Notes and Solved Problems for Common Exam 3 (Does not include Induction)


 Adrian Jacobs
 4 years ago
 Views:
Transcription
1 Notes and Solved Problems for Common Exam 3 (Does not include Induction) 8. MULTI LOOP CIRCUITS Key concepts: Multi loop circuits of batteries and resistors: loops, branches and junctions should be distinguished. A general problem is to find currents insides every branch if the resistances and EMF s of the batteries are known. Three rules apply: Branch rule: inside every branch the current is the same. The current may be different in different branches. Junction rule: the sum of all currents coming into the junction is equal to the sum of all currents leaving the junction. Loop rule: the sum of all potential differences in any complete walk through any closed loop of the circuit is equal to zero. The strategy for solving multiloop circuit problems is the following: 1. Find all branches and enumerate currents. 2. Find all junctions and establish the relationships between the currents 3. Apply Kirchoff loop rules to the loops in the circuit. The total number of equations to be written should be equal to the total number of currents, which are unknown. Typical problems related to multi loop circuits: Problem 5. What is the current inside the resistor R 1? What is the power released in the resistor R 1? A. 1A, 27 W B. 2A, 13 W C. 3 A, 27 W D. 4 A, 12 W E. 5 A, 5 W Solution. While this is a multi loop circuit, the answer to that
2 particular question can be found very easily. We see that the first loop contains the battery E 1 and only this resistor. Let us we write the Kirchoff loop rule for this particular loop. The current is assumed to be directed down inside R 1, therefore we walk through the loop clockwise: E 1 i 1 *R 1 =0 From which we figure out that i 1 =E 1 /R 1 =9/3=3 A. The power released in the resistor (energy rate dissipated in the resistor) is P=i 1 *V=i 1 *i 1 *R=i 1 2 *R=3*3*3=27 W. This trick works when you can find such a loop in the multi loop circuit which contains only one resistor and (possibly many) batteries. Then you can apply the Kirchoof loop rule to this particular loop and find the current inside the resistor immediately! Answer C. Problem 6. Order branches of bulbs by brightness, dimmest first: A. I, II, III B. II, III, I C. III, I, II D. III, II, I E. II, I, III Solution. The brightness of each branch is proportional to the total power released in the bulbs of the branch. Assume that the resistance of the bulbs is the same and it is equal to R. Then the total power of branch I is: P 1 = i 1 *V 1 =( i 1 ) 2 *R, the total power of branch II is: P 2 = ( i 2 ) 2 *R + ( i 2 ) 2 *R = 2*( i 2 ) 2 *R, and the total power of branch III is: P 3 = ( i 3 ) 2 *R + ( i 3 ) 2 *R + ( i 3 ) 2 *R =3*( i 3 ) 2 *R. Note that the power is expressed via the current and resistance here. What about the currents i 1, i 2, i 3? We can find currents easily by applying the Kirchoff loop equations to the loops which accounts for the battery and the particular branch. So, that for the branch I and the battery the equation is E i 1 *R=0 For the branch II and the battery, the equation is E i 2 *R i 2 *R =0 For the branch III and the battery the equation is E i 3 *R i 3 *R i 3 *R =0 Therefore i 1 =E/R, i 2 =E/(2R), i 3 =E/(3R)
3 The resulting powers are P 1 = ( i 1 ) 2 *R=E 2 /R P 2 = 2*( i 2 ) 2 *R=2 E 2 /(4R)= E 2 /(2R) P 3 =3*( i 3 ) 2 *R=3E 2 /(9R)= E 2 /(3R) We finally see that the brightest branch is I (has largest power), then II, and finally III. Therefore, the dimmest is III, then II, then I. Answer D. Problem 7. In the previous problem (see figure to the problem 6), if E=12 V and R= 2Ω, find the power released in each bulb? A. 8W for branch I, 18W for branch II, 72W for branch III B. 18W for branch I, 8W for branch II, 72W for branch III C. 8W for branch I, 72W for branch II, 18W for branch III D. 72W for branch I, 18W for branch II, 8W for branch III E. 72W for branch I, 8W for branch II, 18W for branch III Solution. See derivation to the problem 15. The power released in the bulb of the branch I: ( i 1 ) 2 *R=E 2 /R=12*12/2=72 W The power released in each bulb (total two bulbs) of branch II: ( i 2 ) 2 *R= E 2 /(4R)= 12*12/4/2=18 W The power released in each bulb (total three bulbs) of branch III: ( i 3 ) 2 *R= E 2 /(9R)= 12*12/9/2=8 W Answer D. Problem 8. In the circuit shown in the figure find the currents through each resistor if E 1 =10 V, E 2 =5V and all resistors are equivalent with R 1 =R 2 =R 3 =R= 5Ω. A. i 1 = 1A down, i 2 =3A up, i 3 = 2A down
4 B. i 1 = 1A down, i 2 =1A up, i 3 = 0 C. i 1 = 2A down, i 2 =0, i 3 = 3A down D. i 1 = 3A up, i 2 =2A up, i 3 = 1A down E. i 1 = 1A up, i 2 =2A down, i 3 = 0 Solution. We have a real multi loop circuit problem now. We have three branches corresponding to three resistors. Inside each branch there is its own current. Lets enumerate currents in each resistors to be i 1, i 2, i 3 Assume that the direction of the current i1 in the resistor R1 is down, the direction of the current i2 in the resistor R2 is up and the direction of the current in the resistor i3 is down again. From the junction rule we obtain (i) i2=i1+i3 From two inner loops we obtain (ii) E1i1*Ri2*R=0 (iii) E2i3*Ri2*R=0 We have total three equations. Summing equations (ii) and (iii) and using Eq. (i): E1+E2(i1+i3)*R2*i2*R= E1+E23*R*i2= 0 Therefore i2=(e1+e2)/(3r)=15/15=1a (sign is +, therefore direction chosen is OK, i.e up) From equation (ii): i1=(e1i2*r)/r=e1/ri2=10/51=1a (sign is +, therefore direction of the current i2 is OK, i.e. down.) From equation (i): i3=i2i1=11=0a (no current!) We finally have i1=1a down, i2=1a up, i3=0 Answer B.
