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1 Alternating current Book pg
2 Review of induced emf Emf = BLv, v = speed of rod Φ = BA flux density Emf = NΦ rate of change of flux density t Normal to surface Φ = BA cos θ θ = 0 Φ = BA θ = 90 0 θ Φ = 0 B, magnetic flux density
3 Definitions The magnetic flux density is numerically equal to the magnetic field strength Definition A rate of change of flux of 1 Weber per second induces an emf of 1 Volt across a conductor Φ is measured in Weber (Wb) Magnetic flux density = B vector Magnetic flux = Φ = BA scalar One Telsa = 1 Weber per square meter N Φ = magnetic flux linkage
4 Rotating coil Emf in a coil rotated in an uniform magnetic field: Cathode ray oscilloscope used to measure output voltage of an ac source Time base adjusted to obtain a sine curve Frequency can be determined How to use an oscilloscope with an A_C source - YouTube [720p].mp4
5 Example The potentiometer is set on 2 V/division and the time base is set a 5 ms/cm, what is the voltage and frequency of the ac generator? Solution The amplitude of the wave is 3 divisions and each division is 2 V. Therefore, the emf would be 6 V. Between the two dots there are 6 divisions. λ 2 = 6 divisions λ = 12 divisions There are 5 milliseconds/ cm. 12 x 5 = 60ms T = 60 ms. f = 1 T = 1 60x10 3 = 16.67~17Hz
6 Generator A generator produces electrical energy from mechanical energy Consists of a coil with a large number of turns Coil rotates relative to a magnetic field
7 Experimental setup Assume a fixed coil between two poles of a U shaped magnet at the center of a rotating turntable Possible changes: - turntable different angular speed - coil different # of turns - coil different cross sectional are
8 Observations: When magnetic flux Φ in the coil is at maximum, induced emf (current) is at minimum(0) and vise versa Changing the speed of the turntable changes the frequency and amplitude of induced emf Increasing the number of turns or increasing the area of the coil will - increase the emf - frequency is unchanged if the velocity of the turntable is constant Emf varies as turntable rotates: - this depends on θ, the angle between the normal to the coil and the field lines
9 Flux Zero flux: θ = 90 0 Φ = BA cos 90 = 0 field lines lie in plane of coil Normal to surface Maximum flux: θ = 0 Φ = BA cos 0 = BA max value Field lines are at 90 0 to plane of coil Normal to surface
10 Graphical representation Emf = - NΦ t rate of change of flux density / linkage Φ = BA cos θ flux density / linkage Emf = Φ negative of gradient of t Φ t graph take derivative, y = sinθ y =cos θ, negative = cosθ Flux linkage Observation: Rate of change of flux linkage and induced emf When flux linkage is at maximum, emf is at minimum θ = 90 0 max flux, min emf θ = 0 0 min flux, max emf
11 Mathematically Flux linkage is given by Φ = NBA cos θ Emf = Φ cos θ = NAB t t Using differentiation: Emf = NAB( sin θ) θ t = NAB sin θ θ t But θ t = ω = angular velocity in rads 1 = 2πf And θ = ωt = 2πft Hence emf= NAB sin ωt 2πft t emf = 2πfNAB sin 2πft or emf = ωnab sin ωt
12 emf = ωnab sin ωt When plane of coil is parallel to field lines (normal is at 90 0 ) then θ = 90 0 and sin ωt will have maximum value: sin 90 = 1 This is called peak voltage ε 0 ε 0 = ωnba Hence emf = ε 0 sin ωt From here we can see Flux Φ=BA cos θ cos max at 0 emf ε = ε 0 sin ωt sin max at 90 0
13 AC generator Slip rings: rotate with coil about the same axis Carbon brush: charge flows out into circuit One complete rotation of the coil Through gives one cycle of alternating current Essential requirements rotating coil Magnetic field Relative motion between coil and field Connection to outside world
14 Changes to rotation Increasing rotation speed of coil without changing any other part of coil or field: This results in more cycles every second Hence frequency increases Time between maximum and minimum flux linkage will decrease Since flux linkage Φ is constant, rate of change increases Peak emf or amplitude increases
15 Max emf when θ = 90 0 The maximum value of the e.m.f (E o ) is when θ (= ωt) = 90 o that is, the coil is in the plane of the field, Figure 2 At this point the wires of the coil are cutting through the flux at right angles they chop through the field lines rather than slide along them.
16 N N θ = 90 0 Zero flux Φ = BA cos θ = 0 Max emf ε = ε 0 sin ωt = 1 θ = 0 0 Max flux Φ = BA cos θ = 1 Zero emf ε = ε 0 sin ωt = 0
17 Example Calculate the peak voltage of a simple generator if the square armature has sides of 5.40cm and it contains 120 loops. It rotates in a magnetic field of 0.80 T at the rate of 110 revolutions per second. Solution l = 0.054m N = 120 B = 0.80T f = 110Hz ε = 2πfNAB sin 2πft Peak voltage when sin ωt = 1 ε 0 = 2πfNAB = 2πfN l 2 B = 2π = ~193V
18 Example A coil with 1200 turns has an area of m 2 and is rotating at 50 revolutions per second in a magnetic field of magnitude 0.50T. Draw graphs to show how magnetic flux, emf and current changes as a function of time. Assume current flows in a circuit with R = 25Ω Solution Magnetic flux Φ N = 1200 A = m 2 f = 50Hz B = 0.50T revs in 1 s 1revs in x s x = 1 = 0.02s = 20ms 50 Φ= NBA = = 12Wb
19 ..continued Emf max emf when sin ωt = 1 ε 0 = 2πfNAB = 2π = 100π 12 = 3770V Emf(V) Current I = ε R = ε 0 sin ωt R we want ε 0 when sin ωt= 1 I 0 = ε 0 R = = 150.8~150A Same graph with max and min I 0 at 150A
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