lecture12 - 12. Lecture 12 12.1 LR circuit, comparison with...

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Unformatted text preview: 12. Lecture 12 12.1 LR circuit, comparison with RC An inductor stores energy in the form of a magnetic field similarly as a capacitor stores energy in the form of an electric field. In both cases we can ”charge” the device by loading energy into it taking it it for example from a battery. In the case of a capacitor we did that through a resistor in an RC circuit. The charging time is given by τ = RC . See the fig.68 to remind yourself how the current and voltage behave as a function of time in such case. Figure 68: AS seen before, energy from a battery can be stored in a capacitor using an RC circuit. The time constant of the circuit is given by τ = RC . Compare with the RL circuit below. We can do the same for an inductor by using an RL circuit, a resistor in series with an inductor. See fig.69. Initially, when we turn on the switch, the battery tries to increase the current in the circuit and that will occur almost instantaneously if the inductor is replaced by a cable. However such large changes in current are opposed by the inductor which creates between its terminals a voltage that opposes the increase in current. For that reason the current increases at a slow rate until it reaches its final value given by: 1 If = ∆Vbattery (12.1) R at that time the current does no change anymore and the inductor behaves as a cable, namely no voltage cross it terminals. However the inductor is generating a magnetic field so it has energy stored in it. To see how long it takes to reach such a stationary state, consider the voltage across the inductor Va − Vb = L – 80 – ∆I ∆t (12.2) Initially, as we said, the inductor opposes the battery and does not allow any current to circulate. This means that Va − Vb = ∆Vbattery and the initial slope with which the current start increasing is then: ￿ ∆I ￿ ￿ = 1 ∆V (12.3) ∆t ￿ L battery As the current increases, this slope decreases as can be see in the plot in fig.69. However, we get a good estimate of the time it takes to reach the steady state if we assume the slope to be constant and ask how long it would take with that slope to reach the current If . It is ∆Vbattery If 1 L ∆ t = ∆I = = (12.4) ∆V R R battery ∆t L So we find the characteristic time for an RL circuit is τ= L R (12.5) The larger the inductance the longer it takes to charge. If the resistance is small it also takes longer since we want to reach a larger current in the end. Figure 69: Energy from a battery can be loaded in an inductor using an RL circuit. The time constant of the circuit is given by τ = R . Pay attention to the different voltage and L current curves compared to the capacitor case. 12.2 Using the oscilloscope A very useful tool to check what we discussed in the previous section is the oscilloscope (sometimes called simply ”scope”). It measures the voltage across its terminals and plots it as a function of time. One terminal is usually grounded and the other, which is called a probe, has a conducting pin that you can touch to different parts of the circuit to measure the voltage. Digital oscilloscopes have integrated circuits called analog to – 81 – digital converters which take the voltage and covert it into a number which they store in a memory. The instrument then displays the voltage as a function of time using an LCD display. On the screen the horizontal scale is time and the vertical scale is voltage. The scale can be set manually or can be left automatic (easier). In the past, and also for high frequencies oscilloscopes work with a vacuum tube where an electron beam was directed toward a screen. The path of the beam was changed by a potential applied to two parallel plates. In the horizontal direction is was moved back and forth at a constant rate (backward much faster than forward). In the vertical direction it was moved according to the input voltage. The resulting point on the screen traces the voltage as a function of time. Both type of oscilloscopes are perfectly good for looking at the RL circuit. In the demo we do that and check the voltage and current as a function of time. notice that to measure the current we can use Ohm’s law: ∆V = IR (12.6) So we can just measure the voltage cross the resistor to know the current. The humble resistor can be though as a current to voltage converter!. Figure 70: Demo. An RL circuit is analyzed with the aid of an oscilloscope. 