5 9. RC CIRCUITS The RC circuit consists of the capacitor C, the resistor R, the battery E and the switch S. When the switch S is closed in RC circuit, capacitor is charged. Application of the Kirchoff loop equation for time moment t to RC circuit leads to differential equation. The solution of the equation reads as follows: (1.1) q t EC e ( ) = (1 ) is the charge on the capacitor as a function of time. Current in the RC circuit: (1.2) dq E E i t e e dt R R t / τc ( ) = = = where time constant τ C = RC shows the characteristic time for the charging process. The voltage across the capacitor can be found q( t) (1.3) V ( t) = = E(1 e ) C In the reverse situation, when the switch is opened, the capacitor with charge q0 gets discharged. The functions for the charge, current, and voltage are the following: (1.4) (1.5) q( t) = q e 0 dq q0 i( t) = = e dt RC q( t) q0 (1.6) V ( t) = = e C C Problem 9. What is the voltage across the capacitor in the RC circuit with R= 3 Ohm, C= 5 pf and E=9V just after the switch is closed. A. 0 V B. 1 V C. 3 V D. 5 V E. 9 V
6 Solution. After the switch is closed the capacitor begins to charge. The charge on the plates as a function of time is given by q t EC e ( ) = (1 ) The voltage across the capacitor is thus q( t) V t E e C ( ) = = (1 ) The time constant for the circuit RC is 5pF*3 Ohms = 15 picoseconds. Even though this is a very short time, it is not zero. Therefore, just after the switch is closed, t = 0 in the expression above makes the quantity in parentheses go to zero). Thus, the voltage V(t=0) = 0. Answer A. Problem 10. How long does it take for an RC circuit with R= 3 Ohm, C= 5 pf and E=9V to charge the capacitor to 45 pc. A. 1 second B. 1 hour C. 1 day D. 1 year E. infinity Solution. After the switch is closed the capacitor is beginning to charge. The charge on the plates as a function of time is: q t EC e ( ) = (1 ) We solve this for t with q known. Note that q is the limiting value = EC. t = RC q EC == = 12 ln 1 / 3*5*10 *ln 1 45/(9*5) inf Answer E
7 10. MAGNETIC FIELDS PLEASE NOTE THAT IN ALL FORMULAS THE VECTOR PRODUCT (CROSS PRODUCT) IS DENOTED WITH THE SQUARE BRACKETS: c=[axb] Key concepts: A moving charged particle creates a magnetic field around it. Do not confuse magnetic fields with electric fields: the electric fields are associated with the charges; the magnetic fields are associated with the moving charges. Therefore, a moving charged particle creates both an electric field and a magnetic field. If a moving charged particle is placed inside a magnetic field it experiences magnetic force. This force is proportional to the charge of the particle, its velocity and the strength of the magnetic field. Again, it is essential that the particle is moving: the faster the particle moves, the greater the force it feels from the external magnetic field. When a particle of charge q moves with velocity v inside a magnetic field B it feels a magnetic force F given by the formula: F =q*[vxb] which was found as the result of experimental measurement. A cross product [vxb] is a vector which is perpendicular to the plane made by vectors v and B. The direction of the cross product [vxb] is fixed by the right hand rule: rotate vector v towards vector B using your right hand (use the smallest angle between them). The orientation of your right hand shows the direction of the cross product. The direction of the magnetic force F can be different from the vector product [vxb] since the charge q may be negative. In that case the direction of the force F is opposite to the direction of the vector product since F = q *[vxb]. Typical problems: Problem 11. Find the value and the direction of the force acting on a positively charged particle with q=2 µc moving with velocity v=5 km/s along x direction which is placed inside constant magnetic field B=1 kt oriented along y direction: A. 10 mn, z direction B. 10 mn, +z direction C. 10 N, z direction D. 10 N, +z direction E. 1 kn, +x direction
8 Solution. The vector product [vxb] is oriented along +z direction according to the right hand rule. Since the charge is positive the direction of the force is also lying along +z direction. The value for the force is (angle φ between v and B is 90 degrees) F=q*v*B* sinφ =q*v*b=2*106 *5*10 3 *1*10 3 =10 N Answer D Problem 12. Find the value and the direction of the force acting on a positively charged particle with q=2 mc moving with v=5 mm/s along x direction which is placed inside constant magnetic field B=1.41 kt oriented 45 degrees in xy plane A. 10 mn, z direction B. 10 mn, + z direction C. 10 N,  z direction D. 10 N, +z direction E. 1 kn, +x direction Solution. The vector product [vxb] is oriented along +z direction according to the right hand rule. Since the charge is positive the direction of the force is also lying along +z direction. The value for the force is (angle φ between v and B is 45 degrees) F=q*v*B*sin φ =q*v*b sin45=2*103 *5*103 *1.41*10 3 /1.41=10 mn Answer B. Problem 13. Find the value and the direction of the force acting on a negatively charged article with q=0.5 C moving with v=10 m/s along x direction which is placed inside constant magnetic field B=(10,20,0) T Y y A. 10 N, z axis B. 10 N, +z axis C. 100 N, z axis D. 100 N, +z axis E. 1 N, xaxis Z z
9 Solution. Since vector B has zcomponent equal zero, it is completely oriented within xy plane. Therefore, the vector product [vxb] is oriented along +z direction according to the righthand rule. Since the charge is negative the direction of the force is lying along z direction. Let us find the direction and the value for the force using vector algebra. First, let us represent the vector B in the form: B=10* i +20*j, where i is a unit vector along x axis and j is a unit vector along yaxis. Vector v=10*i is lying completely along x axis. The vector product is: [vxb]= [ 10*i*(10* i +20*j)]=10*10*[i*i]+10*20*[i*j] Since vector product [ixi]=0 and [ixj]=k, the unit vector along zaxis, the vector product becomes: [vxb]=200* k. Now the force is the vector product time the charge. Since the charge is negative: F=q*[vxB]=0.5*200* k=100*k, N, i.e 100 N oriented along vector k which is z axis. Answer C. Problem 14. An electron moving with the constant speed in the direction from top to bottom enters the region of a constant magnetic field. The electron trajectory deviates to the left. What is the orientation of the magnetic field? A. top B. bottom C. left D. right E. inside the page Solution. If electron deviates to the left, it experiences the force directed to the left. This is the magnetic force described as F =q*[vxb]. We need to figure out such direction of B so that to be multiplied by vector v and taking account the sign of charge for the electron, we obtain the direction of the force to the left. To do this, we first note that if the force is directed to the left, the vector product [vxb] should be directed to the right since electron is negatively charged particle (q<0). Second, if v is directed to the bottom, vector B should be directed inside the page, in this way vector product [vxb] is directed to the right, and the force will be directed to the left. Answer E. Problem 15. An electron is circulating within the plane of the page due to magnetic field directed inside the page. The radius of its circulating motion is 9 cm and its speed is 1.6 km/s. Find the value of magnetic field and the direction of the circulating motion,
10 A. 0.1 µt, clockwise B. 0.1 µt, counterclockwise C. 0.1 mt, clockwise D. 0.1 mt, counterclockwise E. 0.1 T, clockwise Solution. Circular motion assumes that there is force acting on the electron, which is directed towards the center of the orbit. This force is of magnetic origin, i.e. F =e*[vxb]. At the same time, the force associated with the circular motion is equal to m*w where centrifugal acceleration w= v 2 /R. Since velocity vector v is always perpendicular to magnetic field vector B, we can write F=e*v*B=m*v 2 /R from which B=m*v/(R*e)=9*1031 *1.6*10 3 /(9*102 *1.6*1019 )=107 T. To understand whether the electron circulates clockwise or counterclockwise, we need to look at the electron at some of its position. Assume that at some moment t it is positioned at 12pm (the topmost part of its trajectory). If its velocity is directed to the right, it is going clockwise, if its velocity is directed to the left, it is going counterclockwise, At the topmost point of its trajectory the electron should experience the force directed to the bottom since it is going around the circle, and the force of the circular motion is always directed to the center. Therefore F =e*[vxb] is directed to the bottom, and we know that e<0 and B is directed inside the page. The vector product [vxb] should be directed to the top since F is directed to the bottom, and e is less than zero. If vector product is directed to the top and B is directed inside the page, the velocity vector v should be directed to the right. In this way, vector product [vxb] will be directed to the top, and multiplied by negative charge, the force will be directed to the bottom. We therefore found that the velocity in topmost point is directed to the right. Therefore, the electron is going around the circle clockwise. We can fix any other of its position (say leftmost) and come to the same conclusion. Answer A.
