FISI30 Física Universitaria II Professor J.. ersosimo hapter 8 Alternating current circuits- Series circuits 8- Introduction A loop rotated in a magnetic field produces a sinusoidal voltage and current. This is the basis of alternating current generators. An external agent, such as falling water or steam, is used to rotate the loop of wire in a magnetic field thus generating a sinusoidal, or alternating, voltage and current. This alternating current, ac for short, has two basis advantages over direct current. Figure : ) Fist, it is easy to increase the voltage for transmission over long distances and later decrease the voltage for distribution to individual users or for specific applications. ) Second, an alternating current in a loop placed in a magnetic field rotates, providing rotational power for all sorts of machines. Figure : The basic circuit for the study of alternating current in circuits is shown in Figure. The source is an alternating voltage: v = v cosωt [] 0 8- The resistive circuit The simple way to start the study of ac circuits is with a resistor and ac source as shown in Figure 3. The current tracks in phase with the voltage and v0 i = cosωt []
FISI30 Física Universitaria II Professor J.. ersosimo The individual vector-like arrows are called phasors. The diagram is started by drawing a phasor I at some arbitrary angleω t. The phasor rotate counterclockwise with the length proportional to I and the projection to the horizontal axes proportional to the instantaneous current. An increase in ωt toward the right on the graph corresponds to an increase in ωt in the counterclockwise direction on the phasor diagram. When an entire cycle o a sine wave is completed the phasor will have rotated through 360 o. Next the phasor representing the voltage is drawn at the same arbitrary angle ω t. In the resistor the instantaneous voltage and current are in phase. The length is proportional to, and the projection along the horizontal axis is proportional to the instantaneous voltage across. 8-3 The capacitive circuit Figure 3 shows a capacitive circuit driven by an ac source. The time varying voltage v = cosωt. The phase relation between this voltage and i is different for a resistor. When an alternating voltage is applied to a capacitor the current altenates (flow in one direction, then in the opposite direction) but does not track with (is not in phase with) the voltage across the capacitor. When the voltage reaches a maximum the capacitor is fully charged and the current is zero! When the voltage reaches a maximum in the other direction the capacitor is again fully, but oppositely, charged and the current is again zero. The current then must be a maximum when the Figure 3: urrent and voltage graphic, resistive circuit and phasor diagram (alternating) voltage is passing through zero. The charge is in phase with the voltage: q = v = cosωt Figure 4: urrent and voltage graphic, capacitive circuit and phasor diagram. The current leads the emf voltge. The current is i = ( dq / dt) = ωv sinωt c
FISI30 Física Universitaria II 3 Professor J.. ersosimo The voltage and current are plotted as a function of time in Figure 4 along with the fasor diagram. The phasor diagram is drawn starting with at an arbitrary angle wt (starting with i c produces the same result.) The hard part in drawing the I phasor is to figure out how to orient it with respect to. The easiest way to do this is to look at the graph of v and i versus time and ask the question Which quantity leads the other and by how much? By looking at adjacent peaks note that i reaches its maximum 90 0 before v.therefore we say i leads v by 90 0 in a capacitive circuit. The I phasor is 900 ahead (rotated counterclockdwise) of. The maximum current is: I = ω or = i ( / ω) = IX The / ω term plays the role of resistance and is called capacitive reactance X. In a purely capacitive ac circuit, the current leads the voltage by 90 0, or ¼ cycle. Example : A 0 μf capacitor is connected to a variable frequency ac source with maximum voltage 30. What is the capacitive reactance at 60 Hz, and 60 khz. For For s f = 60 Hz ; X = = = 30Ω π 60 0 0 6 F f = 60 khz ; = 0. 3Ω X ω ; 8.4 The Inductive ircuit Figure 5 shows an inductor driven by an ac source. The time varying voltages is v = cosωt Again the phase relationship between v and i is different from either the resistor or capacitor. The maximum voltage across an inductor is proportional to the rate of change of current. Therefore the maximum Figure 5: urrent and voltage graphic, inductive circuit and phasor diagram. The current lags the emf voltage
FISI30 Física Universitaria II 4 Professor J.. ersosimo voltage corresponds not to maximum current but to maximum rate of change of current. A quick look at a cine curve indicates that the maximum rate of change (sloe) is when the curve crosses the axis, so we expect the current to be 90 0 out of phase with the voltage. di The kirchhoff-type voltage statement for this circuit is cosω t = [3] dt This statement is easily integrated cosω tdt = di this allow write: i = sinωt ω [4] The voltage and current are plotted as a function of time in Figure 5 along with the phasor diagram. The phasor diagram is drawn by starting with. Now look at the adjacent peaks in the graph of v and i. There fore we say v leads i by 90 0 in an inductive circuit. Notice how, as the phasors rotate at this fixed 90 0 difference, the voltage phasor traces out the cosine function on the horizontal axis and the current phasor traces out the sine function. The maximum current is: I = = [5] ω X The ω term plays the role of resistance and is called inductive reactance X. In a purely inductive ac circuit, the current lags the voltage by 90 0, or ¼ cycle. Example : A 0 mh inductor is connected to a variable frequency ac source with maximum voltage 0. What is the inductive reactance a 00 Hz, and.0 MHz? X = = π 00Hz 0. H = 75Ω ω for 00Hz, and 5 X = 75 0 Ω for.0 Mz
FISI30 Física Universitaria II 5 Professor J.. ersosimo Example 3: At what frequency do a 65 mh inductor and 0μF capacitor have the same reactance? When X = X or ω =, the frequency then is: ω ω = or f = π Then f=40 Hz. The -- circuits The phasor diagram are most helpful in understanding -- circuits like is shown in Figure. There are two important points to keep in mind in the analysis of these circuits. First, the sum of the instantaneous voltages must equal the source voltage cosω t = v + v + v [6] Second, since there is only one current path, the current is everywhere the same. oltages on the various components have different phase relationships, but the current is the same everywhere in the circuit. The phasor diagram for a typical circuits is shown in Figure 6. Do not try to take this in all at once. Follow along the steps in the construction of the diagram. ) Place the I phasor at the arbitrary angle wt ) Place the phasor over I. The voltage and current in the resistor are in phase. Figure 6 3) Add = IX leading by 90 0. 4) Add = IX lagging by 90 0.
