To the internal source voltage V s , the impedance appears as V^/Ii, or V v Z n Z 22 + a> 2 M 2 a> 2 M 2 J" = Am = 7 = Zu + -^ (9.63) l l ^-22 ^22 The impedance at the terminals of the source is Z int — Z s , so co 2 M 2 „ . r a> 2 M 2 2«b = Zu + -Z Z s = R { + ja>L x + — ——. (9.64) Z 22 (R 2 + J<oL 2 + Z L ) Note that the impedance Z ab is independent of the magnetic polarity of the transformer. The reason is that the mutual inductance appears in Eq. 9.64 as a squared quantity. This impedance is of particular interest because it shows how the transformer affects the impedance of the load as seen from the source. Without the transformer, the load would be con- nected directly to the source, and the source would see a load impedance of Z L ; with the transformer, the load is connected to the source through the transformer, and the source sees a load impedance that is a modified version of Z L , as seen in the third term of Eq. 9.64. Reflected Impedance The third term in Eq. 9.64 is called the reflected impedance (Z r ), because it is the equivalent impedance of the secondary coil and load impedance transmitted, or reflected, to the primary side of the transformer. Note that the reflected impedance is due solely to the existence of mutual induc- tance; that is, if the two coils are decoupled, M becomes zero, Z r becomes zero, and Z ab reduces to the self-impedance of the primary coil. To consider reflected impedance in more detail, we first express the load impedance in rectangular form: Z L = R L + jX L , (9.65) where the load reactance X L carries its own algebraic sign. In other words, X L is a positive number if the load is inductive and a negative number if the load is capacitive. We now use Eq. 9.65 to write the reflected imped- ance in rectangular form: Z r = a?M 2 R 2 + R L + j{coL 2 + X L ) _ OJ 2 M 2 [(R 2 + R L ) - ;(o>L 2 + X L )] (R 2 + R L ) 2 + (o>L 2 + X L f 2 A,/2 1^221 The derivation of Eq. 9.66 takes advantage of the fact that, when Z L is written in rectangular form, the self-impedance of the mesh containing the secondary winding is -22 R 2 + R h + j(coL 2 + X L ). (9.67) Now observe from Eq. 9.66 that the self-impedance of the secondary circuit is reflected into the primary circuit by a scaling factor of (wM/|Z 22 |) 2 , and that the sign of the reactive component (wL 2 -I- X L ) is reversed. Thus the linear transformer reflects the conjugate of the self- impedance of the secondary circuit (Zj 2 ) into the primary winding by a scalar multiplier. Example 9.13 illustrates mesh current analysis for a cir- cuit containing a linear transformer. 9.10 The Transformer 337 Example 9.13 Analyzing a Linear Transformer in the Frequency Domain The parameters of a certain linear transformer are /?! = 200 ft, R 2 = 100 ft, L { = 9H, L 2 = 4H, and k = 0.5. The transformer couples an impedance consisting of an 800 O resistor in series with a 1 (x¥ capacitor to a sinusoidal voltage source. The 300 V (rms) source has an internal impedance of 500 + /100 ft and a frequency of 400 rad/s. a) Construct a frequency-domain equivalent circuit of the system. b) Calculate the self-impedance of the primary circuit. c) Calculate the self-impedance of the secondary circuit. d) Calculate the impedance reflected into the pri- mary winding. e) Calculate the scaling factor for the reflected impedance. f) Calculate the impedance seen looking into the primary terminals of the transformer. g) Calculate the Thevenin equivalent with respect to the terminals c,d- Solution a) Figure 9.39 shows the frequency-domain equiva- lent circuit. Note that the internal voltage of the source serves as the reference phasor, and that Vj and V 2 represent the terminal voltages of the transformer. In constructing the circuit in Fig. 9.39, we made the following calculations: ;wL, = /(400)(9) = /3600 ft, ja>L 2 = /(400)(4) = /1600 ft, M = 0.5V(9)(4) = 3 H, joM = /(400)(3) = /1200 ft, 1 10 6 b) The self-impedance of the primary circuit is Z n = 500 + /100 + 200 + /3600 = 700 + /3700 ft. c) The self-impedance of the