This preview has intentionally blurred parts. Sign up to view the full document

View Full Document

Unformatted Document Excerpt

6:51 MasteringPhysics 10/19/08 PM Assignment Display Mode: View Printable Answers [P Physys 202 Fall08 HW5 Due at 11:00pm on Wednesday, October 15, 2008 View Grading Details Problem 26.36 Description: (a) Determine the currents I_1, I_2, and I_3 in the figure. Assume the internal resistance of each battery is 1.0Omega . (b) What is the terminal voltage of the 6.0-V battery? Part A Determine the currents , , and in the figure. Assume the internal resistance of each battery is 1.0 . Express your answer using two significant figures. Enter your answers numerically separated by commas. ANSWER: , , = Part B What is the terminal voltage of the 6.0battery? Express your answer using two significant figures. ANSWER: = Equivalent Resistance Description: Find the equivalent resistance of a network of resistors with series and parallel connections. The network geometry gets progressively more complicated by adding more resistors. Consider the network of four resistors shown in the diagram, where . The resistors are connected to a constant voltage of magnitude . = 2.00 , = 5.00 , = 1.00 , and = 7.00 Part A Find the equivalent resistance Hint A.1 of the resistor network. How to reduce the network of resistors http://session.masteringphysics.com/myct Page 1 of 12 MasteringPhysics 10/19/08 6:51 PM The network of resistors shown in the diagram is a combination of series and parallel connections. To determine its equivalent resistance, it is most convenient to reduce the network in successive stages. First compute the equivalent resistance of the parallel connection between the resistors and , and imagine replacing the connection with a resistor with such resistance. The resulting network will consist of three resistors in series. Then find their equivalent resistance, which will also be the equivalent resistance of the original network. Part A.2 Find the resistance equivalent to and and . Find the equivalent resistance of the parallel connection between the resistors Hint A.2.a Two resistors in parallel Consider two resistors of resistance resistance and that are connected in parallel. They are equivalent to a resistor with , which satisfies the following relation: . Express your answer in ohms. ANSWER: = If you replace the resistors consist of three resistors , and , and with an equivalent resistor with resistance , the resulting network will connected in series. Their equivalent resistance is also the equivalent resistance of the original network. Hint A.3 Three resistors in series Consider three resistors of resistance resistance , which is given by . Express your answer in ohms. ANSWER: = , , and that are connected in series. They are equivalent to a resistor with Part B Two resistors of resistance = 3.00 = 3.00 and = 3.00 are added to the network, and an additional resistor of resistance is connected by a switch, as shown in the diagram.. of the new resistor network Find the equivalent resistance when the switch is open. Hint B.1 How to reduce the extended network of resistors , which can be ignored then. As you did in Part A, reduce is in series with the resistors and , while the new Since the switch is open, no current passes through the resistor the network in successive stages. Note that the new resistor resistor Part B.2 is in series with . , , and Find the resistance equivalent to Find the resistance equivalent to the resistor connection with and and , , and . Part B.2.a Find the resistance equivalent to Find the resistance equivalent to the connection between . Hint B.2.a.i Two resistors in series Consider two resistors of resistance resistance , which is given by , and that are connected in series. They are equivalent to a resistor with http://session.masteringphysics.com/myct Page 2 of 12 MasteringPhysics 10/19/08 6:51 PM . Express your answer in ohms. ANSWER: = and with their equivalent resistor (of resistance ), the resistor will result in If you replace the resistors parallel with . Hint B.2.b Two resistors in parallel Consider two resistors of resistance resistance and that are connected in parallel. They are equivalent to a resistor with , which satisfies the following relation: . Express your answer in ohms. ANSWER: = If you replace the resistors consist of four resistors , , , and , with an equivalent resistor with resistance , and , the resulting network will all connected in series. Their equivalent resistance is also the equivalent resistance of the original network. Hint B.3 Four resistors in series Consider four resistors of resistance resistance , which is given by . Express your answer in ohms. ANSWER: = , , , and that are connected in series. They are equivalent to a resistor with Part C Find the equivalent resistance Hint C.1 of the resistor network described in Part B when the switch is closed. How to reduce the network of resistors when the switch is closed ; therefore the resistor must be included in the calculation is no longer connected in series with the is in parallel with and their equivalent When the switch is closed, current passes through the resistor of the equivalent resistance. Also when the switch is closed, the resistor resistors and , as was the case when the switch was open. Instead, now and . and resistor will be in series with Part C.2 Find the resistance equivalent to Find the equivalent resistance of the parallel connection between the resistors and . Hint C.2.a Two resistors in parallel Consider two resistors of resistance resistance and that