Monday, October 13, 2014

How To Solve Physics Problems Series R-L-C Circuits and Phasors problems and solutions

Series R-L-C Circuits and Phasors

The previous chapter was concerned with the natural oscillations of L-C circuits. This chapter is devoted to the study of R's, L's and C's, individually and in combination, in response to sinusoidal voltages.

Alternating Currents

Problems 37-3 through 37-6 in Chapter 37, Faraday's Law, describe how 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 basic advantages over direct current. First, 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. (See the “Transformers” section in this chapter.) Second, an alternating current in a loop placed in a magnetic field rotates, providing rotational power for all sorts of machines. See the “A Current-Carrying Loop in a B Field” section in Chapter 34, Magnetic Forces. The basic circuit for the study of alternating current in R-L-C circuits is shown in Fig. 41-1.

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Fig 41-1

 The circuit is driven by the ac source. This is in contrast with the circuit of Fig. 40-2 where the capacitor is charged, placed in the circuit, and the transient response studied. The Kirchhoff voltage differential equation for this circuit is like equation 40-5 except that the right-hand side contains a driving voltage Vocosω. This dramatically complicates the mathematics beyond the level for most people taking their first physics course. For this reason we take another, less mathematical, view of this circuit but one that is very helpful in understanding how the circuit operates.

The Resistive Circuit.

The simplest way to start the study of ac circuits is with a resistor and ac source as shown in Fig. 41-2. The time varying voltage is v = VRcosωt.

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 Fig. 41-2

The current tracks (is in phase with) the voltage and is i = (VR/R)cosωt = IR cosωt.

 Fig. 41-3 shows the voltage and current as a function of time and (on the right) what is known as a phasor diagram, an important analysis tool in the study of R-L-C circuits.

image

 Fig. 41-3

 The individual vector-like arrows are called phasors. The diagram is started by drawing a phasor IR at some arbitrary angle ωt. The phasor rotates counterclockwise with the length proportional to IR, and the projection on the horizontal axis 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 of a sine wave is complete the phasor will have rotated through 360°. Next the phasor representing the voltage is drawn at the same arbitrary angle ωt. In a resistor the instantaneous voltage and current are in phase. The length is proportional to VR, and the projection along the horizontal axis is proportional to the instantaneous voltage across R.

The Capacitive Circuit

Figure 41-4 shows a capacitive circuit driven by an ac source.

image

Fig. 41-4

The time varying voltage is v = VC cosωt. The phase relation between this v and iC is different than for a resistor. When an alternating voltage is applied to a capacitor the current alternates (flows 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 (alternating) voltage is passing through zero.

The charge is in phase with the voltage q = Cv = CVC cosωt.

The current is iC = (dq/dt) = -ωCVC sinωt

 The voltage and current are plotted as a function of time in Fig. 41-5 along with the phasor diagram.image

 Fig. 41-5

 The phasor diagram is drawn starting with VC at an arbitrary angle ωt. (Starting with IC produces the same result.) The hard part in drawing the IC phasor is to figure out how to orient it with respect to VC. 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° before v. Therefore we say “i leads v by 90° in a capacitive circuit.” The IC phasor is 90° ahead (rotated 90° further counterclockwise) of VC.

The maximum current is IC = ωCVC or VC = IC(1/ωC) = ICXC.

The 1/ωC term plays the role of resistance and is called capacitive reactance XC.

 41-1 A 20μF capacitor is connected to a variable frequency ac source with maximum voltage 30V. What is the capacitive reactance at 60 Hz, 600 Hz, and 60 kHz?

Solution:image The Inductive Circuit

Figure 41-6 shows an inductor driven by an ac source. The time varying voltage is v = VLcosωt.