secondary circuit is Z 22 = 100 + /1600 + 800 - /2500 = 900 - /900 ft. d) The impedance reflected into the primary winding is Z r = 1200 |900 - /900| (900 + /900) = -(900 + /900) = 800 + /800 ft. e) The scaling factor by which Z 22 is reflected is 8/9. f) The impedance seen looking into the primary ter- minals of the transformer is the impedance of the primary winding plus the reflected impedance; thus Z ab = 200 + /3600 + 800 + /800 = 1000 + /4400 ft. g) The Thevenin voltage will equal the open circuit value of V cd . The open circuit value of \ cd will equal /1200 times the open circuit value of l h The open circuit value of I] is I, = 300 /0 jtoC /400 = -/250() ft. 700 + /3700 = 79.67/-79.29° mA. Therefore V Th = /1200(79.67/-79.29°) x 10~ 3 = 95.60/10.71° V. 300 /0° V 500 n /ioon a 2ooa . innn IOOH soon _onrv^ * ^/^ ] I ZOO /^V • "WV— + • V, /3600O- • + /16000 V 2 -/2500 a Figure 9.39 • The frequency-domain equivalent circuit for Example 9.13. 338 Sinusoidal Steady-State Analysis The Thevenin impedance will be equal to the imped- ance of the secondary winding plus the impedance reflected from the primary when the voltage source is replaced by a short-circuit. Thus / 1200 V Z Th = 100 + /1600 + ( |700 + J . 37()()| j (700 - /3700) = 171.09 +/1224.26 n. The Thevenin equivalent is shown in Fig. 9.40. /1224.26 li 95.60/10.71°/H V 171.09 O •Wv— Figure 9.40 A The Thevenin equivalent circuit for Example 9.13. ^ASSESSMENT PROBLEM Objective 4—Be able to analyze circuits containing linear transformers using phasor methods 9.14 A linear transformer couples a load consisting of a 360 O resistor in series with a 0.25 H inductor to a sinusoidal voltage source, as shown. The voltage source has an internal impedance of 184 + /0 Cl and a maximum volt- age of 245.20 V, and it is operating at 800 rad/s. The transformer parameters are Ri = 100 fl, L t = 0.5 H, R 2 = 40 O, L 2 = 0.125 H, and k = 0.4. Calculate (a) the reflected impedance; (b) the primary current; and (c) the secondary current. NOTE: Also try Chapter Problems 9.76 and 9.77. Source Transformer d Load Answer: (a) 10.24 - /7.68 H; (b) 0.5 cos(800f - 53.13°) A; (c) 0.08 cos 800f A. 9.11 The Ideal Transformer An ideal transformer consists of two magnetically coupled coils having N\ and N 2 turns, respectively, and exhibiting these three properties: 1. The coefficient of coupling is unity (k = 1). 2. The self-inductance of each coil is infinite (Lj = L 2 = oo). 3. The coil losses, due to parasitic resistance, are negligible. Understanding the behavior of ideal transformers begins with Eq. 9.64 which describes the impedance at the terminals of a source connected to a linear transformer. We repeat this equation below and examine it further. Exploring Limiting Values A useful relationship between the input impedance and load impedance, as given by Z ah in Eq. 9.68, emerges as L\ and L 2 each become infinitely large and, at the same time, the coefficient of coupling approaches unity: a> 2 M 2 Ri + joL, + o> 2 M 2 (/¾ + jcoL 2 + Z L )' (9.68) Transformers wound on ferromagnetic cores can approach this condition. Even though such transformers are nonlinear, we can obtain some useful information by constructing an ideal model that ignores the nonlinearities. To show how Z ab changes when k = 1 and L\ and L 2 approach infin- ity, we first introduce the notation Z 22 = R 2 + RL + KoL 2 + X L ) = R 22 + jX 22 and then rearrange Eq. 9.68: to 2 M 2 R 72 . ( <o 2 M 2 X 22 Z a |, = Ri -\—5 Y + 11 oi ^\ 2 2~ «*22 "^ ^22 ^ -**22 "•" -^-22 = R, b + jX. Ar (9.69) At this point, we must be careful with the coefficient of ;* in Eq. 9.69 because, as L\ and L 2 approach infinity, this coefficient is the difference between two large quantities. Thus, before letting L] and L 2 increase, we write the coefficient as where we recognize that, when k = 1, M 2 = L^L 2 . Putting the term mul- tiplying wLy over a common denominator gives (R\y + OiUX V + X}\ X ab = coL x — , - L , . (9.71) ab l \ Rl 2 + X\ 2 J ' Factoring o»L 2 out of the numerator and denominator of