are connected in parallel. They are equivalent to a resistor with , which satisfies the following relation: . Express your answer in ohms. ANSWER: = If you replace the resistors and with their equivalent resistor (of resistance ), and the resistors , and with their equivalent resistor (of resistance resistors , , , and ), calculated in Part B, the resulting network will consist of four all connected in series. Their equivalent resistance is also the equivalent resistance of the original network. Hint C.3 Four resistors in series http://session.masteringphysics.com/myct Page 3 of 12 MasteringPhysics 10/19/08 6:51 PM Consider four resistors of resistance resistance , which is given by , , , and that are connected in series. They are equivalent to a resistor with . Express your answer in ohms. ANSWER: = Problem 26.24 Description: A R_1 and a R_2 resistor are connected in parallel; this combination is connected in series with a R_3 resistor. (a) If each resistor is rated at P (maximum without overheating), what is the maximum voltage that can be applied across the whole network? A 2.6 Part A If each resistor is rated at 0.6 whole network? Express your answer using two significant figures. ANSWER: = (maximum without overheating), what is the maximum voltage that can be applied across the and a 3.8 resistor are connected in parallel; this combination is connected in series with a 1.6 resistor. Problem 26.25 Description: Consider the network of resistors shown in the figure. Answer qualitatively: (a) What happens to the voltage across each resistor when the switch S is closed? (b) ... (c) ... (d) ... (e) What happens to the current through each when the switch is ... Consider the network of resistors shown in the figure. Answer qualitatively: Part A What happens to the voltage across each resistor when the switch is closed? ANSWER: increases. decreases. remains the same. Part B ANSWER: increases. decreases. remains the same. Part C ANSWER: increases. decreases. remains the same. Part D ANSWER: increases. http://session.masteringphysics.com/myct Page 4 of 12 MasteringPhysics 10/19/08 6:51 PM decreases. remains the same. Part E What happens to the current through each when the switch is closed? ANSWER: increases. decreases. remains the same. Part F ANSWER: increases. decreases. remains the same. Part G ANSWER: increases. decreases. remains the same. Part H ANSWER: increases. decreases. remains the same. Part I What happens to the power output of the battery when the switch is closed? ANSWER: The power output increases. The power output decreases. The power output remains the same. Part J Let the switch. Enter your answers numerically separated by commas. ANSWER: , , = 126 and = 22.6 . Determine the current through each resistor before and after closing Part K Enter your answers numerically separated by commas. ANSWER: , , , = Part L Are your qualitative predictions confirmed? ANSWER: Yes No Problem 26.12 Description: Eight identical bulbs are connected in series across a V_tot line. (a) What is the voltage across each bulb? (b) If http://session.masteringphysics.com/myct Page 5 of 12 MasteringPhysics 10/19/08 6:51 PM the current is I, what is the resistance of each bulb? (c) What is the power dissipated in each? Eight identical bulbs are connected in series across a 160 Part A What is the voltage across each bulb? Express your answer using two significant figures. ANSWER: = line. Part B If the current is 0.43 , what is the resistance of each bulb? Express your answer using two significant figures. ANSWER: = Part C What is the power dissipated in each? ANSWER: = Power Dissipation in Resistive Circuit Conceptual Question Description: Short conceptual problem related to power dissipation in resistive circuits. A single resistor is wired to a battery as shown in the diagram below. Define the total power dissipated by this circuit as . Now, a second identical resistor is wired in series with the first resistor as shown in the second diagram to the left . Part A What is the power, in terms of Hint A.1 , dissipated by this circuit? How to find the power dissipated by a circuit The power dissipated by a circuit (or by an element in a circuit) is defined by the relation . If the circuit consists of resistors, we can combine this relation with Ohm's law, , http://session.masteringphysics.com/myct Page 6 of 12 MasteringPhysics 10/19/08 6:51 PM to yield two alternate versions of the power formula: and . Because several circuit parameters can be changing simultaneously, it is easiest to use the formula in which only one of the terms is changing for your situation. This makes it much easier to determine the power dissipated in a resistive circuit. Hint A.2 Effect of adding a resistor in series Adding a resistor in series affects both the total resistance and total current in a circuit. Combining an understanding of these changes with the appropriate version of the power formula should allow you to answer this question. Express your answer in terms of ANSWER: . The second resistor is now removed from the circuit and rewired in parallel with the original resistor as shown in the schematic to the left Part B What is the power, in terms of Hint B.1 , dissipated by this circuit? How to find the power dissipated by a circuit The power dissipated by a circuit (or by an element in a circuit) is defined by the relation . If the circuit consists of resistors, we can combine this relation with Ohm's law, , to yield two alternate versions of the power formula: and . Because several circuit parameters can be changing simultaneously, it is