Physics 142 AC Circuits Page 1. AC Circuits. I ve had a perfectly lovely evening but this wasn t it. Groucho Marx

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1 Physics 142 A ircuits Page 1 A ircuits I ve had a perfectly lovely evening but this wasn t it. Groucho Marx Alternating current: generators and values It is relatively easy to devise a source (a generator ) which produces a sinusoidally varying emf. A rotating coil in a magnetic field gives an important example; another is the sinusoidally varying potential across the inductor in an L- oscillator, used in many situations to provide an emf for another circuit. This kind of oscillating source is called an A (alternating current) generator. Its output is described by E = E max sinωt. Here E max is the maximum emf during a cycle of the oscillation. Shown here is a prototype A generator ( alternator ). The rotating coil ( armature ) has contact with two slip rings, which in turn pass the current to the load through conducting brushes that the rings slide across. The alternator used in most automobiles is a somewhat more sophisticated version of this device. Meters to measure the strength of sinusoidally varying voltages and currents do not usually measure the maximum value. Instead they measure the (root-mean-square) value. This is a statistical measure, defined for any varying quantity by: The value of a quantity f that varies over a MS values distribution is given by f = ( f 2 ) av, where the average is taken over the distribution. In words: square the quantity, take the average of that, then take the square root. In the case of sinusoidal time variation, the average is taken over the time for one cycle. Since the average of sin 2 ωt over a cycle is 1/2, we find a simple rule: E = E max / 2. This is what an A voltmeter, placed across the generator s terminals, will read. For example, the ordinary household outlets in the USA deliver an A voltage with maximum about 170 V, oscillating at frequency 60 Hz. What a voltmeter will read, placed across the terminals, is about 120 V. This is the value.

2 Physics 142 A ircuits Page 2 Example: the series L circuit We will analyze a circuit containing a capacitor, an inductor, a resistor and an A generator, in series as shown. We apply the loop rule at an instant when the current is running clockwise, putting positive charge on the lower capacitor plate. Then we find E L di Q/ I = 0. dt Using I = dq/dt and rearranging, we have L E d 2 Q dt 2 + L dq dt + 1 L Q = 1 L E max sinωt. This differential equation has the same mathematical form as the equation describing a driven oscillator with damping, so its solutions have the same form. The "steady-state" solution (describing the situation where the energy dissipated per cycle is exactly replaced by the energy input per cycle) has the form Q(t) = Q max cos(ωt φ ) where Q max and φ are determined by requiring Q(t) to satisfy the differential equation. We choose the cosine here so that the current, like the emf, will involve the sine. After substitution into the equation and much algebra we find the answer. It is conventionally written in te of quantities determined by the parameters of the circuit, all of which have the dimensions of resistance and are measured in ohms: 1. apacitive reactance: X = 1/ω eactances and impedance for series L circuit 2. Inductive reactance: X L = ωl 3. Impedance: = 2 + (X L X ) 2 In te of these quantities the constants in the solution we seek are: Q max = E max /ω, tanφ = X L X We are usually more interested in the current in the circuit than the capacitor s charge; taking the time derivative of Q we find: urrent in series L circuit I(t) = E max sin(ωt φ) The relations among these quantities is conveniently displayed in the impedance triangle shown. The current "lags" the emf (its peak occurs later in time) by the phase angle φ. If X L > X then φ is positive and the φ X L X

