hapter 31: A ircuits A urrents and Voltages In this chapter, we discuss the behior of circuits driven by a source of A. Recall that A means, literally, alternating current. An alternating current is a current that changes periodically (alternates) in sense of flow (half of the time flowing W and the other half of the time flowing W, e.g.) Also recall that we use A in a broader sense to refer to anything that changes with time. hus, an A voltage is a voltage that varies with time. here are three particular types of time-dependent currents (or voltages) that occur frequently in applications: 1. sinusoids (sine wes or cosine wes). triangle wes 3. square wes 1
We will focus on sinusoids in this course. urrents and voltages of this particular shape arise naturally from A generators driven at constant speeds and are consequently ubiquitous in power distribution systems (the 10-V A power line, e.g.)
Root-Mean-Square (rms) urrents and Voltages onsider an A source supplying a sinusoidal current it ( ) given by: it ( ) Icosωt Note that there is no loss of generality in choosing it ( ) to be given by a cosine function rather than a sine function. hoosing it ( ) to be given by = (1) the expression in Eq. (1) amounts simply to choosing when to start timing; we start timing (at 0 it is at its maximum value, I. t = ) when ( ) he root-mean-square (rms) current supplied by this source is defined to be the square-root of the mean of the square of it: ( ) I i () t rms () From calculus, the erage value of any function ( ) is defined to be: 1 b () () f t b a a f t dt f t from t = a to t = b 3
herefore, the erage of i ( t ) over one complete cycle (from t = 0 to t =, in which is the period) is given by: 1 1 i () t = i () t dt = I cos ωtdt 0 0 o do this integral, I change the variable of integration by defining a new function, u() t : ωt (3) u( t) By the definition of the total differential: du du dt dt du = ω dt or: du dt = ω Substituting u and dt into the expression for i ( t) ω I () cos ω 0 i t = udu above, I get: 4
he limits of integration are determined from the definition of u( t ). At the lower limit ( t = 0), u( 0) = ω ( 0) = 0. At the upper limit (t = ), u ( ) = ω. o perform the integration, we make use of the trig identity: 1+ cosu cos u = Using this, I get: ω ω ω I 1+ cosu I i () t = du du cos udu ω = + ω 0 0 0 I 1 ω I 1 i () t = ω + [ sin u] ω [ sin ω sin 0] 0 = + ω ω But ω = π, so: I 1 I 1 i () t = [ sin 4 sin 0] [ 0 0] ω ω + π = ω ω + I i () t = 5
So: I Irms = i () t = I I rms = (4) Similarly, the rms voltage is defined to be: () (5) Vrms v t For a sinusoidal A source supplying a time-varying voltage vt ( ) = Vcosωt, the rms voltage would be (by a procedure exactly similar to that used to calculate I rms ): V V rms = (6) One reason why the rms voltage and current are important is that they arise naturally in discussing the erage power dissipated in a resistor, as we will see very shortly. 6
Resistors in A ircuits onsider a resistor R driven by an A source that supplies a sinusoidal it = Icosωt, as shown in Figure 1. Let the voltage current of the form ( ) supplied by the source be called vt ( ) and the voltage across the resistor v t, as in Fig. 1. What will v ( t ) be? (hat is, what function be called R () of time?) Ohm s law must be obeyed at every instant (i.e., at any time t). herefore, from Fig. 1: vr ( t) = i( t) R vr ( t) = IRcosωt vr( t) = VRcosωt, (7) v t evidently being: with the amplitude of ( ) R VR R = IR (8) 7
By Kirchhoff s voltage rule: vt ( ) vr ( t) = 0 herefore the voltage supplied by the source, vt, () is given by: vt = v t = IR ωt = V ωt, (9) ( ) ( ) cos cos R in which the amplitude of the voltage supplied by the source is: V = IR (10) 8