easiest to use the formula in which only one of terms the is changing for your situation. This makes it much easier to determine the power dissipated in a resistive circuit. Hint B.2 Effect of adding a resistor in parallel Adding a resistor in parallel affects both the total resistance and current in a circuit. Combining an understanding of these changes with the appropriate version of the power formula should allow you to answer this question. Express your answer in terms of ANSWER: . Kirchhoff's Current Rule Ranking Task Description: Short conceptual problem about currents through resistors in various circuits. (ranking task) The placement of resistors in a circuit is one factor that can determine the current passing through the resistor. You will be given three circuits, and for each circuit you will be asked to compare the current through the various resistors. http://session.masteringphysics.com/myct Page 7 of 12 MasteringPhysics 10/19/08 6:51 PM In each of the circuits in Parts A to C, all resistors are identical. Part A Rank the resistors in the figure below (A to C) on the basis of the current that flows through them. Hint A.1 Kirchhoff's current rule for circuit junctions Kirchhoffs current rule states that the current flowing into a junction (a point at which the number of paths available for current flow changes) must equal the current flowing out of the junction. The portion of the current that flows through each available path depends on the resistance of each path. Paths with less resistance will receive a larger share of the current. Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: View Part B Rank the resistors in the figure below (A to C) on the basis of the current that flows through them. Hint B.1 Kirchhoff's current rule for circuit junctions Kirchhoffs current rule states that the current flowing into a junction (a point at which the number of paths available for current flow changes) must equal the current flowing out of the junction. The portion of the current that flows through each available path depends on the resistance of each path. Paths with less resistance will receive a larger share of the current. Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: View Part C Rank the resistors in the figure below (A to D) on the basis of the current that flows through them. http://session.masteringphysics.com/myct Page 8 of 12 MasteringPhysics 10/19/08 6:51 PM Hint C.1 Kirchhoff's current rule for circuit junctions Kirchhoffs current rule states that the current flowing into a junction (a point at which the number of paths available for current flow changes) must equal the current flowing out of the junction. The portion of the current that flows through each available path depends on the resistance of each path. Paths with less resistance will receive a larger share of the current. Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: View Magnetic Force on Charged Particles Conceptual Question Description: Simple conceptual questions about the direction of the magnetic force vector acting on various charges moving through a magnetic field. For each of the situations below, a charged particle enters a region of uniform magnetic field. Determine the direction of the force on each charge due to the magnetic field. Part A Determine the direction of the force on the charge due to the magnetic field. Hint A.1 Determining the direction of a magnetic force A charged particle moving through a region of magnetic field experiences a magnetic force, unless the velocity and magnetic field are parallel. If the velocity is parallel to the magnetic field, then the force is zero. Otherwise, the direction of the force can be found by using the right-hand rule. To employ the right hand rule: 1. Open your hand so that it is completely flat, and point the fingers of your right hand in the direction of the velocity vector. 2. Rotate your wrist until you can bend your fingers to point in the direction of the magnetic field. 3. The direction of your outstretched thumb is the direction of the magnetic force on a positive charge. ANSWER: points into the page. points out of the page. points neither into nor out of the page and . . Part B Determine the direction of the force on the charge due to the magnetic field. http://session.masteringphysics.com/myct Page 9 of 12 MasteringPhysics 10/19/08 6:51 PM Hint B.1 Determining the direction of a magnetic force A charged particle moving through a region of magnetic field experiences a magnetic force, unless the velocity and magnetic field are parallel. If the velocity is parallel to the magnetic field, then the force is zero. Otherwise, the direction of the force can be found by using the right-hand rule. To employ the right hand rule: 1. Open your hand so that it is completely flat, and point the fingers of your right hand in the direction of the velocity vector. 2. Rotate your wrist until you can bend your fingers to point in the direction of the magnetic field. 