Eq. 9.71 yields L, X L + (R 2 22 + Xl)lo>L 2 X. xb = — = r. 9.72 L 2 (R 22 /<oL 2 ) 2 +[\ + (XJwL 2 )] 2 As k approaches 1,0, the ratio L]/L 2 approaches the constant value of (N1/N1) 2 , which follows from Eqs. 6.54 and 6.55. The reason is that, as the coupling becomes extremely tight, the two permeances 57^ and SP 2 become equal. Equation 9.72 then reduces to ^ = (¾) 2 ¾. (9.73) as Lj -* 00, L 2 —> 00, and k —• 1.0. The same reasoning leads to simplification of the reflected resistance in Eq. 9.69: urM 2 R- >2 L. /iV,V Applying the results given by Eqs. 9.73 and 9.74 to Eq. 9.69 yields Zah = *, + (^) 2 ¾ + (^) 2 («L + /¾. (9.75) Compare this result with the result in Eq. 9.68. Here we see that when the coefficient of coupling approaches unity and the self-inductances of the coupled coils approach infinity, the transformer reflects the secondary winding resistance and the load impedance to the primary side by a scaling 340 Sinusoidal Steady-State Analysis jcoM j(oL A j(oL 2 IN, v, (a) jcoM jcoL 2 \N-> (b) Figure 9.41 A The circuits used to verify the volts- per-turn and ampere-turn relationships for an ideal transformer. factor equal to the turns ratio (M/A^) squared. Hence we may describe the terminal behavior of the ideal transformer in terms of two characteristics. First, the magnitude of the volts per turn is the same for each coil, or ft N 2 (9.76) Second, the magnitude of the ampere-turns is the same for each coil, or i, MI = My LJiVl (9.77) We are forced to use magnitude signs in Eqs. 9.76 and 9.77, because we have not yet established reference polarities for the currents and voltages; we discuss the removal of the magnitude signs shortly. Figure 9.41 shows two lossless (R^ = R 2 = 0) magnetically coupled coils. We use Fig. 9.41 to validate Eqs. 9.76 and 9.77. In Fig. 9.41(a), coil 2 is open; in Fig. 9.41(b), coil 2 is shorted. Although we carry out the following analysis in terms of sinusoidal steady-state operation, the results also apply to instantaneous values of v and L Determining the Voltage and Current Ratios Note in Fig. 9.41(a) that the voltage at the terminals of the open-circuit coil is entirely the result of the current in coil 1; therefore The current in coil 1 is From Eqs. 9.78 and 9.79, V 2 = ja>Ml\. jcoL ] (9.78) (9.79) (9.80) For unity coupling, the mutual inductance equals VL ] L 2 , so Eq. 9.80 becomes V, = (9.81) For unity coupling, the flux linking coil 1 is the same as the flux linking coil 2, so we need only one permeance to describe the self-inductance of each coil. Thus Eq. 9.81 becomes V, = /V?SP A^ V] (9.82) or Voltage relationship for an ideal transformer • N 2 (9.83) Summing the voltages around the shorted coil of Fig. 9.41(b) yields 0 = -j(x)M\ { + jcoL 2 l 2 , (9.84) 9.11 The Ideal Transformer from which, for k = 1, II h M u VUL 2 L 2 (9.85) Equation 9.85 is equivalent to I,^i - l 2 N 2 . (9.86) -4 Current relationship for an ideal transformer Figure 9.42 shows the graphic symbol for an ideal transformer. The vertical lines in the symbol represent the layers of magnetic material from which ferromagnetic cores are often made. Thus, the symbol reminds us that coils wound on a ferromagnetic core behave very much like an ideal transformer. There are several reasons for this. The ferromagnetic material creates a space with high permeance. Thus most of the magnetic flux is trapped inside the core material, establishing tight magnetic coupling between coils that share the same core. High permeance also means high self- inductance, because L = N 2 V. Finally, ferromagnetically coupled coils efficiently transfer power from one coil to the other. Efficiencies in excess of 95% are common, so neglecting losses is not a crippling approximation for many applications. Determining the Polarity of the Voltage and Current Ratios We now turn to the removal of the magnitude signs from Eqs. 9.76 and 9.77. Note that magnitude signs did not show up in the derivations of Eqs. 9.83 and 9.86. We did not need them there because we had established reference