of a light dimmer, what is being changed in the electric circuit? Dimmer a) the power b) the current c) the voltage d) both a) and b) e) both b) and c)

ConcepTest 18.4 When you rotate the knob of a light dimmer, what is being changed in the electric circuit? Dimmer a) the power b) the current c) the voltage d) both a) and b) e) both b) and c) 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?

ConcepTest 18.5a 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 a) the 25 W bulb b) the 100 W bulb c) both have the same d) 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?

ConcepTest 18.5b 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 a) heater 1 b) heater 2 c) 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?

ConcepTest PowerPoints Chapter 19 Physics: Principles with Applications, 7 th edition Giancoli 2014 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.

ConcepTest 19.1a 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. a) 12 V b) zero c) 3 V d) 4 V e) 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

ConcepTest 19.1b Series Resistors II a) 12 V In the circuit below, what is the voltage across R 1? b) zero c) 6 V d) 8 V e) 4 V R 1 = 4 W R 2 = 2 W 12 V

ConcepTest 19.1b Series Resistors II a) 12 V In the circuit below, what is the voltage across R 1? b) zero c) 6 V d) 8 V e) 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, 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?

ConcepTest 19.2a Parallel Resistors I a) 10 A In the circuit below, what is the current through R 1? b) zero c) 5 A d) 2 A e) 7 A R 2 = 2 W R 1 = 5 W 10 V

ConcepTest 19.2a Parallel Resistors I a) 10 A In the circuit below, what is the current through R 1? b) zero c) 5 A d) 2 A e) 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

ConcepTest 19.2b 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 a) increases b) remains the same c) decreases d) 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?

ConcepTest 19.3a Short Circuit Current flows through a lightbulb. If a wire is now connected across the bulb, what happens? a) all the current continues to flow through the bulb b) half the current flows through the wire, the other half continues through the bulb c) all the current flows through the wire d) 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?

ConcepTest 19.3b 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: a) glow brighter than before b) glow just the same as before c) glow dimmer than before d) go out completely e) 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?

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

ConcepTest 19.4b The three light bulbs 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 a) twice as much b) the same c) 1/2 as much d) 1/4 as much e) 4 times as much A 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 B C 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?

ConcepTest 19.5a More Circuits I What happens to the voltage across the resistor R 1 when the switch is closed? The voltage will: a) increase b) decrease c) stay the same R 1 V S R 2 R 3

ConcepTest 19.5a More Circuits I What happens to the voltage across the resistor R 1 when the switch is closed? The voltage will: a) increase b) decrease c) 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?

ConcepTest 19.5b More Circuits II What happens to the voltage across the resistor R 4 when the switch is closed? a) increases b) decreases c) stays the same

ConcepTest 19.5b More Circuits II What happens to the voltage across the resistor R 4 when the switch is closed? a) increases b) decreases c) stays the same We just saw that closing the switch causes an increase in the voltage across R 1 (which is V AB ). The voltage of the battery is constant, so if V AB increases, then V BC must decrease! A B C Follow-up: What happens to the current through R 4?

ConcepTest 19.6 Which resistor has the greatest current going through it? Assume that all the resistors are equal. Even More Circuits a) R 1 b) both R 1 and R 2 equally c) R 3 and R 4 d) R 5 e) all the same The same current must flow through left and right combinations of resistors. On the LEFT, the current splits equally, so I 1 = I 2. On the RIGHT, more current will 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?

ConcepTest 19.7 What is the current in branch P? Junction Rule a) 2 A b) 3 A c) 5 A d) 6 A e) 10 A 5 A 8 A P 2 A

ConcepTest 19.7 What is the current in branch P? Junction Rule a) 2 A b) 3 A c) 5 A d) 6 A e) 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. S 5 A 8 A junction 2 A P 6 A

ConcepTest 19.8 Kirchhoff s Rules The lightbulbs in the a) both bulbs go out circuit are identical. When b) intensity of both bulbs increases the switch is closed, what c) intensity of both bulbs decreases happens? d) A gets brighter and B gets dimmer e) nothing changes

ConcepTest 19.8 The lightbulbs in the circuit are identical. When the switch is closed, what happens? Kirchhoff s Rules a) both bulbs go out b) intensity of both bulbs increases c) intensity of both bulbs decreases d) A gets brighter and B gets dimmer e) 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

ConcepTest 19.9 Wheatstone Bridge An ammeter A is connected a) I between points a and b in the b) I/2 circuit below, in which the four c) I/3 resistors are identical. The current d) I/4 through the ammeter is: e) 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. a b V I

ConcepTest 19.10 Which of the equations is valid for the circuit below? More Kirchhoff s Rules a) 2 I 1 2I 2 = 0 b) 2 2I 1 2I 2 4I 3 = 0 c) 2 I 1 4 2I 2 = 0 d) I 3 4 2I 2 + 6 = 0 e) 2 I 1 3I 3 6 = 0 1 W I 2 2 W 2 V 6 V 4 V I I 1 3 1 W 3 W

ConcepTest 19.10 Which of the equations is valid for the circuit below? More Kirchhoff s Rules a) 2 I 1 2I 2 = 0 b) 2 2I 1 2I 2 4I 3 = 0 c) 2 I 1 4 2I 2 = 0 d) I 3 4 2I 2 + 6 = 0 e) 2 I 1 3I 3 6 = 0 Eqn. 3 is valid for the left loop: The left battery gives +2V, then there is a drop through a 1W resistor with current I 1 flowing. Then we go through the middle battery (but from + to!), which gives 4V. Finally, there is a drop through a 2W resistor with current I 2. I 2 1 W 2 W 2 V 6 V 4 V I I 1 3 1 W 3 W

ConcepTest 19.11a What is the equivalent capacitance, C eq, of the combination below? Capacitors I a) C eq = 3/2 C b) C eq = 2/3 C c) C eq = 3 C d) C eq = 1/3 C e) C eq = 1/2 C o C eq C C C o

ConcepTest 19.11a What is the equivalent capacitance, C eq, of the combination below? Capacitors I a) C eq = 3/2 C b) C eq = 2/3 C c) C eq = 3 C d) C eq = 1/3 C e) C eq = 1/2 C The 2 equal capacitors in series add up as inverses, giving 1/2 C. These are parallel to the first one, which add up directly. Thus, the total equivalent capacitance is 3/2 C. o C eq o C C C

ConcepTest 19.11b How does the voltage V 1 across the first capacitor (C 1 ) compare to the voltage V 2 across the second capacitor (C 2 )? Capacitors II a) V 1 = V 2 b) V 1 > V 2 c) V 1 < V 2 d) all voltages are zero The voltage across C 1 is 10 V. The combined capacitors C 2 +C 3 are parallel to C 1. The voltage across C 2 +C 3 is also 10 V. Since C 2 and C 3 are in series, their voltages add. Thus the voltage across C 2 and C 3 each has to be 5 V, which is less than V 1. Follow-up: What is the current in this circuit?

ConcepTest 19.11c How does the charge Q 1 on the first capacitor (C 1 ) compare to the charge Q 2 on the second capacitor (C 2 )? Capacitors III a) Q 1 = Q 2 b) Q 1 > Q 2 c) Q 1 < Q 2 d) all charges are zero We already know that the voltage across C 1 is 10 V and the voltage across C 2 and C 3 each is 5 V. Since Q = CV and C is the same for all the capacitors, then since V 1 > V 2 therefore Q 1 > Q 2.