Vectors and Phasors. A supplement for students taking BTEC National, Unit 5, Electrical and Electronic Principles. Owen Bishop

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1 Vectors and phasors Vectors and Phasors A supplement for students taking BTEC National, Unit 5, Electrical and Electronic Principles Owen Bishop Copyrught 2007, Owen Bishop 1 page 1

2 Electronics Circuits and Systems Vectors and phasors Physical quantities are of two kinds: Scalars - specified by their magnitude (= size). Examples are mass, electric charge, temperature, speed. Vectors - specified by their magnitude and their direction. Examples are weight (its direction is down ), electric field, velocity, acceleration. Scalars of the same kind can be added together, using the ordinary rules of arithmetic. Example: Distance is a scalar; if I walk 3 km to the shopping centre and then walk 2 km to the swimming pool, the total distance I walk is = 5 km. Note that I may change direction several time on my walk. Vectors of the same kind can be added by techniques that include both magnitude and direction. Two ways of adding vectors are by: making an accurate scale drawing. using trig equations. As the boat steers north-east across the river (vector AB)), it is carried downstream by the current (vector AC). Its resultant velocity relative to the banks is AD, The drawing has sides AB and AC at the correct angles. The lengths of the sides are proportional to the magnitudes of the velocities. The drawings below shows the stages in solving the problem. Adding vectors by drawing Problem:: A boat is headng north-east at 1.4 m/s across a river that flows from west to east at 2 m/s. Find the boat s resultant velocity, the sum of these two velocities. Solution: The drawing above shows the addition of the two velocities: AB, the velocity of the boat relative to the water and AC the velocity of the water relative to the banks. Their sum, or resultant, is a vector AD, obtained by completing a parallelogram with sides AB and AC, and then drawing the diagonal. 2 page 2

3 Vectors and phasors To solve this by drawing, first decide on a scale, for example let 25 m = 1 m/s. We do not need to draw the whole parallelogram because side CD is the same length as AB and at the same angle. Instead, draw a triangle ACD as shown. To scale, AC is 200 mm and CD is 140 mm The angle between AC and the extension of AB is 45. Summing up: Draw the vectors start to end, to scale and at the given angles. The resultant is from the start of the first vector to the end of the second (or last - Yes, you can add three or more vectors by this method). Important note: All vectors must be of the same kind. In the example we added velocities and the resultant is a velocity. You can not, for example, add a velocity to a weight. Adding vectors by trigonometry The drawing method is easy to understand but its accuracy depends on how good you are at drawing and measuring. For absolute accuracy we rely on trig(onometical) relationshops. A few essention ones are listed in the box below. Sketch for the trig solution, with known values written on it. In the triangle ADE: DE = 0.99 AE = AC + CE = = 2.99 angle A = tan 1 DE/AE = tan /2.99 = 18.3 This example shows how use the equations below to calculate the size and direction of the resultant. There are more examples in the following pages. Trig facts For the trig technique, sketch a diagram of the vectors. There is no need to draw to scale or plot angles accurately.. You need to complete a right-angled triangle, obtained by dropping a perpendicular DE from D down to the extension of AC. Mark on it the actual velocities (no scales in this method), as shown at top right. In the triangle CED: sin C = DE/DC DE = DC sin C = 1.4 sin 45 = 0.99 In the triangle CED: cos C = CE/DC CE = DE cos C = 1.4 cos 45 = 0.99 In this example CE = DE because the angle is 45. This is not so when C has some other value. In a right-angled triangle: sin B = b/a cos B = c/a tan 1 b/c = B b 2 + c 2 = a 2 3 page 3