Note that the current through a resistor and the voltage across the resistor both look like some amplitude times cosω t. his implies that it ( ) and vr () t are in phase, meaning that they reach their maximum values at the same times. As we will see, this is not true for capacitors and inductors connected to A sources. Note also that, as shown in Eq. (8), the amplitudes of the voltage across the resistor and the current through the resistor obey Ohm s law: VR = IR Writing the amplitudes I and V R in terms of the rms current through the resistor and the rms voltage across it, Eq. (8) becomes: V = I R ( ) ( ) R rms ( ) rms VR = I rms rmsr, (11) so the rms voltage and current also obey Ohm s law. 9
Average Power Dissipated in a Resistor Recall (hapter 5) the power supplied by a D source supplying voltage V and current I : P= VI For an A source, the voltage and the current are both functions of time. herefore, the power supplied by the source will also depend on time. At any time t, the instantantaneous power supplied will be: Pt ( ) = vtit ( ) ( ) (1) From calculus, the erage power supplied over one cycle is: 1 1 P = P() t dt v()() t i t dt = 0 0 We just found that, for a resistor connected across an A source, vt ( ) and it ( ) are given by vt ( ) = Vcosωt and it ( ) = Icosωt. So: 1 VI P = ( Vcos t)( Icos t) dt cos tdt ω ω = ω 0 0 We ve already done the erage of cos ω t over one cycle in the context of finding the rms voltage and current. By comparison with Eq. (3) and 10
the analysis following it, the erage of cos ω t over one cycle is evidently: 1 1 cos tdt ω = 0 herefore, the erage power supplied by the source to the resistor is: 1 P = VI (13) We usually write this in terms of V rms and I rms as follows: V I P = P = VrmsIrms (14) Note that this has exactly the same form as the equation for the power supplied by a D source. hus, as far as the erage power supplied to the resistor is concerned, the A source behes as though it were a D source supplying a fixed voltage V rms and a fixed current I rms. 11
he power supplied by the source to the resistor must go somewhere. It is dissipated as heat in the resistor. Because for a resistor, the rms voltage and current obey Ohm s law, we can write the erage power dissipated in R as: P = ( IrmsR) Irms P = IrmsR (15) or as: Vrms P = Vrms R Vrms P = (16) R Eqs. (15) and (16) are exactly similar in form to the corresponding results for the power dissipated in a resistor connected to a D source. (See h. 5.) 1
apacitors in A ircuits onsider an A source connected across a capacitor, as shown in Figure. As before, we let the current supplied by the A source be it () = Icosωt. Let the voltage supplied by the source be vt ( ) and the v t. voltage across the capacitor be ( ) By Kirchhoff s voltage rule: vt ( ) v ( t) = 0 vt ( ) v ( t) We would like to know what v ( ) it?) = (17) t is. (hat is, what function of time is o answer this question, we recognize that: dq it () = dt dq = i( t) dt Let the capacitor be uncharged at t = 0. Integrating from t = 0 to any qt), then, I get: later time t (when the charge is ( ) 13
t ( ) qt t 0 0 () dq = i t dt I qt () = Icosωtdt= ωcosωtdt ω 0 0 Letting u = ωt, this becomes: ωt I I qt () = cosudu sinωt ω = ω 0 he voltage across the capacitor is, by the defining equation for the capacitance: qt ( ) v () t = I v () t = sinωt ω or: v( t) = Vsinωt, (18) with: t 14
I V = (19) ω Notice that the voltage across the cap is not in phase with the current to it! he current goes like cosω t but the voltage goes like sinω t. his means that the voltage reaches its peak one-quarter cycle after the current reaches its peak. We say: he voltage across a capacitor lags the current to the capacitor by 90. Also note that the amplitudes of v ( t ) and ( ) it obey the relation: 1 V = I ω his looks like Ohm s law: V = IX, (0) with: 1 X (1) ω X is called the capacitive reactance. It is the effective resistance that the cap presents to an A source. 15