12.3 Energy contained in a solenoid We mentioned several times that a solenoid or inductor contain energy. We are going to derive that such energy is given by 1 U = LI 2 2 (12.7) where L is the inductance and I the current circulating through the solenoid. To see how this works recall that for a solenoid of area A, length ￿ and number of turns N , – 82 – Figure 71: Demo. A vacuum tube where an electron beam is directed toward a screen producing a bright spot. The beam can be deflected by electric and magnetic fields. This allows to build an oscilloscope by using external input to deflect the beam. the inductance is given by L = µ0 N 2A ￿ (12.8) The energy therefore should be given by 1 N 2A 2 U = µ0 I 2 ￿ (12.9) To verify this result we are going to is a procedure analogous to what we did for a capacitor. In that case we computed how much energy was required to separate the parallel plates by a small distance ∆d. This allowed us to compute the energy in the capacitor. Here we can see how much energy is required to expand the solenoid so that the area is increased by ∆A. The reason we do that is that, from eq.(12.9) we expect that the energy is linear in A so it should be easy to compute the energy necessary to change A. The first consideration is that if we increase A, the flux through the solenoid changes and therefore a voltage is induced: Va − Vb = 2 ∆Φ ￿ ∆ A = − µ0 N I ∆ A = −N |B | ∆t ∆t ￿ ∆t – 83 – (12.10) This means that as we do this, we need to push the current against such potential difference. The power necessary to do that is P = I ( V a − V b ) = µ0 N 2 2 ∆A I ￿ ∆t (12.11) Since power is energy consumed in unit time, to get the total energy necessary to do this we multiply the power by ∆t obtaining ∆U1 = P ∆t = µ0 N2 2 I ∆A ￿ (12.12) This energy we loose, so it is incorporated into the solenoid. If we compare with eq.(12.9) we see that there is a factor of two. So we must be gaining energy somewhere else because we overestimated the energy we needed to do this. The reason is that the magnetic field exerts a force on the solenoid that tends to expand it. This forces does work when we expand the solenoid and therefore we get energy form there. Let us see how much this is. The force on a small portion of the cable is radial and per unit length equal to ￿ |F | ￿ = |B ￿ |I (12.13) λ ￿ The magnetic field is indicated as B ￿ because we should not consider the magnetic field created by the small portion of cable itself. To see what this is we can use Ampere’s law to compute the magnetic field generated by a small piece of the solenoid (see fig.72). It gives: N ￿ 2 | B | ∆ ￿ = µ0 ∆ ￿ I (12.14) ￿ implying that µ0 N ￿ |B | = I (12.15) 2￿ which is precisely half the magnetic field of the solenoid!. What that from the magnetic field acting on a small portion of the cable, half is created by the rest of the solenoid and half is created by itself and should be excluded (because the small portion of cable cannot make a force on itself). We conclude that µ0 N I 2￿ (12.16) ￿ |F | µ0 N 2 = I l 2￿ (12.17) ￿ |B ￿ | = We can now compute the force – 84 – The length of each turn is 2π R where R is the radius of the coil. The displacement if we increase the radius by ∆R is precisely ∆R. We can the compute the work as W= ￿ |F | µ0 N 2 2π R ∆ R = I ∆A l 2￿ (12.18) where we used that, if we change the radius by ∆R, the area changes precisely by ∆A = 2π R∆R (assuming ∆R ￿ R). This is work done by the solenoid, so we gain energy. Therefore the total change in the energy of the solenoid is ∆ U = ∆ U1 − W = µ0 N 2 I ∆A 2￿ (12.19) If we now assume that U = 0 when A = 0 because when A = 0 there is no magnetic field to speak of, we obtain that U= µ0 N 2 1 I A = LI 2 2￿ 2 (12.20) as we wanted to show. Admittedly this derivation is a bit lengthy but is a very useful exercise if you want to understand several properties of magnetic field, work, energy, etc. If you go through it carefully, the basic idea should appear quite simple and then one has to do the calculations with great care. – 85 – Figure 72: Energy contained in a coil. There is a force trying to expand the coil. It is created by the magnetic field acting on the current. However the magnetic field created by the small piece of coil considered should be excluded. If we expand the coil this force does work and the energy of the coil is reduced. On the other hand, a voltage is induced and we need to do work to keep the current circulating. In total the energy of the coil increases. It is given by U = 1 LI 2 . 2 – 86 – ...
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This note was uploaded on 12/07/2011 for the course PHY 219 taught by Professor Na during the Fall '11 term at Purdue University-West Lafayette.

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