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DC Electrical Circuits

DC Electrical Circuits. Chapter 28 (Continued) Circuits with Capacitors. The loop method is based on two laws devised by Kirchoff:. Kirchhoff’s Laws. I 2. I 1. I 3 = I 1 + I 2. 1. At any circuit junction, currents entering must equal currents leaving. I.

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DC Electrical Circuits

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  1. DC Electrical Circuits Chapter 28 (Continued) Circuits with Capacitors

  2. The loop method is based on two laws devised by Kirchoff: Kirchhoff’s Laws I2 I1 I3= I1+ I2 1. At any circuit junction, currents entering must equal currents leaving. I 2. Sum of all DV’s across all circuit elements in a loop must be zero. r R + E - E - Ir - IR = 0

  3. R + E C - open closed RC Circuits: Charging I R VR=IR + +++ E - - - - VC=q/C C When the switch closes, at first a high current flows: VR is big and VC is small.

  4. R + E C - open closed RC Circuits: Charging I R VR=IR + +++ E - - - - VC=q/C C When the switch closes, at first a high current flows: VR is big and VC is small. As q is stored in C, VC increases. This fights against the battery so I decreases.

  5. closed I R VR=IR + +++ E - - - VC=q/C - C RC Circuits: Charging Apply the loop law:E – IR - q/C = 0 Take the derivative of this with respect to time: Now use dq/dt = I and rearrange: This is a differential equation for an unknown function I(t). It is solved subject to the initial condition I(0) = E / R.

  6. closed I R VR=IR + +++ E - - - VC=q/C - C RC Circuits: Charging  E And I(0) = I0 = E / R 

  7. closed I R VR=IR + +++ E - - - VC=q/C - C RC Circuits: Charging E E E E From this we get: q = VC C =E C (1 – e–t/RC)

  8. Charging E/R Current I t/RC E VC Potential Drop VR t/RC

  9. R VR=IR q I C VC=q/C -q After closing switch Discharging an RC Circuit R q C VC=V0 -q Open circuit Current will flow through the resistor for a while. Eventually, the capacitor will lose all its charge, and the current will go to zero. Power P = IV = I2R will be dissipated in the resistor (as heat) while the current flows.

  10. Discharging an RC Circuit Loop equation: q/C - IR = 0  I = q / (RC) R VR=IR +q I C VC=q/C Take d/dt -q [Note that I = - dq/dt] Here the current at t=0 is given by the initial voltage on the capacitor: I(0) = V0/R = q0 /RC This equation is solved very much like the other (charging case):

  11. Discharging an RC Circuit R VR=IR q I C VC=q/C -q The charge on the capacitor is given by: q/C - IR = 0 so q = C IR [q = C V]

  12. Discharging E/R Current t/RC VR 0 Potential Drop VC t/RC

  13. Example: A capacitor C discharges through a resistor R. (a) When does its charge fall to half its initial value ? Charge on a capacitor varies as R Q C I

  14. Example: A capacitor C discharges through a resistor R. (a) When does its charge fall to half its initial value ? Charge on a capacitor varies as R Q C Find the time for which Q=Q0/2 I

  15. Example: A capacitor C discharges through a resistor R. (a) When does its charge fall to half its initial value ? Charge on a capacitor varies as R Q C Find the time for which Q=Q0/2 I

  16. Example: A capacitor C discharges through a resistor R. (a) When does its charge fall to half its initial value ? Charge on a capacitor varies as R Q C Find the time for which Q=Q0/2 I RC is the “time constant”

  17. Example: A capacitor C discharges through a resistor R. (b) When does the energy drop to half its initial value? The energy stored in a capacitor is We seek the time for U to drop to U0/2:

  18. Example: A capacitor C discharges through a resistor R. (b) When does the energy drop to half its initial value? The energy stored in a capacitor is We seek the time for U to drop to U0/2:

  19. Magnetic FieldsChapter 29 Permanent Magnets & Magnetic Field Lines The Magnetic Force on Charges

  20. Magnetism • Our most familiar experience of magnetism is through permanent magnets. • These are made of materials which exhibit a property called ferromagnetism - i.e., they can be magnetized. • Depending on how we position two magnets, they will attract or repel, i.e. they exert forces on each other. • Just as it was convenient to use electric fields instead of electric forces, here too it is useful to introduce the concept of the magnetic field B. • There are useful analogies between electric and magnetic fields, but the analogy is not perfect: while there are magnetic dipoles in nature, there seem to be no isolated magnetic charges (called “magnetic monopoles”). And the force laws are different. • We describe magnets as having two magnetic poles: North (N) and South (S). • Like poles repel, opposite poles attract.

  21. N S Field of a Permanent Magnet Shown here are field lines. The magnetic field B at any point is tangential to the field line there.

  22. N N S S Field of a Permanent Magnet The south pole of the small bar magnet is attracted towards the north pole of the big magnet. Also, the small bar magnet (a magnetic dipole) wants to align with the B-field. The field attracts and exerts a torque on the small magnet.

  23. Magnetism • The origin of magnetism lies in moving electric charges. Moving (or rotating) charges generate magnetic fields. • An electric current generates a magnetic field. • A magnetic field will exert a force on a moving charge. • A magnetic field will exert a force on a conductor that carries an electric current.