11 11. MAGNETIC FIELDS FROM CURRENTS Key concepts: In general, the magnetic field due to a moving charged particle is given by: B(r)= µ 0 /4π *q*[v*r]/r 3 where constant µ 0 /4π=107. Note that the direction of the magnetic field is tricky: if the particle is moving from bottom to top, and a point where we look at the field is on the right, the vector product [vxr] is directed inside the page. If the charge of the particle >0, the magnetic field will be pointed inside the page at this point r. If the charge < 0 the magnetic field is pointed outside the page at this point r. All this information is encoded into the above formula. Since current inside the wire is just an array of moving charged particles, every wire carrying current creates magnetic field around it. To find the total field at some point r we need to sum up all the contributions to this field from all moving charges inside the wire. In general, there is an integral, which gives the answer. For simple configurations like straight wire or circular wire the answers are: 1. Infinite straight wire carrying current creates magnetic field which varies with the distance d from the wire as follows: B(d)=µ 0 *i/(2π d), where i is the value of the current. The direction of the field can be found using the right hand rule: point your large finger along with the current, the four other fingers show the direction of the magnetic field (See figure in the textbook). 2. A coil or a circular wire creates magnetic field at the center of the circle: B= µ 0 *i /(2R). The direction of the field can be found from right hand rule: rotate by your right hand in the direction of the current simulating circular motion, the orientation of the right hand will show the direction of the magnetic field within the plane inside the circle. A general arc defined by angle ϕ creates magnetic field B=µ 0 *I*ϕ /(4πR) which for ϕ=2π transforms to the expression above. 3. A solenoid, which is a cylindrically turned wire with some (usually large) number of turns per unit length, creates nonzero constant magnetic field inside itself and zero field outside itself. The magnetic field inside the solenoid is given by: B=µ 0 *n*i, where n is the number of turns per unit length. The direction of the field inside the solenoid can be found by applying the right hand rule: rotate by your right hand in the direction of the current simulating circular motion, the orientation of the right hand shows the direction of the field inside the solenoid. A wire with the current placed into magnetic field experiences a force since every moving charge inside the wire experiences the magnetic force. The total force on the wire with current i depends on the orientation of the wire and on the orientation of the magnetic field. If the wire is a straight piece of some length L, the force is given by F= i* [ L xb], where vector L has length L an is directed towards the direction of the current.
12 Two pieces of wire, which brought together will repel from or attract to each other if the currents exist inside these wires. This can be simply understood since one wire will create magnetic field around it, and the moving particles inside another wire will experience the magnetic force. For example, if two parallel wires carry currents in the same direction, they will attract to each other. If two parallel wires carry currents in different direction, they will repel from each other. In case of two parallel pieces of wire of length L separated by distance d which carry out the currents i1 and i2, the force between them is given by F=µ 0 i1*i2*l/(2πd). Typical problems: Problem 16. Find the direction and the value for the force on the wire, which carries the current of 2 ma, if it is placed inside the magnetic field of 4 kt. The length of the wire is 3 mm, it is vertically oriented with the current flowing to the bottom. The magnetic field is horizontally oriented and is pointed to the left. A. 24 Ν, left B. 24 mν, right C. 24 kn, top D. 24 N, bottom E. 24 mn, inside the page Solution. The formula to use is: F= i*[ L xb], where L is a vector which points from top to bottom (vertical orientation with the current flowing to the bottom) and B points to the left. Both vectors are perpendicular to each other. The vector product [LxB] points inside the page. We should not care about sign of i in this case, it is always positive, and direction of vector L takes care about direction of the current. The value of the force is F=i*L*B=2*103 *3*103 *4*10 3 =24 mn. Answer E. Problem 17. A wire carries a current from the right to the left. What is the orientation of the magnetic field above and below the wire? A. Above inside the page, Below outside the page B. Above outside the page, Below inside the page C. Above left, Below right D. Above right, Below left E. Above bottom, Below top
13 Solution. Applying the right hand rule, orient your large finger to the left, the four other fingers point to the direction of the field above the wire, they point to inside the page. If you rotate by the right hand around the wire so that your large fingers are located below the wire, the point outside the page. This gives the answer to the problem. Answer A. Problem 18. Find the value of magnetic field of the long straight wire at distance d=3 cm if it carries current of 3 ma. A. 20 pt B. 20 nt C. 20 µt D. 20 mt E. 20 T Solution. The formula is B(d)= µ 0 *i/(2π d), where all numbers are given by the problem. B(d)= µ 0 *i/(2π d)= 2 µ 0 *i/(4π d)=2*107 *3*103 /(3*102 )=2*108 =20 * 109 T=20 nt Answer B. Problem 19. Two straight wires of the length 5 m each and parallel to each other carry currents of 3 A and 5 A in opposite directions. If the distance between the wires is 1 cm what is the force between them? A. 3 mn, repulsion B. 3 mn, attraction C. 1.5 mn, repulsion D. 1.5 mn, attraction E. 3 N, repulsion Solution. The formula to use is: F=µ 0 i1*i2*l/(2πd)= 2* 107 *3*5*5/0.01=150*105 N. Since the directions of the currents are opposite, it is repulsion. Answer C.