3. The direction of your outstretched thumb is the direction of the magnetic force on a positive charge. ANSWER: points out of the page. points into the page. points neither into nor out of the page and . . Part C Determine the direction of the force on the charge due to the magnetic field. Note that the charge is negative. Hint C.1 Effect of a magnetic field on a negative charge You can use the right-hand rule to determine the direction of the force exerted on a positive charge. Once you find the direction of the force that would be exerted on a positive charge, the force on a negative charge will point in the opposite direction. ANSWER: points out of the page. points into the page. points neither into nor out of the page and . . Magnetic Force on a Current-Carrying Wire Description: Practice calculating the magnetic force (both magnitude and direction using the right-hand rule) on a straight current-carrying wire in a uniform external magnetic field. Learning Goal: To understand the magnetic force on a straight current-carrying wire in a uniform magnetic field. Magnetic fields exert forces on moving charged particles, whether those charges are moving independently or are confined to a current-carrying wire. The magnetic force on an individual moving charged particle depends on its velocity and charge . In the case of a current-carrying wire, many charged particles are simultaneously in motion, so the magnetic force depends on the total current and the length of the wire . The size of the magnetic force on a straight wire of length carrying current . in a uniform magnetic field with strength is http://session.masteringphysics.com/myct Page 10 of 12 MasteringPhysics 10/19/08 6:51 PM Here is the angle between the direction of the current (along the wire) and the direction of the magnetic field. Hence . Thus this equation can also be written as refers to the component of the magnetic field that is perpendicular to the wire, . The direction of the magnetic force on the wire can be described using a "right-hand rule." This will be discussed after Part B. Part A Consider a wire of length = 0.30 that runs north-south on a horizontal surface. There is a current of = 0.50 flowing north in the wire. (The rest of the circuit, which actually delivers this current, is not shown.) The Earth's magnetic field at this location has a magnitude of 0.50 (or, in SI units, ) and points north and 38 degrees down from the horizontal, toward the ground. What is the size of the magnetic force on the wire due to the Earth's magnetic field? In considering the agreement of units, recall that . Express your answer in newtons to two significant figures. ANSWER: Because the Earth's magnetic field is quite modest, this force is so small that it might be hard to detect. Part B Now assume that a strong, uniform magnetic field of size 0.55 pointing straight down is applied. What is the size of the magnetic force on the wire due to this applied magnetic field? Ignore the effect of the Earth's magnetic field. Part B.1 Recall that Determining the angle theta is the angle between the magnetic field direction and the direction of the current flow in the wire. In this situation, what is the value of ? ANSWER: 0 degrees 38 degrees 90 degrees 180 degrees Express your answer in newtons to two significant figures. ANSWER: This force would be noticeable if the wire were of light weight. The direction of the magnetic force is perpendicular to both the direction of the current flow and the direction of the magnetic field. Here is a "right-hand rule" to help you determine the direction of the magnetic force. 1. Straighten the fingers of your right hand and point them in the direction of the current. 2. Rotate your arm until you can bend your fingers to point in the direction of the magnetic field. 3. Your thumb now points in the direction of the magnetic force acting on the wire. Part C What is the direction of the magnetic force acting on the wire in Part B due to the applied magnetic field? ANSWER: http://session.masteringphysics.com/myct Page 11 of 12 MasteringPhysics ANSWER: 10/19/08 6:51 PM due north due south due east due west straight up straight down Part D Which of the following situations would result in a magnetic force on the wire that points due north? Check all that apply. ANSWER: Current in the wire Current in the wire Current in the wire Current in the wire Current in the wire flows straight down; the magnetic field points due west. flows straight up; the magnetic field points due east. flows due east; the magnetic field points straight down. flows due west and slightly up; the magnetic field points due east. flows due west and slightly down; the magnetic field points straight down. As you can see, many current/magnetic field configurations can result in the same direction of magnetic force. Part E Assume that the applied magnetic field of size 0.55 is rotated so that it points horizontally due south. What is the size of the magnetic force on the wire due to the applied magnetic field now? Part E.1 Recall that Determining the angle theta is the angle between the magnetic field direction and the direction of the current flow in the wire. In this situation, what is the value of ? ANSWER: 0 degrees 38 degrees 90 degrees 180 degrees Express your answer in newtons to two significant figures. ANSWER: Notice that whenever the current in the wire and the magnetic field point in the same direction ( directions ( ), the sine of ) or in opposite is zero, so there is no magnetic force exerted on the wire. This is consistent with the earlier statement that it is the component of the magnetic field that is perpendicular to the direction of the current that produces the magnetic force. Also notice that for these two special values of (when the current is flowing parallel to or antiparallel to the magnetic field) the steps listed for the right-hand rule suggest a unique direction for the magnetic force. This is another clue that the magnetic force is zero. Summary 0 of 9 items complete (0% avg. score) 0 of 10 points http://session.masteringphysics.com/myct Page 12 of 12 ... View Full Document

End of Preview

Sign up now to access the rest of the document