  24. What Force Does a Magnetic Field Exert on Charges? • If the charge is not moving with respect to the field (or if the charge moves parallel to the field), there is NO FORCE. q

  25. What Force Does a Magnetic Field Exert on Charges? • If the charge is not moving with respect to the field (or if the charge moves parallel to the field), there is NO FORCE. q • If the charge is moving, there • is a force on the charge, • perpendicularto both v and B. • F = q vxB q

  26. Force on a Charge in aMagnetic Field F v q m B (Use “Right-Hand” Rule to determine direction of F)

  27. Units of Magnetic Field Since Therefore the units of magnetic field are: (Note: 1 Tesla = 10,000 Gauss)

  28. The Electric and Magnetic Forces are Different Whereas the electric force acts in the same direction as the field: The magnetic force acts in a direction orthogonal to the field:

  29. The Electric and Magnetic Forces are Different Whereas the electric force acts in the same direction as the field: The magnetic force acts in a direction orthogonal to the field: (Use “Right-Hand” Rule to determine direction of F)

  30. The Electric and Magnetic Forces are Different Whereas the electric force acts in the same direction as the field: The magnetic force acts in a direction orthogonal to the field: (Use “Right-Hand” Rule to determine direction of F) And – the charge must be moving.

  31. Trajectory of Charged Particlesin a Magnetic Field (B field points into plane of paper.) B + + + + + + + + + + + + + + + + + + + + v F

  32. Trajectory of Charged Particlesin a Magnetic Field (B field points into plane of paper.) v B B + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + v F F

  33. Trajectory of Charged Particlesin a Magnetic Field (B field pointsinto plane of paper.) v B B + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + v F F Magnetic Force is a centripetal force

  34. Rotational Motion  = s / r  s =  r  ds/dt = d/dt r  v =  r s  r  = angle,  = angular speed,  = angular acceleration at = r  tangential acceleration ar = v2 / rradial acceleration  at ar The radial acceleration changes the direction of motion, while the tangential acceleration changes the speed. Uniform Circular Motion  = constant  v and ar constant but direction changes ar  KE = (1/2) mv2 = (1/2) mw2r2 ar = v2/r = 2 r v F = mar = mv2/r = m2r

  35. v B + + + + + + + + + + + + + + + + + + + + F r Radius of a Charged ParticleOrbit in a Magnetic Field Centripetal Magnetic Force Force =

  36. Radius of a Charged ParticleOrbit in a Magnetic Field Centripetal Magnetic Force Force = v B + + + + + + + + + + + + + + + + + + + + F r

  37. v B + + + + + + + + + + + + + + + + + + + + F r Radius of a Charged ParticleOrbit in a Magnetic Field Centripetal Magnetic Force Force =

  38. Radius of a Charged ParticleOrbit in a Magnetic Field Centripetal Magnetic Force Force = v B + + + + + + + + + + + + + + + + + + + + F r Note: as , the magnetic force does no work.

  39. v B + + + + + + + + + + + + + + + + + + + + F r Cyclotron Frequency The time taken to complete one orbit is:

  40. v B + + + + + + + + + + + + + + + + + + + + F r Cyclotron Frequency The time taken to complete one orbit is: Hence the orbit frequency, f

  41. v B + + + + + + + + + + + + + + + + + + + + F r Cyclotron Frequency The time taken to complete one orbit is: Hence the orbit frequency, f - known as the “cyclotron frequency” T = 2/ = 1/ƒ  ƒ = /2

  42. The Electromagnetic Force If a magnetic field and an electric field are simultaneously present, their forces obey the superposition principle and must be added vectorially: The Lorentz force + + + + + + + + + + + + + + + q

  43. Exercise electron B v v’ • In what direction does the magnetic field point? • Which is bigger, v or v’ ?

  44. Exercise: answer electron B v v’ F • In what direction does the magnetic field point ? • Into the page [F = -e vxB] • Which is bigger, v or v’ ? • v = v’ [B does no work on the electron, Fv]

  45. What is the orbital radius of a charged particle (charge q, mass m) having kinetic energy K, and moving at right angles to a magnetic field B, as shown below?. x x x B x x x K • q m

  46. What is the orbital radius of a charged particle (charge q, mass m) having kinetic energy K, and moving at right angles to a magnetic field B, as shown below?. F = q v x B = m a and a = v2 / r q v B = m v2 / r x x x B x x x q B = m v / r  r q B = m v r r = m v / (q B) K = (1/2) mv2 • q m r2 = m2 v2 / (q B)2 (1/2m)r2 = K / (q B)2  r = [2mK]1/2 / (q B)

  47. x x x B E x x x v • q m What is the relation between the intensities of the electric and magnetic fields for the particle to move in a straight line ? The magnetic field points into the picture. The direction of the electric field is not yet specified.

  48. x x x B E x x x v • q m v = E / B What is the relation between the intensities of the electric and magnetic fields for the particle to move in a straight line ? FE = q E and FB = q v B If FE = FB the particle will move following a straight line trajectory q E = q v B FB FE •

  49. v = E / B What is the relation between the intensities of the electric and magnetic fields for the particle to move in a straight line ?. x x x B E FE = q E and FB = q v B x x x If FE = FB the particle will move following a straight line trajectory v • q m q E = q v B FB FE • So need E pointing to the right.

  50. Trajectory of Charged Particlesin a Magnetic Field What if the charged particle has a velocity component along B? unchanged Circular motion in xy plane. x z y

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