FISI30 Física Universitaria II 6 Professor J.. ersosimo 5) On an axis perpendicular to and I, and, add in a vector manner to produce >.. In this example If the load is resistive, voltage is in phase with current. If the load is entirely inductive or entirely capacitive the voltage is 90 0 out of phase with current. In this situation, with all elements present, the voltage is the vector-like cum of and. In equation form: = + ( ) = I + ( IX IX ) = I + ( X + X ) [ + ( ) ] = I X + X [7] [ ] This suggest another expression [ + ( ) ] Z = X + X [8] This is called impedance. We now have the numeric relations between voltage, current, and the values of, and. The phase relation between and I is seen from the phasor diagram as tg = φ [9] To obtain a better picture of what is going on here imagine measuring the ac voltages of the source, resistor, capacitor, and inductor, and the current in the circuit. The voltages across the resistor, capacitor, and inductor do not add up to the source voltage! They are not in phase! These voltages will satisfy equation. Example 4: An circuit with =000Ω, =0.4 H, and =3.0 μf is driven by an ac source of 0 maximum and frequency 00 Hz. Find the reactance, impedance, and maximum current and voltage across each of the components. circuits problems can be confusing. The key to successfully solving them is to follow a logical path through the problem. The current is everywhere the same and is determined by the source voltage and the impedance. The impedance is determined by the resistance and the reactance, and the reactance are
FISI30 Física Universitaria II 7 Professor J.. ersosimo frequency dependent. As you proceed through this problem be aware of the logic in the calculations. The schematic of the circuit is shown in Figure. X = ω = π 00Hz 0. 40H = 5Ω X 6 0 = = = 530Ω ω π 00Hz 3. 0F ( X X ) = 77800Ω [ + ( X + ) ] = 00 X + 77800 = Ω Z = 343 I = / Z = 0. 058A = I =. 6 = I X = I X = 4. 6 = 30. 7
FISI30 Física Universitaria II 8 Professor J.. ersosimo esonance The impedance in a series ac circuits is a minimum when X = X ; under these circumstances Z = and I = /. The resonant frequency f 0 of a circuit is that frequency at which X = X ; π f0 = πf 0 f = 0 π [0] esonant circuits have a variety of applications. One common application is in the tuning mechanism of a radio. Each radio station has an assigned broadcast frequency at which its radio waves are transmitted. When the waves are received at the antenna, their oscillating electric and magnetic fields set the electrons in the antenna into regular back-and-forth motion. In other words, they produce an alternating current in the receiver circuit just as a regular ac voltage source would do. Example 5: In the antenna circuit of a radio receiver that is tuned to a particular station, =5Ω, =5 mh, and =5pF. (a) Find the frequency of the station. (b) If the potential difference applied to the circuit is find the current that flows. a) f0 = = 006kH π b) At resonance X = X and Z=, then I = / = 0. ma 5 0 4, Power in ac circuits The instantaneous power in the resistive circuit of Figure 3 is p = vi = ( cosωt)( I cosωt) = I cos ωt The average value of the cos function over one cycle. The sin function and have the same shape (area under the curve), and sin θ + cos θ =. The only way for cos functions sin to equal cos and their sum equal is for cos θ = /. Therefore the average value of the cosine squared function over one cycle is ½, and the average power is
FISI30 Física Universitaria II 9 Professor J.. ersosimo These values of I and I I P = = [] used to compute the average power are equivalent to and I used to compute power in a dc circuit. D voltmeters and ammeters measure and I with the product being power, P = I. A voltmeters and ammeters must measure I and, the time average of these quantities, so power calculations in ac and dc will be the same. The and I measurements are called the rms (root mean square) values of voltages and current. max rms = and Imax I rms = [] Example 5: alculate the power loss in each of the components of the circuit described in example 4. Inductance: the energy goes to build up the magnetic field and is released in the collapse of the magnetic field. The product of vi averaged over one cycle is zero. A similar argument can be made for a capacitor.. 6 0. 058 P = = 0. 34W
FISI30 Física Universitaria II 0 Professor J.. ersosimo Transformer A transformer consist of two coils (called primary and secondary) wound one over the other with usually a soft iron core to enhance the magnetic field or an arrangement with two coils wound on a soft iron core as illustrated in Figure 7. Suppose an ac voltage is applied to the primary. Since we assume no resistance in the coils, the applied voltage will equal the induced back EMF. If the flux through one coil is dφ N dt = Similarly the voltage across the secondary: Φ B, then = N dφ dt Figure 7: From these equations we obtain: p = s [3] N p N s By varying the relative number of windings we can make either a step-up or step-down (voltage) transformer. The relative currents in the primary and secondary are determined with a simple statement that the power in equals the power out. I I = N I I = s s p p = s s p s or p p Np Np N I Example 6: A power distribution transformer steps voltage down from 8.5 k to 0 (both rms values). If the 0 side of the transformer supplies 500 A to a resistive load, what current is taken from the primary (high) side of the transformer and what is the turns ratio? First find the turns ratio Figure 8:
FISI30 Física Universitaria II Professor J.. ersosimo 8. 5k 0 Np 8500 = or = = 7 N N 0 p N s s The current is: Ns Ip = Is = 500A = 7. 0A N 7 p