14 Problem 20. A circular wire of radius 6.28 cm is oriented within the page and carries a current of 2 A. The current flows clockwise. What are the value and the direction for the magnetic field created at its center? A. 20 µt, outside the page B. 20 µt, inside the page C. 20 mt, to the left D. 20 mt to the right E. 20 T, to the top Solution. The formula to use is: B=µ 0 *i /(2R)=4*3.14 * 107 * 2 / (2 * 6.28 * 102 )= 2*105 T. Let us apply the right hand rule: to simulate the current flow by rotating the right hand, the orientation of the right hand should be inside the page. Answer B. Problem 21. A solenoid is oriented perpendicular to the page, and carries a current in the counterclockwise direction. What are the value and the direction of the magnetic field inside it? Assume that the current is 1 A, and the number of turns per centimeter is 25. A mt, outside the page B mt, inside the page C T, to the left D T to the right E kt, to the top Solution. The formula to use is: B=µ 0 *i*n=4*3.14*107 *1*25*10 2 =3.14 mt. The direction is outside the page, since simulation of the current flow by rotating your right hand gives us the answer. Answer A.
15 Problem 22. Find the direction of the force on each side of rectangular frame. A. Top side  top, bottom side  bottom, left side  left, right side right. B. Top side  bottom, bottom side  top, left side  right, right side left. C. Top side  left, bottom side  right, left side  bottom, right side top. D. Top side  right, bottom side  left, left side  bottom, right side top. E. Top side  left, bottom side  right, left side  top, right side bottom. Solution. Parallel wires attract to each other if the currents flow in the same direction. They repel from each other if the currents flow in opposite direction. For perpendicular wires the situation is more complicated. It can be figured out by noting that the wire below itself creates a magnetic field directed inside the page. Since the current flow in the left side of the frame is on top, this is the flow of positively charged particles. They will experience the force F =q*[vxb], pointed to the left (q>0, v is on top, B is inside the page). Therefore the entire left part of the frame will experience the force pointed to the left. Using similar argument, the entire right part experiences the force pointed to the right. Answer A. Problem 23. Find the total force on the frame in the above problem if the current inside the wire is 30 A, and inside the frame is 20 A. The horizontal dimension of the frame is 30 cm and the vertical dimension is 7 cm. The top side of the frame is located at the distance 1 cm from the wire. A N, top B N, bottom C mn, top D mn, bottom E kn, left Solution. First note that the forces on left side and right side are the same and directed oppositely. Therefore they do not contribute to the total force. The force on the top side is directed to the top and it is given by: F top =µ 0 *i1*i2*l/(2πd) with d=1cm. Evaluating it: 2*107 *30*20*30*102 /102 =36*104 N. The force on the bottom part is directed to the bottom and it is given by the same formula with d = 8 cm. Therefore, the force F bottom =µ 0 *i1*i2*l/(2πd)= 2*107 *30*20*30*102 /(8*102 ) = 4.5*104 N. The total force is directed on top and it is 36* *104 =31.5*104 N.= 3.15 mn.
PHYS 2326 University Physics II Class number
PHYS 2326 University Physics II Class number HOMEWORK SET #1 CHAPTERS: 27,28,29 (DUE JULY 22, 2013) Ch. 27.======================================================= 1. A rod of 2.0m length and a square
More informationPHYS 1102 EXAM  II. SECTION: (Circle one) 001 (TH 9:30 AM to 10:45AM) 002 (TH 3:30 PM to 4:45 PM) You have 1 hr 45 minutes to complete the test
PHYS 1102 EXAM  II SECTION: (Circle one) 001 (TH 9:30 AM to 10:45AM) 002 (TH 3:30 PM to 4:45 PM) Your Name: Student ID: You have 1 hr 45 minutes to complete the test PLEASE DO NOT START TILL YOU ARE INSTRUCTED
More informationThe next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d:
PHYS 102 Exams Exam 2 PRINT (A) The next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d: It is connected to a battery with constant emf V.
More informationLouisiana State University Physics 2102, Exam 3 April 2nd, 2009.
PRINT Your Name: Instructor: Louisiana State University Physics 2102, Exam 3 April 2nd, 2009. Please be sure to PRINT your name and class instructor above. The test consists of 4 questions (multiple choice),
More informationPhysics 212 Midterm 2 Form A
1. A wire contains a steady current of 2 A. The charge that passes a cross section in 2 s is: A. 3.2 1019 C B. 6.4 1019 C C. 1 C D. 2 C E. 4 C 2. In a Physics 212 lab, Jane measures the current versus
More informationCircuits Capacitance of a parallelplate capacitor : C = κ ε o A / d. (ρ = resistivity, L = length, A = crosssectional area) Resistance : R = ρ L / A
k = 9.0 x 109 N m2 / C2 e = 1.60 x 1019 C ε o = 8.85 x 1012 C2 / N m2 Coulomb s law: F = k q Q / r2 (unlike charges attract, like charges repel) Electric field from a point charge : E = k q / r2 ( towards
More informationLouisiana State University Physics 2102, Exam 3, November 11, 2010.
Name: Instructor: Louisiana State University Physics 2102, Exam 3, November 11, 2010. Please be sure to write your name and class instructor above. The test consists of 3 questions (multiple choice), and
More informationPHYS 241 EXAM #2 November 9, 2006
1. ( 5 points) A resistance R and a 3.9 H inductance are in series across a 60 Hz AC voltage. The voltage across the resistor is 23 V and the voltage across the inductor is 35 V. Assume that all voltages
More informationExam 2 Solutions. ε 3. ε 1. Problem 1
Exam 2 Solutions Problem 1 In the circuit shown, R1=100 Ω, R2=25 Ω, and the ideal batteries have EMFs of ε1 = 6.0 V, ε2 = 3.0 V, and ε3 = 1.5 V. What is the magnitude of the current flowing through resistor
More informationP202 Practice Exam 2 Spring 2004 Instructor: Prof. Sinova
P202 Practice Exam 2 Spring 2004 Instructor: Prof. Sinova Name: Date: (5)1. How many electrons flow through a battery that delivers a current of 3.0 A for 12 s? A) 4 B) 36 C) 4.8 10 15 D) 6.4 10 18 E)
More informationExam 2, Phy 2049, Spring Solutions:
Exam 2, Phy 2049, Spring 2017. Solutions: 1. A battery, which has an emf of EMF = 10V and an internal resistance of R 0 = 50Ω, is connected to three resistors, as shown in the figure. The resistors have
More informationLouisiana State University Physics 2102, Exam 2, March 5th, 2009.