4 Electronics Circuits and Systems Phasors We are used to representing an alternating voltage by plotting a graph showing how the voltage varies with time (see pp. 48 and the diagram below). Another way of representing the voltage is to draw a phasor diagram. In a phasor diagram, the voltage is represented by a point (A) moving in a circular path, as on the right at the bottom of this page. At any instant, the vertical distance of A above the horizontal axis is the instantaneous value of the voltage. The angle marked in the diagram is the phase angle. This increases with time at a constant rate as A moves round the circle. The radius of the circle equals the peak voltage, the peaks occuring when A is directly above and below the centre of the circle. In other words, the radius of the circle equals the amplitude V of the signal. The vertical distance of A from the horizontal axis is alternately positive and negative the voltage is alternating. Using the trig equation for the sine of an angle, the vertical distance of A at any instant is: v = V sin φ Where v is the instantaneous voltage, V is the amplitude and φ is the phase angle. The graph on the left, which shows v plotted against t, is therefore a sine curve. The signal is a sinusoid. In the equation above we usually know V and want to find v. We need to know φ, which increases steadily with time t. The rate of change of the phase angle is called the angular velocity, symbol ω. This is related to the frequency of the signal by the equation: ω. = 2πƒ Where ω is the angular velocity in radian per second and ƒ is the frequency in cycles per second. The constant value 2π refers to the fact that 2π radian equals one complete cycle of 360. If the phase angle is zero at zero seconds, its value at t s is: ωt = 2πƒt Entering this into the equation for v, we find that: v = V sin ωt = V sin 2πƒt If the signal leads or lags by a given angle, ϕ, the value of v is calculated from: v = V sin (ωt + ϕ) = V sin (2πƒt + ϕ) Remember that all angles must be expressed in radian. An alternating voltage signal (left) can be represented by a phasor diagram (right). Think of point A moving anti-clockwise round the circle. At zero time it is at the extreme right of the dagram, and moes once around the circle in one period of the signal 4 page 4

5 Vectors and phasors Adding phasors The discussion on the opposite page shows that an alternating voltage can be represented as a phasor and that a phasor has two properties: Magnitude: depends on the amplitude V of the signal. Direction: depends on time t. Vectors have the same two properties, so a phasor can be treated in the same way as a vector. In particular, two or more phasors can be added together using the same techniques (drawing or trig) as we use with vectors. This provides a way of analysing circuits in which there are two or more alternating signals at the same time. We can add them and obtain their resultant. The only requirement is that all the signals have the same frequency. Problem Using a phasor diagram, find the sum of two alternating voltages, represented by the equations: v 1 = 60 sin 500t v 2 = 40 sin (500t + π/5) Solution Both voltages have the same frequency so they can be summed by vector addition of their phasors (see diagram). The phasor for v 1 has no phase lead or lag, so it is drawn as a horizontal vector pointing right. Its length is scaled to represent a peak amplitude of 60 V. The phasor for v 2 leads v 1 by π/5 rad (= 36 ), so this is drawn as shown, rotated by π/5 rad anticlockwise of v 1, and with a length scaled to represent an amplitude of 40 V. Completng the parallelogram (or drawing a triangle as on p. 2), the length of the resultant is equivalent to 95 V. The phase angle of the resultant is (= 0.24 rad). The equation for the resultant phasor is: v = 95 sin (500t ) Note that all the signals have the same frequency. In this problem the frequency of the signals is represented by the value 500. Referring to the equation on the opposte page: 2πƒ = 500 ƒ = 500/2π = 80 Hz Inductive impedance This is an example of the kind of electrical problem that can be solved by using phasors. Problem An alternating voltage with the equation v = 100 sin 1600πt, is applied to a coil with inductance 200 mh and resistance 500 Ω. Calculate the impedance of the coil and the current that passes through it. Solution The standard equation for a sinusoidal alternating voltage is: v = V sin (2πƒt + φ) where: v = the instantaneous voltage V = the peak voltage 2πƒ = ω = angular velocity in rad/s φ = phase lead or lag, in radian Comparing the data supplied in the problem with this equation, the peak voltage is 100 V and the frequency is 1600π/2π = 800 Hz. The phasor diagram is drawn overleaf. At 800 Hz, the impedance of the coil is: X L = 2πƒL = 2π = 1005 Ω In the diagram, the resultant reperesents the impedance of the whole circuit, the vector sum of the resistive and inductive phasors. 5 page 5

