Unit 3 AC Fundamentals PDF

Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ Unit III: AC Fundamentals Syllabus Representation of sinusoidal...

Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ Unit III: AC Fundamentals Syllabus Representation of sinusoidal waveforms, peak and rms values, phasor representation, Analysis of single-phase ac circuits consisting of R, L, C load. 3.1 Introduction Electrical energy is used in two forms : (i) Direct current form (D.C. energy) (ii) Alternating current form (A.C. energy) Except for a few cases the electrical energy now days is used in A.C. form only 3.1.1 Difference between D.C. and A.C. A direct current is that current which flows continuously in one direction and has constant magnitude with respect to time. It is shown in the figure 3.1. On the other hand an alternating current is that current which changes periodically in magnitude & direction. It is also shown in the figure given. Fig 3.1: (a) Direct Current (b) Alternating Current An alternating current, as shown in the figure consists of a definite cycle. Each cycle consists half- cycles, a positive half cycle and a negative half cycle. Current flows in one direction in positive half cycle and in another direction in negative half cycle. While flowing in one particular direction, the magnitude of the current changes in some regular manner with respect to time. 3.1.2 Advantages of Electrical Energy in A.C. form Over D.C. Form : 1. It is possible to built high voltage, high speed ac generators with reduce cost per kw. This not possible in case of d.c. generators. 2. A.C. voltage can stepped up or stepped down very easily with the help of the transformer. Such type raising or lowering of d.c. voltage is not easy and economical 3. Due to adoption of high voltage, a.c. transmission is always economical and efficient. This is so because, higher the voltage, lesser the current flowing through the transmission line conductors for the given power. Hence, the volume of conductor and the power loss in the line are reduced. 4. A.C. motors are simple in construction, cheaper and more efficient than the d.c. motors. 5. A.C. can be converted into d.c. by using the rectifiers, for various applications like traction, arc lamps etc. generation of d.c. energy for such a small applications at ordinary voltages is uneconomical. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 1 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ 3.2 Generation of Alternating Current and Voltages A generator is used to produce electrical energy, whether it is d.c. or a.c.. the generators converts mechanical energy into electrical energy. The operation of all electrical generators whether dc or ac is based on the fact that when a conductor is moved in a magnetic field or magnetic field is moved w.r.t. to conductor, an electromotive force is setup in the conductor as per faraday’s law of electromagnetic induction. Thus as long as there is a relative motion between the conductor and the magnetic field, an e.m.f. or voltage is always produced in the conductor ac generators are normally known as alternators 3.2.1 Elementary Alternator: Fig. 3.2: Elementary Alternator The above figure illustrates an elementary alternator. It consists of a single turn rectangular coil made up of some conducting material like copper or aluminum. The coil can be rotated about its own axis at a constant speed in a uniform magnetic field provided by the north and south poles of the magnet. The coil is known as armature of the alternator. The armature coils are connected to rings R1and R2, called slip rings which rotate with the armature.Two carbon brushes, B1and B2 pressed against the slip rings collect the current induced in the coil carry it to the external resister. Operation: Assume that the coil is rotating in the anticlockwise direction. The conductors A and B of the coil cut through the magnetic field; an e.m.f is induced in them, according to the Faraday’s law of electromagnetic induction. This causes a current to flow in the external resister R. The magnitude of this e.m.f. depends on the position of the armature coil in relation to the magnetic field. Let us consider some selected position as shown below in figure 3.3. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 2 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ Fig. 3.3: The end views of the armature coil in different positions Position -1:this is the initial position of the coil. The plane of the coil being perpendicular to the magnetic field, the conductors A&B move parallel to the magnetic field. Since there is no flux cutting, the e.m.f. generated in the conductor and therefore, no current flows through the external circuit. Position – 2 : As the coil rotates from position 1 (θ = 0˚) to position 2 (θ = 90˚), more & more lines of forces are cut by the conductors. In position 2, both conductors of the coil move at the right angles to the magnetic field and cut through maximum number of lines of force. Hence e.m.f. induced will be maximum. Hence, the e.m.f. builds up from zero to maximum between θ = 0˚to θ = 90˚. The resultant e.m.f. in the coil is two times that induced one conductor. This is so as both the conductors are in series. The current through the external circuit varies just as the e.m.f. varies. It is zero as θ = 0˚ and gradually increase to maximum value at θ = 90˚. At position 2 the induce current in conductor A is towers the observer and is away from the observer in conductor B. this according to Fleming’s right hand side rule. Due to these the current in the external circuit flows from C to D. Position -3:The lines of force cut by the conductor gradually reduce as the coil rotates from position 2 to position 3 i.e. from θ = 90˚ to θ = 180˚. The e.m.f. induced in the coil gets reduced gradually. In position 3, the conductors again move parallel to the field and no e.m.f. is induce in them. Therefore the current through the external circuit also becomes zero. Thus, we observe that during the rotation of the coil from 0˚ to 180˚ (during the first half revolution), the conductors move in the same direction through the magnetic field. Hence the polarity of the induced e.m.f. is same through this period and current flows through the external circuit from C to D only. Position-4 :As the coil rotates from position 3 to position 4 i.e. from θ= 180˚ to θ = 270˚ the conductor moves in the opposite direction as compared to the first half revolution. Hence the direction of e.m.f. induced and the current also get reversed. Now, in position 4, the current in conductor A is away from the observer and that in B is towers the observer. Thus the current flows from terminal D to terminal C in the external circuit. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 3 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ Thus the during the entire second half revolution i.e. when the coil rotates from position 3 to position 4, and back to the position 1, the current flows in the opposite direction to path in first in first half revolution. Also, the e.m.f. is maximum in position 4. In general, the variations in the magnitude of the e.m.f. of the alternator when the coil rotates from θ = 180˚ to θ = 360˚ are exactly similar to those in the first half revolution. The graph of e.m.f. or current for the complete revolution of the coil is shown in the figure 3.4. Fig. 3.4 Graphical representation of e.m.f. or current We observe from the graph that the voltage or current supplied by the alternator is alternating in nature and follow sine curved In practice, an alternator has a number of poles and number of coils suitably connected to obtain necessary e.m.f. 3.2.2 Important Definitions Related to AC Waveform Fig. 3.5 Cycle: One complete set of positive and negative values of an alternating quantity is known as cycle. Periodic Time / Time Period: The time taken by alternating quantity to complete one cycle. Denoted by T, Unit Sec. Frequency: The number of cycles per second of a waveform is defined frequency. Frequency is denoted by the lower-case letter f. In the SI system, its unit is the hertz (Hz) 1 Hz = 1 cycle per sec JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 4 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ Angular frequency (ω) : Rate of change of angle θ with respect to time is called angular frequency. ω =𝒅θ/ dt= 2𝝅/T = 2𝝅𝒇 unit = radian per sec Amplitude / Peak Value/ maximum value: The amplitude of ac waveform is the maximum value attained by waveform. It is distance from its average to its peak. Denoted by Vm or Im Peak-to-Peak Value: It is measured between minimum and maximum peaks. Denoted by Vpp or Ipp. For symmetric wave Vpp = 2 Vm Instantaneous value: The value of ac quantity at any particular instant on the waveform is referred to its instantaneous value. Mathematical and Graphical representation of AC waveform Equation for instantaneous value V = Vm Sin θ Amp = Vm Sin (ωt) = Vm Sin (2πft) = Vm Sin ((2π/T)*t) Note: Same equations can be written for instantaneous current. Important Positions on AC Waveform Fig. 3.6 3.3 Average Value Defination:The average value of an AC quantity is defined as the arithmetic mean of all instantaneous values over a period of half cycle. We know, instantaneous value of current is i = Im sinθ JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 5 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ Taking average of instantaneous values over half cycle, ∫ ∫ [ ] [ ] [ ] [ ] 𝐼𝑚 𝐼 𝑎𝑣𝑔.637𝐼 𝑚 π Average value of a sine wave over a complete cycle is zero (due to equal positive and negative half cycle). 3.4 Effective or RMS Value Effective value or RMS value of an ac waveform is an equivalent dc value. Definition :It is the value of DC current which is required to produce same amount of heat as produced by AC provided the resistance and time for which these currents flow are identical. It is the component of AC current which produces heating effect. All equipment’s are rated in terms of RMS value. Alternating current can be expressed as i = Imsinθ Taking square of current i , i2 = Im2 sin2θ To find the area under the curve an interval dθ is considered at a distance θ over half cycle The length of base is π JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 6 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ Average value of squared curve over half cycle = (Area of squared curve over half cycle / length of base over half cycle) ∫ ∫ ∫ [ ] [ ] √ 𝐼 𝑟𝑚𝑠.7 7 𝐼𝑚 3.5 Peak Factor and Form Factor These parameters are used to relate the maximum, average and r.m.s. values of an alternating quantity. Peak Factor: It is defined as the ratio of maximum value of an alternating quantity to its r.m.s. value. This is also known as crest factor or amplitude factor. For sinusoidal voltage or current,.7 7.... Application: The knowledge of this value is useful in the applications like insulating testing and measurement of iron losses. Form Factor: It is defined as the ratio of the r.m.s. value to average value of an alternating quantity. For sinusoidal voltage or current,....7 7.7 7.637.637. Application: The knowledge of this value is useful in determining the r.m.s. value of an alternating quantity. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 7 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ 3.6 Concept of Phase and Phase Difference Phase: Fractional part of time period or cycle through which the AC quantity has advanced from the selected reference zero position. Phasor: A line of definite length, rotating in anticlockwise direction at a constant angular velocity ω. A phasor is a rotating line whose projection on a vertical axis can be used to represent sinusoidally varying quantities. Length represents magnitude of AC & angle with X axis represents phase angle. Length of arrow generally used to represent rms value. Fig. 3.8 Voltages and Currents with Phase Shifts If a sine wave does not pass through zero at t = 0, it has a phase shift a) Leading phase difference b) Lagging phase difference Fig. 3.9 For a voltage waveform shifted to left v = Vm sin(ω t + θ) For a voltage waveform shifted to right v = Vm sin(ω t - θ) Phase Difference: When two alternating quantities of same frequency have different zero position of reference, they are said to have a phase difference. Illustrating phase difference, in these examples, voltage is taken as reference Fig. 3.10 JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 8 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ Example : Consider current and voltage equations , there waveforms and phasor diagram shown below. v = Vm sin ω t I leads v by angle θ i= Im sin(ω t + θ) Out of phase quantities – when phase difference is π. Equation for instantaneous values are, V1 = vm1 sin ωt V2 = vm2 sin (ωt – π) Then, V1 and V2 are out of phase. When using a calculator - set to radian mode when working with angles measured in radians. 3.7 Rectangular & polar representations of Phasors A phasor can be represented in two forms In polar form I=r _θ In rectangular form I = x +jy Fig. 3.12 Addition and Subtraction of two phasors Consider two voltages V1 = X1 +jY1 and V2 = X2 +jY2 Addition of voltages = V1 + V2 = (X1 + X2 ) + j (Y1 +Y2) Subtraction of voltages = V1 + V2 = (X1 - X2 ) + j (Y1 -Y2) Multiplication and Division of Phasors Consider voltage V = r1 ∕_ θ1 and current I = r2 ∕_ θ2 Multiplication V. I = r1. r2 ∕_(θ1+ θ2) Division V/ I = (r1 / r2) ∕_ (θ1- θ2). JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 9 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ 3.8 Pure Resistive, Pure Inductive and Pure Capacitive Circuit Before starting the discussion let us discuss two terms, reactance and impedance as applied to a.c. circuits Reactance : It is the opposition to flow of ac current by the inductance and capacitance parameters in the a.c. circuit it is of two types : i) Inductance reactance ii) Capacitive reactance Inductive reactance (XL) : It is the opposition , to flow of ac current , offered by an inductance. It is denoted by XL & is given by XL = 2πfL Where f = frequency of supply voltage and L =inductance of the inductive coil. The unit of inductive reactance is ohm (Ω) Capacitive reactance (Xc) : It is the opposition, to flow of a.c. current, offered by a capacitance. it is denoted by Xc and is given by Xc = 1 / 2πfc Where f = frequency of supply voltage and C = capacitance of the capacitor The unit of capacitive reactance is ohm (Ω). Impedance (Z) : It is the opposition to a.c. current , offered by the combination of resistance, inductance and capacitance in a.c. circuits. It is denoted by Z and is given by Z = R + jX Where R = resistance in the circuit X = reactance in the in the circuit Its unit is Ohms (Ω) Let us start to understand the behavior of circuits involving pure resistance, inductance and Capacitance. 3.8.1 A.C. Through Pure Ohmic Resistance Alone: The circuit is shown in Fig.4.13. Let the applied voltage be given by the equation. v = Vm sin θ = Vm sin ωt (1) Let R = ohmic resistance ; i = instantaneous current Obviously, the applied voltage has to supply ohmic voltage drop only. Hence for equilibrium v = iR JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 10 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ Putting the value of ‘v’ from above, we get Vm sin ωt = iR (2) Current ‘i’ is maximum when sin ωt is unity ∴ Im = Vm/R Hence, equation (2) becomes, i= Im sin ωt (3) Comparing (1) and (2), we find that the alternating voltage and current are in phase with each other as shown in Fig. 16. It is also shown vectorially by vectors VR and I in Fig. 3.14. Fig.3.13 Fig.3.14 Power. Instantaneous power, p = vi = VmIm sin 2ωt Power consists of a constant part and a fluctuating part of frequency double that of voltage and current waves. For a complete cycle. the average value of Is zero. OR P = V × I watt where V = r.m.s. value of applied voltage. I = r.m.s. value of the current. It is seen from