PRINT Your Name: Instructor: Louisiana State University Physics 2102, Exam 2, March 5th, 2009. Please be sure to PRINT your name and class instructor above. The test consists of 4 questions (multiple choice),
More informationReview. Spring Semester /21/14. Physics for Scientists & Engineers 2 1
Review Spring Semester 2014 Physics for Scientists & Engineers 2 1 Notes! Homework set 13 extended to Tuesday, 4/22! Remember to fill out SIRS form: https://sirsonline.msu.edu Physics for Scientists &
More informationa. Clockwise. b. Counterclockwise. c. Out of the board. d. Into the board. e. There will be no current induced in the wire
Physics 1B Winter 2012: Final Exam For Practice Version A 1 Closed book. No work needs to be shown for multiplechoice questions. The first 10 questions are the makeup Quiz. The remaining questions are
More informationPhys102 Final163 Zero Version Coordinator: Saleem Rao Tuesday, August 22, 2017 Page: 1. = m/s
Coordinator: Saleem Rao Tuesday, August 22, 2017 Page: 1 Q1. A 125 cm long string has a mass of 2.00 g and a tension of 7.00 N. Find the lowest resonant frequency of the string. A) 2.5 Hz B) 53.0 Hz C)
More informationGeneral Physics (PHYC 252) Exam 4
General Physics (PHYC 5) Exam 4 Multiple Choice (6 points). Circle the one best answer for each question. For Questions 13, consider a car battery with 1. V emf and internal resistance r of. Ω that is
More informationPRACTICE EXAM 1 for Midterm 2
PRACTICE EXAM 1 for Midterm 2 Multiple Choice Questions 1) The figure shows three identical lightbulbs connected to a battery having a constant voltage across its terminals. What happens to the brightness
More informationPhysics 42 Exam 2 PRACTICE Name: Lab
Physics 42 Exam 2 PRACTICE Name: Lab 1 2 3 4 Conceptual Multiple Choice (2 points each) Circle the best answer. 1.Rank in order, from brightest to dimmest, the identical bulbs A to D. A. C = D > B > A
More informationPHYS 1444 Section 02 Review #2
PHYS 1444 Section 02 Review #2 November 9, 2011 Ian Howley 1 1444 Test 2 Eq. Sheet Terminal voltage Resistors in series Resistors in parallel Magnetic field from long straight wire Ampére s Law Force on
More informationMaterial World: Electricity
17. Coulomb s Law The force, F, between two objects with charge q1 and q2, is given by: k q q 1 2 F , where r = distance between the two charges in meters 2 r k = Coulomb's constant = 9 X 10 9 m 2 /C
More informationPhysics 2020 Exam 2 Constants and Formulae
Physics 2020 Exam 2 Constants and Formulae Useful Constants k e = 8.99 10 9 N m 2 /C 2 c = 3.00 10 8 m/s ɛ = 8.85 10 12 C 2 /(N m 2 ) µ = 4π 10 7 T m/a e = 1.602 10 19 C h = 6.626 10 34 J s m p = 1.67
More informationSolutions to PHY2049 Exam 2 (Nov. 3, 2017)
Solutions to PHY2049 Exam 2 (Nov. 3, 207) Problem : In figure a, both batteries have emf E =.2 V and the external resistance R is a variable resistor. Figure b gives the electric potentials V between the
More informationAP Physics C. Electric Circuits III.C
AP Physics C Electric Circuits III.C III.C.1 Current, Resistance and Power The direction of conventional current Suppose the crosssectional area of the conductor changes. If a conductor has no current,
More informationPhysics 2B Winter 2012 Final Exam Practice
Physics 2B Winter 2012 Final Exam Practice 1) When the distance between two charges is increased, the force between the charges A) increases directly with the square of the distance. B) increases directly
More informationPhysics 106, Section 1
Physics 106, Section 1 Magleby Exam 2, Summer 2012 Exam Cid You are allowed a pencil and a testing center calculator. No scratch paper is allowed. Testing center calculators only. 1. A circular coil lays
More informationExam 2 Solutions. Note that there are several variations of some problems, indicated by choices in parentheses.
Exam 2 Solutions Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1 Part of a long, straight insulated wire carrying current i is bent into a circular
More informationMagnets. Domain = small magnetized region of a magnetic material. all the atoms are grouped together and aligned
Magnetic Fields Magnets Domain = small magnetized region of a magnetic material all the atoms are grouped together and aligned Magnets Ferromagnetic materials domains can be forced to line up by applying
More information/20 /20 /20 /60. Dr. Galeazzi PHY207 Test #3 November 20, I.D. number:
Signature: Name: I.D. number: You must do ALL the problems Each problem is worth 0 points for a total of 60 points. TO GET CREDIT IN PROBLEMS AND 3 YOU MUST SHOW GOOD WORK. CHECK DISCUSSION SECTION ATTENDED:
More informationChapter 28. Direct Current Circuits
Chapter 28 Direct Current Circuits Circuit Analysis Simple electric circuits may contain batteries, resistors, and capacitors in various combinations. For some circuits, analysis may consist of combining
More informationPRACTICE EXAM 2 for Midterm 2
PRACTICE EXAM 2 for Midterm 2 Multiple Choice Questions 1) In the circuit shown in the figure, all the lightbulbs are identical. Which of the following is the correct ranking of the brightness of the bulbs?
More informationLast time. Ampere's Law Faraday s law
Last time Ampere's Law Faraday s law 1 Faraday s Law of Induction (More Quantitative) The magnitude of the induced EMF in conducting loop is equal to the rate at which the magnetic flux through the surface
More informationPhysics 6B Summer 2007 Final
Physics 6B Summer 2007 Final Question 1 An electron passes through two rectangular regions that contain uniform magnetic fields, B 1 and B 2. The field B 1 is stronger than the field B 2. Each field fills
More informationChapter 27. Circuits
Chapter 27 Circuits 1 1. Pumping Chagres We need to establish a potential difference between the ends of a device to make charge carriers follow through the device. To generate a steady flow of charges,
More informationPractice Exam 1. Necessary Constants and Equations: Electric force (Coulomb s Law): Electric field due to a point charge:
Practice Exam 1 Necessary Constants and Equations: Electric force (Coulomb s Law): Electric field due to a point charge: Electric potential due to a point charge: Electric potential energy: Capacitor energy:
More informationFigure 1 A) 2.3 V B) +2.3 V C) +3.6 V D) 1.1 V E) +1.1 V Q2. The current in the 12 Ω resistor shown in the circuit of Figure 2 is:
Term: 13 Wednesday, May 1, 014 Page: 1 Q1. What is the potential difference V B V A in the circuit shown in Figure 1 if R 1 =70.0 Ω, R=105 Ω, R 3 =140 Ω, ε 1 =.0 V and ε =7.0 V? Figure 1 A).3 V B) +.3
More information= e = e 3 = = 4.98%
PHYS 212 Exam 2  Practice Test  Solutions 1E In order to use the equation for discharging, we should consider the amount of charge remaining after three time constants, which would have to be q(t)/q0.