3 Physics 142 A ircuits Page 3 current lags behind the generator voltage; if X L < X then φ is negative and the current leads the generator voltage. The current and the generator voltage are exactly in phase (φ = 0 ) if X L = X. Voltage drops and power in the circuit elements We now examine the voltage drops across the elements and the power delivered to them. The instantaneous voltage drops are obtained from the expressions for I and Q: esistor:! V (t) = I(t) = E max sin(ωt φ) apacitor:v (t) = Q(t) = E max X cos(ωt φ) Inductor:! V L (t) = L di dt = E X L max cos(ωt φ) Note that the capacitor and inductor voltage drops are of opposite sign (i.e., are out of phase with each other by π ), and that they are out of phase with the current by ±π /2 [since cosθ = sin(θ + π /2)]. The values are relatively simple, and all take the form of Ohm s law:!!!! urrent:! I = E /!!!! esistor:! V!!!! apacitor:! V L!!!! Inductor:! V = I = I X = I X L An value is always positive, so the sum of the voltages around the circuit is not zero. (The sum of all the instantaneous potential differences around the circuit at any given time is zero, of course.) Indeed, there are situations where one or more of the voltages across an element exceeds the generator emf. Now we look at the power delivered to each element. Instantaneously, this is the product of the current into the element and the voltage drop across it. This product varies during a cycle, usually being sometimes positive and sometimes negative. We are ordinarily interested in the average power over a cycle. Take as an example the power delivered by the generator. Instantaneously it is P gen (t) = E(t)I(t) = E max 2 sinωt sin(ωt φ) To find the average of this over a cycle we integrate it with respect to t from 0 to T = 2π /ω (the period of the oscillation) and then divide the result by T. We find from this calculation:

4 Physics 142 A ircuits Page 4 Average power from generator P av (from generator) = E 2 cosφ = E I cosφ The last expression shows that the average power supplied by the generator is the product of the readings of an A voltmeter and an A ammeter, multiplied by the power factor cosφ. Applying this procedure to the other elements, we find that P av = I 2, while the average power to the capacitor and inductor are both zero. The capacitor and inductor take in power during half of the cycle and return it to the circuit in the other half, so the average is zero. The net power supplied by the generator thus goes entirely to the resistor, where it is converted to heat. Although the capacitor and inductor consume no net power themselves, they play roles in determining the power actually supplied by the generator, because they help determine the value of, which in turn determines the strength of the current. Series resonance: tuning to a frequency Since the series L circuit is mathematically analogous to a driven mechanical oscillator, there is a resonance phenomenon in it as well. The power supplied to the circuit is a maximum when is a minimum. This happens when X L = X. In this case φ = 0 so the current is exactly in phase with the generator voltage; there is never a time during the cycle when the generator is taking power out of the circuit. This is why the power supplied is as large as possible. The resonance condition is usually written in te of ω: esonant frequency ω = ω 0 = 1/ L The resonant frequency is thus equal to the "free" oscillation frequency of an L- oscillator without resistance. If is small then the average power to the resistor is small except at frequencies near ω 0. This is a "sharp" resonance, corresponding to a high Q-value for the circuit. This kind of resonance is used by the tuner sections of radios and TV sets to select one channel or station and reject others. P av = 10Ω Shown are curves of P av vs. ω for a series L circuit. Two values of are plotted to show its effect on the sharpness and height of the peak. ω 0 = 40Ω ω

5 Physics 142 A ircuits Page 5 Passive filters as A circuits A common use of the elements we have been discussing is in construction of passive filters. (A passive element is one that has in it no sources of e-m energy.) An A input voltage is passed into the device by a pair of terminals. On the other side of the filter are two other terminals, across which is the output voltage. The situation is drawn schematically as shown. V in Filter V out Here V in is an A voltage, which can be considered to be that of an A generator of frequency ω, and V out is also an A voltage at the same frequency, usually the voltage across some circuit element in the filter. Our interest is in the ratio V out /V in, sometimes called the attenuation factor. onsider as an example the circuit shown. Let the instantaneous input voltage be V 0 sinωt. Then V in = V 0 / 2. The impedance of the circuit containing and is = 2 + (1/ω) 2. The V V in out voltage across (which is V out ) is given by V out = I = V in, so we find for this filter V out V in = 2 + (1/ω) 2 = ω. (ω) We see that as ω 0 the ratio goes to zero, while as ω the ratio approaches 1. This type of filter is called high-pass since it leaves high frequency voltages unaffected and suppresses low frequency voltages. If we reverse the locations of and we have V out = I (1/ω) = V in from which we find (1/ω) 2 + (1/ω) 2, V out V in = (1/ω) = (1/ω) 2 (ω) Now the ratio approaches 1 for low frequencies, and it approaches zero for high frequencies. This configuration is a low-pass filter. One can also use inductors and resistors, or combinations of L, and, to make filters. Filters are very common in electronic circuits and in audio systems. The attenuation ratio is often measured in te of db differences.

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