* AP PHYSICS B Circuits and Resistance Teacher Packet AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not involved in the production of this material. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance Objective To review the student on the concepts, processes and problem solving strategies necessary to successfully answer questions on circuits and resistance. Standards Circuits is addressed in the topic outline of the College Board AP* Physics Course Description Guide as described below. III. Electricity and Magnetism C. Electric Circuits 1. Current, Resistance, Power 2. Steady State Direct Current Circuits with batteries and resistors only 3. Capacitors in Circuits a. Steady State AP Physics Exam Connections Topics relating to circuits and resistance are tested every year on the multiple choice and in most years on the free response portion of the exam. The list below identifies free response questions that have been previously asked over circuits. These questions are available from the College Board and can be downloaded free of charge from AP Central. http://apcentral.collegeboard.com. 2007 2003 2002 2001 Free Response Questions Question 3 2007 Form B Question 3 Question 2 2003 Form B Question 2 Question 3 Question 5 ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance What I Absolutely Have to Know to Survive the AP* Exam • • • • • • Draw diagrams for series and parallel circuits using accepted symbols, including proper placement of ammeters and voltmeters. Calculate total resistance and total potential difference for series, parallel, and combination circuits. Determine power in a circuit. Find the charge stored in a capacitor (in steady state). Determine the short term and long term (steady state) behavior of a capacitor. Apply Kirchhoff’s Rules (especially in multi-loop circuits). Key Formulas and Relationships Δq Δt V = IR I= Current is the rate at which charge passes a point in an electrical circuit. (Amperes A) The potential difference across a resistor is equal to the current through the resistor times the size of the resistor. (Volts V) R= ρL Resistance of a resistor is proportional to the length L and inversely proportional to the A Req = R1 + R2 + R3 + ... cross sectional area A. The resistance is also proportional to the resistivity ρ (Ωm) which is a characteristic of the material. The equivalent resistance of a series combination of resistors equals the sum of the individual resistors in the combination. (Ohms Ω) 1 1 1 1 = + + + ... The inverse of the equivalent resistance of a parallel combination of resistors equals Req R1 R2 R3 the sum of the inverse of each of the individual resistors. (Ohms Ω) ΣI int o a node = ΣI out of a node Kirchoff's First law (conservation of charge) the current that flows into a node must equal the current that flows out of a node. Kirchoff's Second Law (conservation of energy) the sum of the voltage changes around any closed path within a circuit must equal zero. Power provided or dissipated to a circuit by a circuit element is equal to the current ΣΔV = 0 P = IV through the element times the voltage across the element. (Watts W or P = I 2R = E = Pt V2 R J ) S Power dissipated by a resistor. When power is dissipated by a resistor, the "lost energy" is converted into heat or light. Energy is power times time. (Joules J) ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance Important Concepts • The charge on one electron (or one proton) is called the elementary charge (e) and is equal to 1.6×10-19C. Hence, it takes 6.25×1018 electrons to generate one coulomb of charge. 1 Ampere is the amount of current that moves one Coulomb of charge through the circuit each second. • A node is a junction where two wires meet within a circuit. A branch is a section of circuitry between two nodes. The current in a particular branch will be the same everywhere within the branch. • Conventional current—tracks the flow of positive charge. Electron current tracks the path of the electron flow and it is opposite to conventional current flow. • Ohm’s Law: V = IR V is the potential difference or voltage, measured in volts. I is current, measured in amps which is equal to coulombs/second. R is resistance, measured in ohms and