More informationPhysics 202 Midterm Exam 2 Oct 30th, 2012
ID CODE: B Physics 202 Midterm Exam 2 Oct 30th, 2012 Name:...Yibin Pan... Student ID:... Section:... TA (please circle): James Buchannan Diptaranjan Das Ross Devol Yutao Gong Minho Kwon Greg Lau Andrew
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A jeweler needs to electroplate gold (atomic mass 196.97 u) onto a bracelet. He knows
More information21 MAGNETIC FORCES AND MAGNETIC FIELDS
CHAPTER 1 MAGNETIC FORCES AND MAGNETIC FIELDS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1 (d) RightHand Rule No 1 gives the direction of the magnetic force as x for both drawings A and B In drawing C, the
More informationVersion 001 CIRCUITS holland (1290) 1
Version CIRCUITS holland (9) This printout should have questions Multiplechoice questions may continue on the next column or page find all choices before answering AP M 99 MC points The power dissipated
More informationName (Print): 4 Digit ID: Section:
Physics 11 Sample Common Exam 3: Sample 5 Name (Print): 4 Digit ID: Section: Honors Code Pledge: As an NJIT student I, pledge to comply with the provisions of the NJIT Academic Honor Code. I assert that
More informationInductance, RL Circuits, LC Circuits, RLC Circuits
Inductance, R Circuits, C Circuits, RC Circuits Inductance What happens when we close the switch? The current flows What does the current look like as a function of time? Does it look like this? I t Inductance
More informationChapter 12. Magnetism and Electromagnetism
Chapter 12 Magnetism and Electromagnetism 167 168 AP Physics Multiple Choice Practice Magnetism and Electromagnetism SECTION A Magnetostatics 1. Four infinitely long wires are arranged as shown in the
More informationStudent number: Question # Mark Maximum Mark. Multiple Choice 20
Name: Student number: Academic Honesty: Cheating in an examination includes the following: 1. the unauthorized sharing of material such as textbooks during an open book examination; 2. concealing information
More informationLecture 31: MON 30 MAR Review Session : Midterm 3
Physics 2113 Jonathan Dowling Lecture 31: MON 30 MAR Review Session : Midterm 3 EXAM 03: 8PM MON 30 MAR in Cox Auditorium The exam will cover: Ch.26 through Ch.29 The exam will be based on: HW07 HW10.
More informationPhysics 2135 Exam 2 October 20, 2015
Exam Total / 200 Physics 2135 Exam 2 October 20, 2015 Printed Name: Rec. Sec. Letter: Five multiple choice questions, 8 points each. Choose the best or most nearly correct answer. 1. A straight wire segment
More informationPhysics 2401 Summer 2, 2008 Exam III
Physics 2401 Summer 2, 2008 Exam e = 1.60x1019 C, m(electron) = 9.11x1031 kg, ε 0 = 8.845x1012 C 2 /Nm 2, k e = 9.0x10 9 Nm 2 /C 2, m(proton) = 1.67x1027 kg. n = nano = 109, µ = micro = 106, m =
More informationPhysics 126 Fall 2004 Practice Exam 1. Answer will be posted about Oct. 5.
Physics 126 Fall 2004 Practice Exam 1. Answer will be posted about Oct. 5. 1. Which one of the following statements best explains why tiny bits of paper are attracted to a charged rubber rod? A) Paper
More informationHW7: Ch. 26 P 34, 36 Ch.27 Q 2, 4, 8, 18 P 2, 8, 17, 19, 37
Fall 12 PHY 122 Homework Solutions #7 HW7: Ch. 26 P 34, 36 Ch.27 Q 2, 4, 8, 18 P 2, 8, 17, 19, 37 Chapter 26 Problem 34 Determine the magnitudes and directions of the currents in each resistor shown in
More informationDirectCurrent Circuits. Physics 231 Lecture 61
DirectCurrent Circuits Physics 231 Lecture 61 esistors in Series and Parallel As with capacitors, resistors are often in series and parallel configurations in circuits Series Parallel The question then
More informationTIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES. PHYS 1112, Exam 2 Section 1 Version 1 April 2, 2013 Total Weight: 100 points
TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1112, Exam 2 Section 1 Version 1 April 2, 2013 Total Weight: 100 points 1. Check your examination for completeness prior to starting.