6 Electronics Circuits and Systems Solution As frequency increases, the voltage waveform across the capacitor has a phase lag that approaches 90. We assume that the circuit is operating in this region. The rms current is the same throughout the circuit, so the rms voltage drop across each component is proportional to its impedance/ resistance. Given that the same current flows through the impedance and resistance of the coil as well as through the circuit as a whole, the sizes of the voltage phasors is proportional to their impedances. The magnitude of the resultant, the total impedance, is found by using the Pythagoras equation (p. 2): Z = ( ) = 1123 Ω From this we obtain the current flowing in the circuit: I = V/Z = 100/1123 = 89 ma and the phase angle: φ = tan 1 X L (Z/R) = 1123/500 = 1.15 rad The current lags V Z by this angle. Draw a triangle to represent the impedance of the capacitor and the resistance of the resistor (below). The triangle is right-angled because of the phase lag between the resistance and the capacitance. We know the resistance and can calculate the impedance of the whole circuit: Z = supply volts/current = 50/0.8 = 62.5 Ω Now we find X C by using the Pythagoras equation (p. 2): X C = ( ) = Ω Capacitative impedance Phasors can also be used when solving problems about impedance. Problem A capacitor 47 µf and resistor 56 Ω are wired in series across an alternating supply of rms voltage 50 V, and draw an rms current of 0.8 A. What is th e frequency of the supply, and what are the voltages across the capacitor and resistor? State the equations for the total voltage and current. We can now calculate the frequency: ƒ = 1/(2πCX C ) = 125 Hz Given the frequency, the angular velocity is: ω = 2πƒ = 785 rad/s To plot the phasor diagram we calculate the magnitudes of the vectors, using the Ohm s Law equation: V R = IR = = 44.8 V V C = IX C = = 21.7 V 6 page 6

7 Vectors and phasors The phasor diagram below shows V C lagging 90 behind V R. The phase angle of the resultant is found by calculating: tan 1 (V C /V R ) = 21.7/44.8 = 25.8 or 0.45 rad This equals the expression in brackets 785t 0.45 = t = ( ) /785 = 1.24 ms v first reaches half-peak after 1.24 ms Phase angle in an RLC series resonant circuit This is an example in which three phasors are added. The problem quotes the supply voltage as 50 V rms, but the value of the resultant must be based on the instantaneous voltage. The peak instanteous voltage is 50 2 = 70.7 V. We now have all the values needed for writing the equations for the total voltage and current. For the current, the instantaneous peak value is: = 1.13 A The current equation is: i = 1.13 sin 785 t A For the voltage: v = 70.7 sin (785 t 0.45) V Problem A 10 mh inductor with coil resistance 25 Ω is connected in series with a 4.7 µf capacitor. As in the first problem, the resistance of the coil can be taken to be in series with its inductance. An allternating signal frequency 1.2 khz is applied to the circuit. What is the impedance of the circuit at t his frequency? Calculate the current and voltage phase angle. Solution Calculate the inductances at the given frequency: X L = 2πƒL = 75.4 Ω X C = 1/2πƒC = 28.2 Ω These two inductances act in opposite directions (inductive leads, capacitive lags) as shown in the left-hand drawing below. These equations can be used to calculate the current or voltage at any given instant. Times are taken from t = 0 when i or v = 0. For instance, when t = 5 µs, v = 70.7 sin ( ) = 30.5 V Another example is to find when the voltage first reaches half the peak, that is, V. This occurs when: 70.7 sin (785t 0.45) = Rearranging terms: sin (785t 0.45) = 35.35/70.7 = 0.5 Find the angle which has 0.5 as its sine: sin = page 7

8 Electronics Circuits and Systems If two phasors have exactly opposite directions the magnitude of their resultant is found by ordinary subtraction. In this case, we replace the two phasors with a single phasor X LC which has magnitude 47.2 Ω and acts in the direction of X L. This is shown in the right-hand drawing. Now there are only two phasors to add together and the drawing shows the resuktant Z of all three original phasors. The magnitude of Z is found by using the Pythagors equation. Having found this we use it to calculate the current: I = V/Z = 100/53.4 = 1.87 A The phase angle of the resultant is: φ = tan 1 (47.2/25) = 1.08 rad The voltage across the circuit leads the current by 1.08 rad. Questions There are problems on phasors on this Companion Site. Click on Questions and then on Phasors. 8 page 8