symbolized by the Greek letter omega, Ω • RESISTORS IN SERIES Note that there is only one path for the current to travel; therefore, the current through each of the resistors is the same. V Requivalent = V1 V2 V3 + + R1 R2 R3 Also, the voltage “drops” across the resistors must add up to the total voltage supplied by the battery: VTotal = V1 + V2 + V3 Ohm’s Law states that V=IR, then VTotal = IR1 + IR2 + IR3 BUT, Ohm’s Law must also be satisfied for the complete circuit: VTotal = IRequivalent Set these two equations equal to each other since they each equal Vtotal, ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance IRequivalent = IR1 + IR2 + IR3 Factor out the current I. IRequivalent = I ( R1 + R2 + R3 ) Since the current is the same through each resistor in the circuit, it divides out, and we see that Requivalent is simply the sum of the resistors connected in series: Req = R1 + R2 + R3 • RESISTORS IN PARALLEL Note that there are now three paths for the current to travel between points A and B. The current is split along these three paths as it travels from point A to point B. Furthermore the current is the same at both points A and B so, the sum of the currents through the three branches is the same as the current at both point A [where the current splits] and at point B [where the current reunites].This means that ITotal = I1 + I 2 + I 3 At both points A and B, the potential difference must be the same, so between points A and B the potential must be the same. That is, each of the three resistors in the parallel circuit must have the same voltage across it: VTotal = V1 = V2 = V3 By Ohm’s Law , V = IR, this is equivalent to V Requivalent ® = V1 V2 V3 + + R1 R2 R3 Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance Factor out the potential difference, V. ⎛ 1 V⎜ ⎜ Requivalent ⎝ ⎞ ⎛ 1 1 1 ⎞ + ⎟ ⎟⎟ = V ⎜ + ⎝ R1 R2 R3 ⎠ ⎠ Since the potential difference, V, is the same across each resistor in the circuit, it divides out, and we see that 1 1 1 1 = + + Requivalent can be found by Req R1 R2 R3 • SUMMARY Series Parallel 1 1 1 1 = + + ... Req R1 R2 R3 Req = R1 + R2 + R3... • VT = V1 + V2 + V3... VT = V1 = V2 = V3... IT = I1 = I2 = I3... IT = I1 + I2 + I3... Req>R1, R2, R3… Req<R1, R2, R3… COMPLEX CIRCUIT DIAGRAMS Complex circuits involve resistors in series in combination with resistors in parallel. Steps in simplifying complex circuits: 1. Calculate the equivalent resistance of each set of resistors in parallel. 2. Calculate the total resistance of the circuit. 3. Calculate the total current using Ohm’s Law. 4. Apply Ohm’s Law to each individual series resistor to determine the individual resistor’s voltage drop. 5. Apply Ohm’s Law to each resistor in the parallel group to find the current in that branch. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance • EXAMPLE Three 5-Ω resistors are connected in parallel. This parallel combination is in series with a 7-Ω, and two 10-Ω resistors as pictured below. Calculate the total resistance in the circuit. Use Ohm’s Law to calculate the total current in the circuit. Calculate the voltage drop across each resistor. o First, calculate the equivalent resistance of the parallel network of resistors. 1 Req ( parallel ) = 1 1 1 3 5 + + = ∴ Req = = 1.67 5 5 5 5 3 RT = Req ( parallel ) + R2 + R3 + R4 RT = 1.67Ω + 7Ω + 10Ω + 10Ω = 28.67Ω o Second, redraw the circuit so that the equivalent resistance is in series with the remaining series resistors. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance Use Ohm's Law to solve for the current: V = IR so I = V R 36V = 1.26A for each resistor in series, but now we need to split this 1.26 A across 28.67 Ω each of the resistors in the parallel branch. This means we first need to know the voltage I= across each resistor in our new drawing. V = IR V = 1.26 A × 1.67 Ω = 2.1V V = 1.26 A × 7 Ω = 8.8V across the 7Ω resistor V = 1.26 A × 10 Ω = 12.6V across EACH of the 10 Ω resistors This gives a total voltage of [2.1 +8.8+2(12.6)]=36.1 V (due to rounding) as expected since we have a 36 V battery. Now we can determine the current through EACH resistor in the parallel branch: V 2.1V I= = = 0.42A across each of the 5Ω resistors R 5Ω The grand total of all of the current through EACH resistor