More informationPhysics 208, Spring 2016 Exam #3
Physics 208, Spring 206 Exam #3 A Name (Last, First): ID #: Section #: You have 75 minutes to complete the exam. Formulae are provided on an attached sheet. You may NOT use any other formula sheet. You
More informationInduction and Inductance
Welcome Back to Physics 1308 Induction and Inductance Michael Faraday 22 September 1791 25 August 1867 Announcements Assignments for Tuesday, November 6th:  Reading: Chapter 30.630.8  Watch Videos:
More information PHYS 2326 University Physics II Class number 
More informationPHYS102 Previous Exam Problems. Induction
PHYS102 Previous Exam Problems CHAPTER 30 Induction Magnetic flux Induced emf (Faraday s law) Lenz law Motional emf 1. A circuit is pulled to the right at constant speed in a uniform magnetic field with
More informationExam 2 Solutions. = /10 = / = /m 3, where the factor of
PHY049 Fall 007 Prof. Yasu Takano Prof. Paul Avery Oct. 17, 007 Exam Solutions 1. (WebAssign 6.6) A current of 1.5 A flows in a copper wire with radius 1.5 mm. If the current is uniform, what is the electron
More informationExam 2 Solutions. PHY2054 Spring Prof. Paul Avery Prof. Pradeep Kumar Mar. 18, 2014
Exam 2 Solutions Prof. Paul Avery Prof. Pradeep Kumar Mar. 18, 2014 1. A series circuit consists of an open switch, a 6.0 Ω resistor, an uncharged 4.0 µf capacitor and a battery with emf 15.0 V and internal
More informationPhysics 2B Spring 2010: Final Version A 1 COMMENTS AND REMINDERS:
Physics 2B Spring 2010: Final Version A 1 COMMENTS AND REMINDERS: Closed book. No work needs to be shown for multiplechoice questions. 1. A charge of +4.0 C is placed at the origin. A charge of 3.0 C
More informationPhysics 227 Final Exam December 18, 2007 Prof. Coleman and Prof. Rabe. Useful Information. Your name sticker. with exam code
Your name sticker with exam code Physics 227 Final Exam December 18, 2007 Prof. Coleman and Prof. Rabe SIGNATURE: 1. The exam will last from 4:00 p.m. to 7:00 p.m. Use a #2 pencil to make entries on the
More informationFinal Exam: Physics Spring, 2017 May 8, 2017 Version 01
Final Exam: Physics2331  Spring, 2017 May 8, 2017 Version 01 NAME (Please Print) Your exam should have 11 pages. This exam consists of 18 multiplechoice questions (2 points each, worth 36 points), and
More informationTactics Box 23.1 Using Kirchhoff's Loop Law
PH203 Chapter 23 solutions Tactics Box 231 Using Kirchhoff's Loop Law Description: Knight/Jones/Field Tactics Box 231 Using Kirchhoff s loop law is illustrated Learning Goal: To practice Tactics Box 231
More informationDC Circuits. Electromotive Force Resistor Circuits. Kirchoff s Rules. RC Circuits. Connections in parallel and series. Complex circuits made easy
DC Circuits Electromotive Force esistor Circuits Connections in parallel and series Kirchoff s ules Complex circuits made easy C Circuits Charging and discharging Electromotive Force (EMF) EMF, E, is the
More information2R R R 2R. Phys Test 1
Group test. You want to calculate the electric field at position (x o, 0, z o ) due to a charged ring. The ring is centered at the origin, and lies on the xy plane. ts radius is and its charge density
More informationExam 2 Solutions. Applying the junction rule: i 1 Applying the loop rule to the left loop (LL), right loop (RL), and the full loop (FL) gives:
PHY61 Eam Solutions 1. [8 points] In the circuit shown, the resistance R 1 = 1Ω. The batter voltages are identical: ε1 = ε = ε3 = 1 V. What is the current (in amps) flowing through the middle branch from
More informationPhysics 8.02 Exam Two Equation Sheet Spring 2004
Physics 8.0 Exam Two Equation Sheet Spring 004 closed surface EdA Q inside da points from inside o to outside I dsrˆ db 4o r rˆ points from source to observer V moving from a to b E ds 0 V b V a b E ds
More informationPhysics Jonathan Dowling. Final Exam Review
Physics 2102 Jonathan Dowling Physics 2102 Final Exam Review A few concepts: electric force, field and potential Electric force: What is the force on a charge produced by other charges? What is the force
More informationElectromagnetism II. (a) enav Na enav Cl (b) enav Na + enav Cl (c) enav Na (d) enav Cl (e) zero
Electromagnetism II 1. Salt water contains n sodium ions (Na + ) per cubic meter and n chloride ions (Cl ) per cubic meter. A battery is connected to metal rods that dip into a narrow pipe full of salt
More informationAP Physics C Mechanics Objectives
AP Physics C Mechanics Objectives I. KINEMATICS A. Motion in One Dimension 1. The relationships among position, velocity and acceleration a. Given a graph of position vs. time, identify or sketch a graph
More informationPhysics 112. Study Notes for Exam II
Chapter 20 Electric Forces and Fields Physics 112 Study Notes for Exam II 4. Electric Field Fields of + and point charges 5. Both fields and forces obey (vector) superposition Example 20.5; Figure 20.29
More informationP114 University of Rochester NAME S. Manly Spring 2010
Exam 2 (March 23, 2010) Please read the problems carefully and answer them in the space provided. Write on the back of the page, if necessary. Show your work where indicated. Problem 1 ( 8 pts): In each
More information1 2 U CV. K dq I dt J nqv d J V IR P VI
o 5 o T C T F 3 9 T K T o C 73.5 L L T V VT Q mct nct Q F V ml F V dq A H k TH TC L pv nrt 3 Ktr nrt 3 CV R ideal monatomic gas 5 CV R ideal diatomic gas w/o vibration V W pdv V U Q W W Q e Q Q e Carnot
More informationExam III Solution: Chapters 18 20
PHYS 1420: College Physics II Fall 2006 Exam III Solution: Chapters 18 20 1. The anode of a battery A) has a positive charge, while the cathode has a negative charge. B) has a negative charge, while the
More information= 8.89x10 9 N m 2 /C 2
PHY303L Useful Formulae for Test 2 Magnetic Force on a moving charged particle F B = q v B Magnetic Force on a current carrying wire F B = i L B Magnetic dipole moment µ = NiA Torque on a magnetic dipole:
More informationHandout 10: Inductance. SelfInductance and inductors
1 Handout 10: Inductance SelfInductance and inductors In Fig. 1, electric current is present in an isolate circuit, setting up magnetic field that causes a magnetic flux through the circuit itself. This
More informationPhysics 42 Exam 3 Spring 2016 Name: M T W
Physics 42 Exam 3 Spring 2016 Name: M T W Conceptual Questions & Shorty (2 points each) 1. Which magnetic field causes the observed force? 2. If released from rest, the current loop will move a. upward
More informationPhysics GRE: Electromagnetism. G. J. Loges 1. University of Rochester Dept. of Physics & Astronomy. xkcd.com/567/
Physics GRE: Electromagnetism G. J. Loges University of Rochester Dept. of Physics & stronomy xkcd.com/567/ c Gregory Loges, 206 Contents Electrostatics 2 Magnetostatics 2 3 Method of Images 3 4 Lorentz
More informationElectromagnetic Induction Practice Problems Homework PSI AP Physics B
Electromagnetic Induction Practice Problems Homework PSI AP Physics B Name Multiple Choice Questions 1. A square loop of wire is placed in a uniform magnetic field perpendicular to the magnetic lines.
More information9. M = 2 π R µ 0 n. 3. M = π R 2 µ 0 n N correct. 5. M = π R 2 µ 0 n. 8. M = π r 2 µ 0 n N
This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 00 0.0 points A coil has an inductance of 4.5 mh, and the current
More informationPhysics Physics 2102
Physics 2102 Jonathan Dowling Physics 2102 Exam 2: Review Session Chapters 24.928.8 / HW0406 Some links on exam stress: http://appl003.lsu.edu/slas/cas.nsf/$content/stress+management+tip+1 http://wso.williams.edu/orgs/peerh/stress/exams.html
More informationName: Class: Date: Multiple Choice Identify the letter of the choice that best completes the statement or answers the question.