 Again the phase relationship between v and iL 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 voltage corresponds not to maximum current but to maximum rate of change of current. A quick look at a sine curve indicates that the maximum rate of change (slope) is when the curve crosses the axis, so we expect the current to be 90° out of phase with the voltage. image

Fig 41-6

 The Kirchhoff-type voltage statement for this circuit is V cosωt = L(di/dt)

This statement is easily integrated imageThe voltage and current are plotted as a function of time in Fig. 41-7 along with the phasor diagram.

image

 Fig. 41-7

 The phasor diagram is drawn by starting with VL. Now look at the adjacent peaks in the graph of v and i versus ωt and note that v reaches its peak earlier in time than iL. Therefore we say “v leads i by 90° in an inductive circuit.” Notice how, as the phasors rotate at this fixed 90° 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 IL = VL/ωL = VL/XL or VL = ILXL.

 The ωL term plays the role of resistance and is called inductive reactance XL.

41-2 A 120mH inductor is connected to a variable frequency ac source with maximum voltage 10V. What is the inductive reactance at 100 Hz, and 1.0 MHz?

Solution:image

The basic relation v = L(di/dt) shows that [H = V .s/A] making XL have the units of Ω.

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 41-3 At what frequency do a 65mH inductor and a 20μF capacitor have the same reactance?

Solution: When XL = XC, orimage The frequency then is imageThis is the condition for resonance or oscillation in the L-C circuit!image 

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 41-4 An ac source of 100Hz and maximum voltage of 20V is connected to a 70mH inductor. What is the maximum current? When the current is maximum, what is the voltage of the source?

 Solution: First calculate the inductive reactance

XL = ωL = 2π.100Hz.0.070H = 44Ω

The maximum current isimage Look at the graphs in Fig. 41-7 and note that when the current is a maximum, the voltage is zero.

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 41-5 For the circuit described in problem 41-4, what is the current when the source voltage is 12 and increasing?

Solution: Look at the graphs of voltage and current, and voltage versus time as shown in Fig. 41-8 and note where on the graph the voltage is 12 V and increasing. This point (12 V and increasing) is between 270° and 360°.

The curve between 270° and 360° is a sine curve so the angle β = sin-1 (12/20) = 37°.

Therefore the point where the voltage is 12 V and increasing is 270° + 37° = 307°.

image

Fig. 41-8

The current at this point isimage This solution using the graphs may appear a bit cumbersome to you. The visual aspect of using the graphs, however, is a considerable help in keeping track of where you are as you proceed through the problem. It is easy to make the mistake of finding the arcsin of 12/20 and putting 37° into the current equation, completely missing that this is the point where i is decreasing, not increasing. After you have done a few problems like this you may be able to shorten the procedure.

The R-L-C Circuit

The phasor diagrams are most helpful in understanding R-L-C circuits as shown in Fig. 41-1. 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 V cosωt = vR + vL + vC. Second, since there is only one current path, the current is everywhere the same. Voltages on the various components have different phase relationships, but the current is the same everywhere in the circuit. The phasor diagram for a typical R-L-C circuit is shown in Fig. 41-9. Do not try to take this in all at once. Follow along the steps in the construction of the diagram. imageFig. 41-9

 Place the I phasor at the arbitrary angle ωt

Place the VR phasor over l. The voltage and current in the resistor are in phase.

Add VL = l XL leading VR by 90°.

Add VC = l XC lagging VR by 90°.

On an axis perpendicular to VR and I, VL and VC, add in a vector manner to produce VL - VC. In this example VL > VC.

If the load is resistive, voltage is in phase with current. If the load is entirely inductive or entirely capacitive the voltage is 90° out of phase with current. In this situation, with all elements present, the voltage is the vector-like sum of VR and VL - VC.

In equation form imageorimage This suggests another resistance-like expressionimage which is called impedance. We now have the numeric relations between voltage, current, and the vaues of R, L, and C.

The phase relation between V and I is seen from the phasor diagram as image

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 41-2. The source voltage divided by the impedance, equation 41-3, will equal the current. Finally the phase angle between the source voltage and current comes from equation 41-4.

41-6 An R-L-C circuit with R = 200Ω, L = 0.40H, and C = 3.0μF is driven by an ac source of 20 V maximum and frequency 100 Hz. Find the reactances, impedance, and maximum current and voltages across each of the components.