in the parallel branch should match our 1.26 A total current: (3 × 0.42A) = 1.26A • Meters You must know how to hook up and use ammeters and voltmeters for the exam. Ammeters must be placed in series with the component you are measuring the current through. In theory anywhere in a circuit that is in series with a component should have identical current, due to approximations like ideal wires. Proper lab procedure, however, is to place the ammeter just before the component you wish to measure the current flowing through it. Voltmeters must be hooked up in parallel. This allows them to touch a circuit at two different points without any current passing through them so that they do not affect the circuit they are measuring. • Resistance of a wire You must understand the relationship between resistance of a material, its length and cross sectional area. Fat wires have less resistance to electric current in a manner similar to wide highways with many lanes having less resistance to the flow of traffic. Shorter wires have less resistance as well. This is typically tested in a multiple choice question of ρL variable manipulation. R= A ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance • Batteries An ideal battery can maintain a constant voltage when the current flowing through it is very small or very large. The exam often assumes batteries to be ideal. Real batteries, however, do not maintain a constant voltage when you begin to draw large currents from them. They act as if there is a small resistor inside them impeding the current. You can tell when a real battery is going bad by looking at the voltage between the terminals when it is not under load and then again when it is under load. If the numbers change a lot then the battery has a large internal resistance. The idea of using a voltmeter on a battery gives us an equation V A − VB = ε − Ir where the left hand side of the equation is the voltage measured under load, ε is the voltage when it is not under load and r is the internal resistance of the battery. A very good battery like a car battery can get ΔV very close to ε so the approximation of an ideal battery is not a bad one. • Kirchhoff’s Rules for Multi-loop Circuits 1. The net (total) current entering a node (junction) must equal the net current out of a node (junction). This is sometimes called the junction rule, and is a statement of conservation of charge. I in = I out 2. The sum of the potential rises and drops (voltage differences) around a closed loop of a circuit must equal zero. This is sometimes called the loop rule, and is a statement of conservation of energy. ΣV = 0 a. If we pass a battery from negative to positive, we say that there is a rise in potential, +ε. If we pass a battery from positive to negative, we say that there is a drop in potential, - ε. If we pass a resistor against the direction of our arbitrarily chosen current, we say there is a rise in potential across the resistor, + IR. If we pass a resistor in the direction of our arbitrarily chosen current, we say there is a drop in potential across the resistor, - IR. b. c. d. • Capacitors 1. An empty capacitor does not resist the flow of current, and thus acts like a wire. 2. A capacitor which is full of charge will not allow current to flow, and thus acts like a broken wire. Capacitors in DC circuits are used to store energy. They do this when a certain amount of charge is placed on one plate while an equal amount of opposite charge is placed on the other plate. This creates an electric field between the plates. The energy is stored in the electric field. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance CP = C1 + C2 + C3 + ... Capacitors in parallel add like resistors in series. 1 1 1 1 = + + + ... Cs C1 C2 C3 Capacitors in series add like resistors in parallel. q V 1 1 1 Q2 2 U E = CV = QV = 2 2 2 C q = CV or C = Capacitance is the ratio of charge to voltage. (Farads F or C ) V Energy stored in a capacitor. (Joules J) Capacitance can be expressed as C = Q/V or C = KA/d where: • • • • • Q is the charge held on the capacitor. V is the voltage across the conductive plates. A is the area of the conducting plate. d is the distance between the plates. K represents the dielectric constant (measures how well the insulator does its job). On the AP Exam k=1 since it is normally air or vacuum. As you can see from looking at the formula C = κA d , if you pull the two plates of a capacitor apart you decrease its capacitance. It takes work to accomplish this if there is charge on the capacitor since you are moving the plates against the electric field. If you decrease the area of the plates it has a similar effect of decreasing the capacitance. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance Free Response Question 1 (15 pts) S _ 2Ω R + 12 V 4Ω The circuit shown above includes a 12-volt battery, a resistor of unknown resistance R, an open switch, a 2-Ω resistor, and a 4-Ω resistor. The switch is then closed so that a total current of 3.0 A flows in the circuit. A. Determine the current passing through the unknown resistor. (4 points max) V 12 volts I 2Ω and 4Ω = = = 2.0 A R 2Ω + 4Ω I R = I total − I 2 Ω and 4Ω = 3.0 A − 2.0 A =1.0 A 1 point for the application of Ohm’s law 1 point for finding the current through the 2Ω and 4Ω resistors by dividing by 6Ω 1 point for determining the difference between the total current and the current through the two resistors 1 point for the correct answer with correct units ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance B. Determine the value of the unknown resistor R. (1 point max) V 12 volts R= = = 12 Ω 1.0 A I 1 point for the correct answer with correct units consistent with the answer in part (A) C. Determine the heat dissipated by resistor R in one minute. 1 point for a correct equation for power (4 points max) V 2 144 volts = = 12 W P = IV = I R = 12Ω R 2 P = IV = (1.0 A )(12 volts ) = 12 watts = 12 J⎞ ⎛ Heat energy = Pt = ⎜ 12 ⎟ ( 60 s ) = 720 J s⎠ ⎝ ® J s 1 point for the correct value for power 1 point for the correct relationship between power and heat energy 1 point for the correct answer for heat energy with correct units and reasonable number of significant digits Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance The resistor R is removed and replaced with a 6μF parallel-plate capacitor, as shown on the figure below. The switch is then closed. S _ – – – – – – – 2Ω + 12 V ++ + + + + + 4Ω D. Determine the total current in the 2-Ω and 4-Ω resistors after the switch has been closed for a long time. (3 points max) After a long time, the capacitor is full, so no current flows through it, and the current flows through the 4Ω and 2Ω resistors. V 12V Io = = =2A Rtot 6Ω 1 point for recognition that no current flows through a full capacitor 1 point for the correct application of Ohm’s law 1 point for the correct answer including correct units E. Determine the total amount of energy stored in the capacitor. (3 points max) 1 1 2 U = CV 2 = ( 6 μ F )(12 V ) = 432 μ J 2 2 1 point for recognition that the voltage drop across the capacitor is 12 V 1 point for the correct application of the equation for the energy stored in a capacitor 1 point for the correct answer including correct units ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance Question 2 (15 pts) A student is given three identical resistors, five 1.5 Volt “D” cell batteries, a voltmeter, an ammeter, and some wires. She is instructed to complete a circuit using one resistor such that batteries can be added in series one at a time and in which both the current and voltage can be properly measured. A. In the box below, draw a schematic diagram of the circuit when 2 batteries are connected in series with one resistor, and with both the ammeter and voltmeter connected correctly. 1 point For two batteries connected correctly in series (3 points max) 1 point For a complete series circuit with resistor 1 point For proper placement of both the voltmeter and ammeter After completing the circuit with one battery, the student measures the voltage and current for the resistor, then continues adding batteries one at a time and makes the following measurements: ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance B. On the grid below, plot the necessary data that illustrates the mathematical relationship between voltage and current. Use appropriate graphing techniques. 