Name: Class: _ Date: _ w9final Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. If C = 36 µf, determine the equivalent capacitance for the
More informationName. Physics 111. Winter Exam #2. February 10, Part Multiple Choice / 12 Problem #1 / 24 Problem #2 / 32 Problem #3 / 32 Total / 100
Physics 111 Winter 010 Exam # February 10, 010 Name In keeping with the Union College policy on academic honesty, it is assumed that you will neither accept nor provide unauthorized assistance in the completion
More informationPhysics 2401 Summer 2, 2008 Exam II
Physics 2401 Summer 2, 2008 Exam II e = 1.60x1019 C, m(electron) = 9.11x1031 kg, ε 0 = 8.845x1012 C 2 /Nm 2, k e = 9.0x10 9 Nm 2 /C 2, m(proton) = 1.67x1027 kg. n = nano = 109, µ = micro = 106, m
More information4. An electron moving in the positive x direction experiences a magnetic force in the positive z direction. If B x
Magnetic Fields 3. A particle (q = 4.0 µc, m = 5.0 mg) moves in a uniform magnetic field with a velocity having a magnitude of 2.0 km/s and a direction that is 50 away from that of the magnetic field.
More informationDiscussion Question 7A P212, Week 7 RC Circuits
Discussion Question 7A P1, Week 7 RC Circuits The circuit shown initially has the acitor uncharged, and the switch connected to neither terminal. At time t = 0, the switch is thrown to position a. C a
More informationExam 2 Fall 2014
1 95.144 Exam 2 Fall 2014 Section instructor Section number Last/First name Last 3 Digits of Student ID Number: Show all work. Show all formulas used for each problem prior to substitution of numbers.
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Spring 2014 Final Exam Equation Sheet. B( r) = µ o 4π
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2014 Final Exam Equation Sheet Force Law: F q = q( E ext + v q B ext ) Poynting Vector: S = ( E B) / µ 0 Force on Current Carrying
More information2) As two electric charges are moved farther apart, the magnitude of the force between them.
) Field lines point away from charge and toward charge. a) positive, negative b) negative, positive c) smaller, larger ) As two electric charges are moved farther apart, the magnitude of the force between
More information(a) zero. B 2 l 2. (c) (b)
1. Two identical coaxial circular loops carry equal currents circulating in the same direction: (a) The current in each coil decrease as the coils approach each other. (b) The current in each coil increase
More informationMagnetic Fields; Sources of Magnetic Field
This test covers magnetic fields, magnetic forces on charged particles and currentcarrying wires, the Hall effect, the BiotSavart Law, Ampère s Law, and the magnetic fields of currentcarrying loops
More informationPHY 131 Review Session Fall 2015 PART 1:
PHY 131 Review Session Fall 2015 PART 1: 1. Consider the electric field from a point charge. As you move farther away from the point charge, the electric field decreases at a rate of 1/r 2 with r being
More informationDe La Salle University Manila Physics Fundamentals for Engineering 2 Quiz No. 3 Reviewer
De La Salle University Manila Physics Fundamentals for Engineering 2 Quiz No. 3 Reviewer Multiple Choice: 1. Which of the two arrangements shown has the smaller equivalent resistance between points a and
More informationFirst Name: Last Name: Section: n 1. March 26, 2003 Physics 202 EXAM 2
First Name: Last Name: Section: n 1 March 26, 2003 Physics 202 EXAM 2 Print your name and section clearly on all five pages. (If you do not know your section number, write your TA s name.) Show all work
More informationASSOCIATE DEGREE IN ENGINEERING TECHNOLOGY RESIT EXAMINATIONS SEMESTER 2 JUNE 2011
ASSOCIATE DEGREE IN ENGINEERING TECHNOLOGY RESIT EXAMINATIONS SEMESTER 2 JUNE 2011 COURSE NAME: PHYSICS 2 CODE: GROUP: ADET 1 DATE: JUNE 29 TIME: 1:00 DURATION: 2 HOUR INSTRUCTIONS: 1. This paper consists
More informationChapter 26 DirectCurrent Circuits
Chapter 26 DirectCurrent Circuits 1 Resistors in Series and Parallel In this chapter we introduce the reduction of resistor networks into an equivalent resistor R eq. We also develop a method for analyzing
More informationLenz s Law (Section 22.5)
Lenz s Law (Section 22.5) : Thursday, 25 of February 7:00 9:00 pm Rooms: Last Name Room (Armes) Seats A  F 201 122 G  R 200 221 S  Z 205 128 20160221 Phys 1030 General Physics II (Gericke) 1 1) Charging
More information2006 #3 10. a. On the diagram of the loop below, indicate the directions of the magnetic forces, if any, that act on each side of the loop.
1992 1 1994 2 3 3 1984 4 1991 5 1987 6 1980 8 7 9 2006 #3 10 1985 2006E3. A loop of wire of width w and height h contains a switch and a battery and is connected to a spring of force constant k, as shown
More information3. In the adjacent figure, E 1 = 6.0 V, E 2 = 5.5 V, E 3 = 2.0 V, R 1 = 1W, and R 2 = 6W. All batteries are ideal. Find the current in resistor R 1.
1. A cylindrical copper rod of length L and crosssectional area A is reformed to twice its original length with no change in volume. If the resistance between its ends was originally R, what is it now?
More informationA) 4 B) 3 C) 2 D) 5 E) 6
Coordinator: Saleem Rao Monday, January 01, 2018 Page: 1 Q1. A standing wave having three nodes is set up in a string fixed at both ends. If the frequency of the wave is doubled, how many antinodes will
More informationQuestions A hair dryer is rated as 1200 W, 120 V. Its effective internal resistance is (A) 0.1 Ω (B) 10 Ω (C) 12Ω (D) 120 Ω (E) 1440 Ω
Questions 441 36. Three 1/ µf capacitors are connected in series as shown in the diagram above. The capacitance of the combination is (A).1 µf (B) 1 µf (C) /3 µf (D) ½ µf (E) 1/6 µf 37. A hair dryer is
More informationChapter 6 DIRECT CURRENT CIRCUITS. Recommended Problems: 6,9,11,13,14,15,16,19,20,21,24,25,26,28,29,30,31,33,37,68,71.
Chapter 6 DRECT CURRENT CRCUTS Recommended Problems: 6,9,,3,4,5,6,9,0,,4,5,6,8,9,30,3,33,37,68,7. RESSTORS N SERES AND N PARALLEL  N SERES When two resistors are connected together as shown we said that
More information