polarities for voltages and reference directions for currents. In addition, we knew the magnetic polarity dots of the two coupled coils. The rules for assigning the proper algebraic sign to Eqs. 9.76 and 9.77 are as follows: If the coil voltages V! and V 2 are both positive or negative at the dot- marked terminal, use a plus sign in Eq. 9.76. Otherwise, use a nega- tive sign. If the coil currents I] and I 2 are both directed into or out of the dot- marked terminal, use a minus sign in Eq. 9.77. Otherwise, use a plus sign. The four circuits shown in Fig. 9.43 illustrate these rules. A", • N, Ideal Figure 9.42 • The graphic symbol for an ideal transformer. A Dot convention for ideal transformers + • JV, N 2 \ • + \ V, I > Ideal v i = v_ 2 /V, /V 2 " Ni\i = ~N 2 l 2 (a) I 2 V 2 + • yv, N 2 \ + f V, I II, V, Ideal I • W,I, = N 2 l 2 (b) + • V, I,) J |N, N 2 \ • + Ideal V, V 2 /V, N 2 A^I, = M,I 2 (c) l/V, N 2 \ V t I, >' I, V 2 Ideal V ] = _V_ 2 Ni N 2 A7,I, = -N 2 l 2 Figure 9.43 • Circuits that show the proper algebraic signs for relating the terminal voltages and currents of an ideal transformer. 342 Sinusoidal Steady-State Analysis AT,- v, = 500 N 2 = •1 i J Ideal L (a) 2500 + V 2 - + n i : 5 nr^- Ideal (b) V, + ni/5 :irr^ V, Ideal (c) Figure 9.44 • Three ways to show that the turns ratio of an ideal transformer is 5. The ratio of the turns on the two windings is an important parameter of the ideal transformer. The turns ratio is defined as either N]/N 2 or M/N,; both ratios appear in various writings. In this text, we use a to denote the ratio N 2 /N], or a = (9.87) Figure 9.44 shows three ways to represent the turns ratio of an ideal transformer. Figure 9.44(a) shows the number of turns in each coil explic- itly. Figure 9.44(b) shows that the ratio Ni/N] is 5 to 1, and Fig. 9.44(c) shows that the ratio ./V 2 //V, is 1 to |. Example 9.14 illustrates the analysis of a circuit containing an ideal transformer. Example 9.14 Analyzing an Ideal Transformer Circuit in the Frequency Domain The load impedance connected to the secondary winding of the ideal transformer in Fig. 9.45 consists of a 237.5 mil resistor in series with a 125 /xH inductor. If the sinusoidal voltage source (-y g ) is generat- ing the voltage 2500 cos 400/ V, find the steady- state expressions for: (a) /,; (b) V\; (c) / 2 ; and (d) v 2 . 0.25 ft 5mH 237.5 mft O 10:1 Ideal ; 125 yaH Figure 9.45 • The circuit for Example 9.14. Solution a) We begin by constructing the phasor domain equivalent circuit. The voltage source becomes 2500/0° V; the 5 mH inductor converts to an impedance of /2 D; and the 125 fxH inductor converts to an impedance of/0.05 ft. The phasor domain equivalent circuit is shown in Fig. 9.46. It follows directly from Fig. 9.46 that 2500/0° = (0.25 + /2)1] + V,, and V, = 10V 2 = 10[(0.2375 + /0.05)I 2 ]. Because I 2 = 101, we have Vi = 10(0.2375 +/0.05)101! = (23.75 + /5)1,. 0.25 II /2 ft AAA, /-YVW- 0.2375 ft -I, -vw /TN 2500/0! 0110:1/0- vv ^~ V 2 /0.05 ft; Ideal Figure 9.46 A Phasor domain circuit for Example 9.14. Therefore 2500 /0° = (24 +/7)1,, or I, = 100/-16.26° A. Thus the steady-state expression for i 1 is /, = 100cos(400r - 16.26°) A. b) V, = 2500/0° - (100 /-16.26" )(0.25 + /2) = 2500 - 80 - /185 = 2420 - /185 = 2427.06/-4.37° V, Hence v ] = 2427.06 cos (400r - 4.37°) V. c) I 2 = 101, = 1000/-16.26° A. Therefore i 2 = 1000 cos (400f - 16.26°) A. d) V 2 = 0.1V, = 242.71 /-4.37° V, giving y 2 = 242.71 cos (400/ - 4.37°) V. 9.11 The Ideal Transformer 343 The Use of an Ideal Transformer for Impedance Matching Ideal transformers can also be used to raise or lower the impedance level of a load.Tlie circuit shown in Fig. 9.47 illustrates this.The impedance seen by the practical voltage source (V v in series with Z v ) is Vi/Ij. The voltage and current at the terminals of the load impedance (V 2 and I 2 ) are related to V] and Ij by the transformer turns ratio; thus and v, = -, I, = a\ 2 . Therefore the impedance seen by the practical source is ZINJ — _ — -T 1 Y> 'IN Ii a 2 h but the ratio V 2 /I 2 is the load impedance Z L , so Eq. 9.90 becomes (9.88) (9.89) (9.90) 1 :a deal • + —0— z L Figure 9.47 A Using an ideal transformer to couple a load to a source. 