1 point For both axes properly labeled with units and using an appropriate scale (3 point max) see graph 1 point For plotting the points correctly 1 point For drawing an appropriate best-fit line through the points C. From your graph, determine the value of the resistor. 1 point For using the slope to find the resistance (3 points max) m= y2 − y1 6.5 V − 2.1 V 4.4 V = = = 10 Ω x2 − x1 0.66 A − 0.22 A 0.44 A 1 point For using points on the best fit line 1 point For the correct answer with units ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance a R1 R2 ε1= 4.5 V R3 b ε2= 3 V D. The student arranges the three identical resistors and the five batteries in the arrangement shown above such that there are neither series nor parallel combinations of resistors. She then measures the current through each resistor. Determine I1, I2, and I3. (6 points max) Junction Equation for A: ΣI in junction a = ΣI out junction a 1 point For stating that the net current into a junction must equal the net current coming out of the junction I1 + I 3 = I 2 Loop Equations Loop I: traversing clockwise from 4.5V battery through R1 , junction A, and R 2 4.5 − 10 I1 − 10 I 2 = 0 ⇒ 10 I1 + 10 I 2 = 4.5 I1 + I 2 = 0.45 Loop II: traversing counterclockwise from 3V battery through R 3 , junction A, and R 2 3 − 10 I 3 − 10 I 2 = 0 ⇒ 10 I 3 + 10 I 2 = 3 I 2 + I 3 = 0.3 1 point For stating that the sum of the voltage differences around a closed loop must equal zero 1 point For writing the potential rises and drops around a closed loop either CW or CCW 1 point For determining three equations with 3 unknowns 1 point For attempting to simultaneously solve the three equations for the values of the three ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance Solve simulataneously I1 + I 3 = I 2 Equation One I1 + I 2 = 0.45 Equation Two I 2 + I 3 = 0.3 Equation Three Substituting Equation one into Equation two I1 + ( I1 + I 3 ) = 0.45 ⇒ 2 I1 + I 3 = 0.45 ⇒ I1 + 0.5 I 3 = 0.225 currents by substituting the values of the resistors and the emfs of the batteries in the equations 1 point For the correct answers including correct units or answers consistent with part C Substituting Equation one into Equation three ( I1 + I3 ) + I 3 = 0.3 ⇒ I1 + 2 I 3 = 0.3 I1 + 0.5 I 3 = 0.225 − ( I1 + 2 I 3 = 0.3) −1.5 I 3 = −0.075 ⇒ I 3 = 0.5 A I 2 + I 3 = 0.3 ⇒ I 2 = 0.25 A I1 + I 2 = 0.45 ⇒ I1 = 0.20 A ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance Multiple Choice Questions 1-3 1. What is the ε of the battery? A) 3.0 V B) 4.0 V C) 12.0 V D) 24.0 V E) 48.0 V ε = IR and I = 4 A Rs = R1 + R2 = 4Ω + 2Ω = 6Ω 1 1 1 1 1 2 = + = + = RP R1 R2 6Ω 6Ω 6Ω C 6Ω = 3Ω 2 ε = IR = ( 4 A )( 3Ω ) = 12 V Req = 2. What is the power dissipated by the 4.0 Ω resistor ? A) B) C) D) E) 6.0 W 16 W 24 W 48 W 64 W At junction X, the current splits in half so that 2 A flows through the 4Ω B P = IV = V2 = I 2R R P = ( 2 A ) ( 4Ω ) = 16 W 2 ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance 3. What is the voltage drop across the 2.0 Ω resistor? A) B) C) D) E) 2.0 V 4.0 V 6.0 V 8.0 V 12.0 V B V = IR = ( 2 A)( 2Ω ) = 4 V 4. Which of the ammeters shown in the diagram above will correctly indicate the current through the 5Ω resistor? A) B) C) D) E) 1 and 2 only 2 and 3 only 2 and 4 only 2 only 3 only C Ammeters must be placed in series with the 5 Ω resistor, hence, 2 and 4 only will measure the current through the resistor. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance 5. Which of the following statements are true concerning the currents in resistors R1, R2, and R3? I. I1 > I2 II. I1 > I3 III. I1< I2 + I3 A) B) C) D) E) I only II only III only I and II only I, II, & III D All the current flows through resistor R1 and portions of the total current flow through resistors R2 and R3, such that I1 =I2+ I3. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance Questions 6-8 refer to the following circuits. Assume all the light bulbs and all the sources are identical. I II III 6. Which of the circuits would draw the most total current? If two or more circuits draw the same total current, and this current is the largest, choose all the circuits in which this is the case. A) B) C) D) E) I only II only III only I and III only I, II, and III C Identical bulbs connected in parallel will have less total resistance than in series, causing more current to be drawn in the circuit. 