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lecture13 - 13 Lecture 13 13.1 Electric generators and...

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Unformatted text preview: 13. Lecture 13 13.1 Electric generators and alternate current We discussed before that a cable moving in a magnetic field acts as a battery since the magnetic field creates a force that moves the charges along the cable. This is the principle of the electric generator. It is more efficient to move the cable in circles so the motion is repetitive. In figs.73, 74, 75 we have a simple generator (see fig.76 for an actual example). A rectangular loop moves inside a magnetic field an a voltage is generated between terminals (a) and (b). We can find the voltage from Faraday’s law or equivalent from the Lorentz force. Figure 73: Schematic AC generator. A cable in the shape of a loop rotates inside a magnetic field. View in perspective. The magnetic flux going through the loop is given by ￿ Φ = A|B | cos φ (13.1) ￿ where A is the area of the loop, |B | is the modulus of the magnetic field (created for example by a permanent magnet) and φ is the angle between the normal to the loop and the magnetic field. If the loop is horizontal the flux is maximum and if it is vertical the flux is zero. If the loop rotates with angular velocity ω we have φ = ω t. The flux changes in time and, according to Faraday’s law a voltage is appears between the terminals: ∆Φ ￿ ∆ cos ω t Va − Vb = = A|B | (13.2) ∆t ∆t The only complication is that we need to compute the variation in time of cos ω t. This can be done by using a mechanical analog. Indeed, if the position of an object is given by x = cos ω t (13.3) – 87 – Figure 74: Schematic AC generator. A cable in the shape of a loop rotates inside a magnetic field. Front view. Figure 75: Schematic AC generator. The variation in flux creates a potential between terminals (a) and (b). The polarity alternates as seen in this figure. The symbols at the end ￿ ￿ of the conductor indicate if the current goes into the page or out of the page . then its velocity is given by vx = ∆x ∆ cos ω t = ∆t ∆t (13.4) We know how to do this, we just need to consider a particle moving in circles as in fig.77. If it rotates with angular velocity omega, the position is given by x = R cos ω t (13.5) y = R sin ω t (13.6) – 88 – Figure 76: Demo. Actual AC generator. Cranking the handle creates enough power to light up a small light bulb. The modulus of the velocity is given by the fact that it goes around the circle in time π T = 2ω . The length of the circle is ￿ = 2π R so the velocity is |￿ | = v 2π R = ωR T (13.7) The direction of the velocity is as in the figure. Projecting over the x and y axis we get for the velocity: vx = −ω R sin ω t vy = ω R cos ω t (13.8) (13.9) We then derived the following very important formulas ∆ cos ω t = −ω sin ω t ∆t ∆ sin ω t = ω cos ω t ∆t – 89 – (13.10) (13.11) Figure 77: For a particle rotating in circle sit is easy to compute the velocity. This gives us same useful formulas that we can apply in other situations. The notation is not very precise but what it means is that is something moves as cos ω t then its velocity is −ω sin ω t and similarly for the sine. This mechanical analog allows us then to compute the voltage as ￿ Va − Vb = −A|B |ω sin ω t (13.12) It is clearly seen form the nature of the motion that the relative polarity of (a) and (b) alternates. For that reason this is called an AC generator. AC stands for alternate current. If connected to a resistor the current circulates in one direction and then the other, that is why it is is called alternate current. The household current alternates 60 times a second. This is called the frequency and is measured in Hertz: 1Hz = 1 1 s (13.13) The household current therefore has a frequency of 60Hz. It should be noticed that the function V = V0 sin ω t (13.14) – 90 – π repeats itself when we shift t → t + 2ω (since sine is a function that repeats itself when the argument is shifted by 2π ). therefore the period and frequency of the current are defined as: 2π 1 ω T= , f= = (13.15) ω T 2π 13.2 AC resistor circuit An AC generator connected to a resistor dissipates power. Notice that power is dissipated no matter the direction of the current. According to Ohm’s law we have ∆V = IR (13.16) ∆V = V0 sin ω t (13.17) 1 V0 sin ω t R (13.18) For the generator we have therefore I= the power dissipated is 12 2 V sin ω t (13.19) R0 It changes in time but we can compute the average power dissipated. To average over one cycle we notice that, if instead of a sine we had a cosine, the power dissipated would be the same. But we have (denoting average with parenthesis): P = I ∆V = ￿sin2 ω t + cos2 ω t￿ = ￿1￿ We get then ￿P ￿ = where we defined ⇒ V02 Vrms = 2R R 1 Vrms = √ V0 2 ￿sin2 ω t￿ = 1 2 (13.20) (13.21) (13.22) which is known as root mean square voltage (and V0 is known as peak voltage). – 91 – ...
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