 Solution: R-L-C circuit 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 reactances, and the reactances are frequency dependent. As you proceed through this problem be aware of the logic in the calculations. The schematic of the circuit is shown in Fig. 41-1. image

41-7 For the circuit described in problem 41-6 construct the phasor diagram and find the angle of the source voltage with respect to the current.

 Solution: The phasor diagram is shown in Fig. 41-10. Again, follow the logic in the construction

image

Fig 41-10

 of the diagram. Draw the phasor representing I = 0.058A. at an arbitrary angle ωt. Draw the phasor representing VR = 11.6 V in the same direction as I. Draw VL = 14.6 V leading VR by 90°. Add VC = 30.7 V lagging VR by 90°. Complete the rectangle with side VL - VC and VR. Since VC > VL, this vector points in the same direction as VC. The source voltage V is the diagonal of this rectangle. The phase angle is determined from equation 41-4.image The load is resistive and capacitive and (from the diagram) the source voltage lags the current by 54°.

After you are clear on the calculations go back over these last two problems and concentrate on the logic. The biggest pitfall in R-L-C circuit problems is losing your way!

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41-8 An R-L-C circuit consists of a 300Ω resistor, 0.15H inductor, and 4.5 μF capacitor driven by an ac source of 800 rad/s. Find the phase angle of the source voltage with respect to the current.

Solution: Notice that the maximum voltage is not given in the problem, yet a phase diagram is to be drawn. The problem is written this way to emphasize that it is not necessary to know the maximum source voltage to find the phase angle. The phase diagrams so far have used voltage.

The voltage across each component is the resistance or reactance times a constant, the current. Thus the diagram could as well be drawn with resistance and reactances as well as voltage.

Calculate the reactances: imageNow draw the phasor diagram as shown in Fig. 41-11. Draw I at some arbitrary angle ωt.image Fig. 41-11

 Draw R along I proportional to 300 Ω. Draw XL leading R by 90° proportional to 120 Ω. Draw XC lagging R by 90° proportional to 278 Ω. Draw XL - XC = -158 Ω. Complete the rectangle formed by R and XL - XC and draw the diagonal, which is proportional to Z. The phase angle is imageThe load is resistive and capacitive, and the phase angle is -28° with the source voltage lagging the current.

Power in ac Circuits

The instantaneous power in the resistive circuit of Fig. 41-2 is image

The average power is the average value of the cos2 function over one cycle. The sin2 function and cos2 functions have the same shape (area under the curve), and sin2 θ+cos2 θ= 1. The only way for sin2 θ to equal cos2 θ and their sum to equal 1 is for cos2 θ to equal 1/2. Therefore the average value of the cosine squared function over one cycle is 1/2, and the average power is imageThe most convenient associations are shown as imageand imageThese values of imageand imageused to compute the average power are equivalent to V and I used to compute power in a dc circuit. DC voltmeters and ammeters measure V and I with the product being power, P = VI AC voltmeters and ammeters must measure and , the time average of these quantities, so power calculations in ac and dc will be the same. The and measurements are called the rms (root mean square) values of voltage and current. image

41-9 For the circuit described in problem 41-6 calculate the power loss in each of the components.

 Solution: There is no energy loss in an inductor. Energy goes in to build up the magnetic field and is released in the collapse of the magnetic field. Look at the curves of Fig. 41-7. The product vi averaged over one cycle adds to zero. A similar argument can be made for a capacitor. For the resistor the voltage is v = (11.6V)cosωt and the current is i = (0.058A)cosωt, so the power is p = 11.6.0.058Wcos2ωt. The average power isimage Using the rms voltages and currentsimage

 Transformers

A transformer consists 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 Fig. 41-12. Based on Faraday's law the voltage induced in a coil is proportional to the number of windings.image Fig. 41-12

image

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. image

41-10 A power distribution transformer steps voltage down from 8.5 kV to 120V (both rms values). If the 120V side of the transformer supplies 500A to a resistive load, what current is taken from the primary (high) side of the transformer and what is the turns ratio?

Solution: First find the turns ratioimage The current isimage