7. In which of the circuits would the light bulbs be the least bright? If two or more circuits are the same brightness, and are also the least bright, choose all the circuits in which this is the case. A) B) C) D) E) I only II only III only I and III only I, II, and III B Each bulb would get the least amount of current in circuit II since they are in series, and the total resistance is the greatest of the three circuits. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance 8. In which of the circuits would the light bulbs be the brightest? If two or more circuits are the same brightness, and are also the brightest, choose all the circuits in which this is the case. A) B) C) D) E) I only II only III only I and III only I, II, and III D In circuits I and III, each bulb has the same voltage (from the battery) across it, and is the same resistance. Thus, each bulb will have the same current passing through it, giving equal brightness, as well as the greatest brightness. 9. A 1.5-volt battery is connected to a circuit in such a way that 4 Coulombs of charge pass a point in the circuit in a time of 2 seconds. The current flowing in the circuit is A) B) C) D) E) 2A 3A 4A 6A 8A Current is defined as the amount of charge that passes a given point per unit time: A ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance 10. The internal resistance of the battery in the circuit shown below is most nearly 2A 6Ω A) B) C) D) E) 20V 4Ω 7Ω 8Ω 10 Ω 12 Ω A By Ohm’s law, 2 A = ® 20V , so the internal resistance R = 4 Ω. (6Ω + R ) Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance 11. In which of the circuits below are the ammeter and voltmeter correctly connected in order to measure current through the light bulb and voltage across the light bulb? A) V A B) V A C) A V D) A V E) V C A The ammeter should be connected in series with the light bulb, and the voltmeter should be connected in parallel with the light bulb. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance Questions 12-14 refer to the circuit diagram shown below. The switch is initially open and the ammeter labeled A2 indicates the current I2 through the 3-Ω light bulb is 3 A. A1 A3 3.0 Ω R2 9.0 V A2 12. The power dissipated in the 3-Ω light bulb is A) B) C) D) E) 1W 3W 9W 18 W 27 W E 13. The switch is now closed. The first ammeter indicates a current of 9.0 A. Calculate the current I3 through the third ammeter, A3: A) B) C) D) E) 1.0 A 3.0 A 6.0 A 8.0 A 9.0 A C The current through the 3-Ω bulb must still be 3.0 A, so the remainder of the 9.0-A total current must pass through the other bulb and the ammeter A3. Thus, the current through A3 is 9.0 A - 3.0 A = 6.0 A. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance 14. The value of the resistance R is A) B) C) D) E) 1.0 Ω 1.5 Ω 3.0 Ω 6.0 Ω 9.0 Ω B 15. The equivalent resistance in the figure below is A) B) C) D) E) less than 2.0 Ω between 2.0 and 4.0 Ω between 4.0 and 6.0 Ω between 6.0 and 10.0 Ω greater than 10.0 Ω C The 6.0-Ω and 10.0-Ω resistors are in series, so their equivalent resistance is 16.0-Ω. This resistance is in parallel with the 4.0-Ω resistor, giving 16/5 Ω for the parallel combination. Adding this value to the 2.0-Ω resistor gives between 4.0-Ω and 6.0-Ω for the total equivalent resistance. ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org Circuits and Resistance 16. A copper wire has a cross-sectional area, A and a length, L. Which of the following would reduce the resistance of the wire by a factor of four? ρ A L A) B) C) D) E) double both the length and the diameter double only the length double only the diameter reduce the length by half reduce both the length and the diameter by half C R= ρL A Doubling the diameter will reduce the resistance of the wire by a factor of four. 17. A parallel-plate capacitor has a capacitance Co. A second parallel-plate capacitor has plates with three times the area and half the separation. The capacitance of the second capacitor is most nearly F) G) H) I) J) ½ Co 3/2 Co Co 3 Co 6 Co E C= Q ε0 A 3 = so C = C 0 = 6C 0 V d .5 ® Copyright © 2009 Laying the Foundation , Inc., Dallas, TX. All rights reserved. Visit: www.layingthefoundation.org