7 IN lZ L . (9.91) Thus, the ideal transformer's secondary coil reflects the load impedance back to the primary coil, with the scaling factor \/a~. Note that the ideal transformer changes the magnitude of Z L but does not affect its phase angle. Whether Z )N is greater or less than Z L depends on the turns ratio a. The ideal transformer —or its practical counterpart, the ferromag- netic core transformer—can be used to match the magnitude of Z L to the magnitude of Z v We will discuss why this may be desirable in Chapter 10. ^ASSESSMENT PROBLEM Objective 5—Be able to analyze circuits with ideal transformers 9.15 The source voltage in the phasor domain circuit in the accompanying figure is 25 /0° kV. Find the amplitude and phase angle of V 2 an d ^2- Answer: V 2 = 1868.15 /142.39° V; I 2 = 125 /216.87° A. NOTE: Also try Chapter Problem 9.83. '•© 4ft ^7125:1 V, Ideal -/14.4 ft As we shall see, ideal transformers are used to increase or decrease voltages from a source to a load. Thus, ideal transformers are used widely in the electric utility industry, where it is desirable to decrease, or step down, the voltage level at the power line to safer residential voltage levels. 344 Sinusoidal Steady-State Analysis 9.12 Phasor Diagrams 2/150°^-^ 8/-170° s^-150t. -170*- 5 30° i / -45° /-45° 10/30° Figure 9.48 A A graphic representation of phasors. Figure 9.49 A The complex number -7 - /3 = 7.62 /-156.80°. When we are using the phasor method to analyze the steady-state sinu- soidal operation of a circuit, a diagram of the phasor currents and voltages may give further insight into the behavior of the circuit. A phasor diagram shows the magnitude and phase angle of each phasor quantity in the complex-number plane. Phase angles are measured counterclockwise from the positive real axis, and magnitudes are measured from the origin of the axes. For example, Fig. 9.48 shows the phasor quantities 10/30°, 12 /150% 5/-45°, and 8/-170°. Constructing phasor diagrams of circuit quantities generally involves both currents and voltages. As a result, two different magnitude scales are necessary, one for currents and one for voltages. The ability to visualize a pha- sor quantity on the complex-number plane can be useful when you are check- ing pocket calculator calculations. The typical pocket calculator doesn't offer a printout of the data entered. But when the calculated angle is displayed, you can compare it to your mental image as a check on whether you keyed in the appropriate values. For example, suppose that you are to compute the polar form of -7 - /3. Without making any calculations, you should anticipate a magnitude greater than 7 and an angle in the third quadrant that is more neg- ative than —135° or less positive than 225°, as illustrated in Fig. 9.49. Examples 9.15 and 9.16 illustrate the construction and use of phasor diagrams. We use such diagrams in subsequent chapters whenever they give additional insight into the steady-state sinusoidal operation of the cir- cuit under investigation. Problem 9.84 shows how a phasor diagram can help explain the operation of a phase-shifting circuit. Example 9.15 Using Phasor Diagrams to Analyze a Circuit For the circuit in Fig. 9.50, use a phasor diagram to find the value of R that will cause the current through that resistor, i R , to lag the source current, („ by 45° when <o = 5 krad/s. h = v m /v_ \V m /90°, -//(5000)(800 X 10~ 6 ) and the current phasor for the resistor is given by I V,„ /0 y = ^ /0°. R R z — Figure 9.50 A The circuit for Example 9.15. Solution By Kirchhoff s current law, the sum of the currents l R , l L , and I c must equal the source current I 5 . If we assume that the phase angle of the voltage V,„ is zero, we can draw the current phasors for each of the components. The current phasor for the induc- tor is given by These phasors are shown in Fig. 9.51. The phasor diagram