With ω in rad/s and in farads, the unit for X will be: 1 1 1 V X = = = =Ω ( rad/s)( F) ( rad/s)( V) A V A So X does he units of resistance. Finally, note that X is a frequency-dependent resistance. It depends on the angular frequency ω of the source. he angular frequency (in rad/s) is related to the frequency f (in Hz) by: ω = π f () So: 1 X = π f (3) At very low frequencies (i.e., as f 0), X. And at very high frequencies (as f ), X 0. herefore, we say: At very low frequencies, a cap acts like an open. At very high frequencies, a cap acts like a short. 16
Average Power Dissipated in a apacitor From the analysis we did before for the resistor, the erage power dissipated in the capacitor would be: 1 ()() P = v t i t dt 0 1 VI P = ( V sin t)( Icos t) dt sin tcos tdt ω ω = ω ω 0 0 Using the trig identity sin θ = sinθcosθ I find: ω VI VI VI VI P = sin tdt sinudu cos cos0 1 1 ω = ω 4ω = = 4ω 4ω 0 0 P = 0 here is, over one cycle, no power dissipated in the capacitor! [ ] [ ] 17
Inductors in A ircuits onsider an A source connected across an inductor (coil) of inductance L, as shown in Figure 3. Let the current supplied by the source be given by it () = Icosωt, as shown in Fig. 3. Let the voltage supplied by the source be vt () and let the voltage across the coil be vl ( t ), as shown in the figure. By Kirchhoff s voltage rule, vt ( ) vl ( t) = 0 vt ( ) vl ( t) As before, we d like to know what v ( ) = (4) t is (i.e., what function of time). o answer this question, we recognize that vl () t is just the induced emf across the coil: vl ( t) = εind Or, by Faraday s law (in terms of the inductance L): di vl () t = L dt L 18
Plugging in our expression for it: ( ) d () = [ cos ] = ( ) vl t L I ωt I ωl sinωt dt or: vl( t) = VLsinωt (5) with: VL = IXL, (6) in which: X L ωl (7) Notice, once again, that the voltage across the inductor and the current through it are not in phase! he current goes like cosω t but the voltage goes like sinωt. his means that the voltage gets to its peak one-quarter cycle before the current gets to its peak. We say: he voltage across an inductor leads the current through the inductor by 90. X L in (7) is called the inductive reactance. It is the effective resistance that the inductor presents to an A source. 19
With ω in rad/s and L in henries, X L has units: X L ( rad/s)( H) = ( rad/s)( Ω s) =Ω So X does he units of resistance. L Finally, note that X L is a frequency-dependent resistance: XL = π f L (8) At very low frequencies (i.e., as f 0), X L 0. And at very high frequencies (as f ), X. herefore, we say: L At very low frequencies, an inductor acts like a short. At very high frequencies, an inductor acts like an open. 0
Average Power Dissipated in an Inductor he erage power dissipated in the inductor would be: 1 ()() P = v t i t dt 0 1 VI L P = ( VL sin t)( Icos t) dt sin tcos tdt ω ω = ω ω 0 0 But, once again: 0 sinωtcosω tdt = 0 So: P = 0 here is, over one cycle, no power dissipated in the inductor! 1
ransformers onsider the situation shown in Figure 4, in which two coils, called the primary coil and the secondary coil, are wound around an iron coil form. Let oil #1 be the primary coil and oil # be the secondary coil. Faraday s law applied to oil #1 says: dφ1 V1 = ε1 = N1 (9) dt Faraday s law applied to oil # says: dφ V = ε = N (30) dt If all of the flux passing through oil #1 also passes through oil #, then dφ1 dφ = dt dt Dividing Eq. (30) by Eq. (9) thus gives: V N = (31) V1 N1 If N > N1, then V > V1 and the transformer is called a step-up transformer. If N < N1, then V < V1 and we he a step-down transformer.
If a load resistor R is connected across the secondary, then the power delivered to the load (the power delivered by the secondary) is P = VI (3) Similarly, the power delivered to the primary coil (by whatever A source is driving it) is P1 = VI 1 1 (33) If negligible power is lost in heating up the windings of the transformer, then P = P1, which leads to: I N1 = (34) I N 1 3