also shows the source current phasor, sketched as a dotted line, which must be the sum of the current phasors of the three circuit components and must be at an angle that is 45 ° more positive than the current phasor for the resistor. As you can see, summing the phasors makes an isosceles triangle, so the length of the current phasor for the resistor must equal 3V,„. Therefore, the value of the resistor is | Cl. Ic = /4V> 1/ = v m /o £ /(5000)(0.2 X 10~ 3 ) V,n /-90°, h nvm» whereas the current phasor for the capacitor is given by > / A h = VJR Figure 9.51 A The phasor diagram for the currents in Fig. 9.50. 9.12 Phasor Diagrams 345 Example 9.16 Using Phasor Diagrams to Analyze Capadtive Loading Effects The circuit in Fig. 9.52 has a load consisting of the parallel combination of the resistor and inductor. Use phasor diagrams to explore the effect of adding a capacitor across the terminals of the load on the amplitude of V s if we adjust \ s so that the amplitude of V L remains constant. Utility compa- nies use this technique to control the voltage drop on their lines. R? Figure 9.52 • The circuit for Example 9.16. For convenience, we place this phasor on the pos- itive real axis. b) We know that I a is in phase with V L and that its magnitude is |V L |/# 2 - (On the phasor diagram, the magnitude scale for the current phasors is independent of the magnitude scale for the volt- age phasors.) c) We know that I b lags behind V L by 90° and that its magnitude is |V L |/wL 2 . d) The line current I is equal to the sum of I a and I b . e) The voltage drop across Ry is in phase with the line current, and the voltage drop across jo)L] leads the line current by 90°. f) The source voltage is the sum of the load voltage and the drop along the line; that is, V s = V L + (R x + jcoL { )\. Solution We begin by assuming zero capacitance across the load. After constructing the phasor diagram for the zero-capacitance case, we can add the capacitor and study its effect on the amplitude of Y,, holding the amplitude of V L constant. Figure 9.53 shows the fre- quency-domain equivalent of the circuit shown in Fig. 9.52. We added the phasor branch currents I, I a , and I b to Fig. 9.53 to aid discussion. rY-v-v>__—» a -—V, V L R 2 i\h J<oL 2 i\l u -• * Figure 9.53 • The frequency-domain equivalent of the circuit in Fig. 9.52. Figure 9.54 shows the stepwise evolution of the phasor diagram. Keep in mind that we are not interested in specific phasor values and positions in this example, but rather in the general effect of adding a capacitor across the terminals of the load. Thus, we want to develop the relative posi- tions of the phasors before and after the capacitor has been added. Relating the phasor diagram to the circuit shown in Fig. 9.53 reveals the following points: a) Because we are holding the amplitude of the load voltage constant, we choose VL as our reference. Figure 9.54 • The step-by-step evolution of the phasor diagram for the circuit in Fig. 9.53. Note that the completed phasor diagram shown in step 6 of Fig. 9.54 clearly shows the amplitude and phase angle relationships among all the currents and voltages in Fig. 9.53. Now add the capacitor branch shown in Fig. 9.55. We are holding V L constant, so we construct the phasor dia- gram for the circuit in Fig. 9.55 following the same steps as those in Fig. 9.54, except that, in step 4, we add the capacitor current I c to the diagram. In so doing, I c leads V L by 90°, with its magnitude being |V L wC|. Figure 9.56 shows the effect of I c on the line current: Both the magni- tude and phase angle of the line current I change with changes in the magnitude of I c . As I changes, so do the magnitude and phase angle of the voltage drop along the line. As the drop along the line changes, the magnitude and phase angle of V ? change. The phasor diagram shown . or decrease voltages from a source to a load. Thus, ideal transformers are used widely in the electric utility industry, where it is desirable to decrease, or step down, the voltage level at