Fig. 3.15 that no part of the power cycle becomes negative at any time. In other words, in a purely resistive circuit, power is never zero. This is so because the instantaneous values of voltage and current are always either both positive or negative and hence the product is always positive. Fig. 3.15 JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 11 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ 3.8.2 A.C. Through Pure Inductance Alone Whenever an alternating voltage is applied to a purely inductive coil*, a back e.m.f. is produced due to the self-inductance of the coil. The back e.m.f., at every step, opposes the rise or fall of current through the coil. As there is no ohmic voltage drop, the applied voltage has to overcome this self-induced e.m.f. only. So at every step ∴ ∫ ( ) ( ) Maximum value of Im= when ( ) is unity. Hence the equation for current becomes ( ) Fig.3.16 Clearly, the current lags behind the applied voltage by a quarter cycle (Fig. 3.16(a)) or the phase difference between the two is π/2 with voltage leading. Vectors are shown in Fig. 3.16 (b) where voltage has been taken along the reference axis. We have seen that Im = Vm/ω L = Vm/XL. Here ‘ωL’ plays the part of ‘resistance’. It is called the (inductive) reactance XL of the coil and is given in ohms if L is in henry and ω is in radian/second. Now, XL = ω L = 2π f L ohm. It is seen that XL depends directly on frequency of the voltage. Higher the value of f, greater the reactance offered and vice-versa. Power: Instantaneous Power = ( ) ∴ ∫ JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 12 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ It is also clear from Fig. 16 (b) that the average demand of power from the supply for a complete cycle is zero. Here again it is seen that power wave is a sine wave of frequency double that of the voltage and current waves. The maximum value of the instantaneous power is VmIm/2. 3.8.3 A.C. Through Pure Capacitance Alone When reference to Fig.3.17, let Fig.3.17 Fig.3.18 When an alternating voltage is applied to the plates of a capacitor, the capacitor is charged first in one direction and then in the opposite direction. v = p.d. developed between plates at any instant q = Charge on plates at that instant. Then q = Cv...where C is the capacitance = C Vm sin ωt...putting the value of v. Now, current i is given by the rate of flow of charge. ( ) ( ) ⁄ The denominator XC = 1/ωC is known as capacitive reactance and is in ohms if C is in farad and ω in radian/ second. It is seen that if the applied voltage is given by v = Vm sin ωt, then the current is given by i = Im sin (ωt + π/2). Power: Instantaneous Power = p=vi= Vm Im sin (ωt + π/2) ∫ JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 13 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ This fact is graphically illustrated in Fig. 3.19. We find that in a purely capacitive circuit, the average demand of power from supply is zero (as in a purely inductive circuit). Again, it is seen that power wave is a sine wave of frequency double that of the voltage and current waves. The maximum value of the instantaneous power is VmIm/2. Fig. 3.19 Numerical Ex. 1) An alternating current is given by i = 14.14 sin 377t , find RMS value of current , frequency and instantaneous value of current when t = 3 ms and also find time taken by current to reach 10 A from first time after passing through zero and increasing positively Solution: i = 14.14 sin 377t RMS value: Irms = Im /√2 =14.14/√2 =10 A Frequency: 2πft = 377t f = 377/2π = 60 Hz Instantaneous value at t = 3 ms: i = 14.14 sin (377 *3*10-3) =12.8 A Time taken to reach 10 A(t1): 10 = 14.14 sin 377t1 377 t1 = Sin-10.707 = 45◦ So, 377 t1 = π /4 t1 = 2.08 m sec. Ex. 2) A sinusoidal wave of frequency 50 Hz has its maximum value of 9.2 A. What will be its value at (a) 0.002 sec after the wave pass to the zero in positive direction (b) 0.0045 sec after the wave passes through positive maximum. Solution: f =50 Hz. Im = 9.2 A. The instantaneous equation for the current wave is given by i = 9.2 Sin(100πt) Current at 0.002 sec after the wave pass to the zero in positive direction i.e. t=0.002 sec. i = 9.2 Sin(100π x 0.002) i = 5.40 A. JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 14 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ Current at 0.0045 sec after the wave passes through positive maximum i.e. t= 0.005 + 0.0045 = 0.0095 sec i = 9.2 Sin(100π x 0.0095) i = 1.439 A. Ex. 3) At t=0, the instantaneous value of 60Hz sinusoidal current is 5A and increases in magnitude further. Its RMS value is 10 A. (a) Write its instantaneous expression. (b) Find current at t = 0.01 and t= 0.015 sec. (c) Sketch waveform for one complete cycle. Solution: f= 60 Hz, I=10 A. Maximum value of current Im= I/0.707=10/0.707 = 14.14 A Instantaneous expression i = Im Sin (wt + Ø) at t=0, i = 5A 5 = 14.14 Sin (120 x π x 0 + Ø) Ø = Sin-1(5/14.14) = 20.7040 = 0.3613 rad. i = 14.14 Sin (120πt + 0.3613) Current at t = 0.01 i = 14.14 Sin (120π x 0.01 + 0.3613) = 11.82 A. Current at t = 0.015 i = 14.14 Sin (120π x 0.015 + 0.3613) = 3.73 A. Waveform Ex. 4) The mathematical expression for instantaneous value of an alternating current is i = 7.071 Sin (157.08t – π/4) Find effective value, periodic time and the instant at which it reaches its positive maximum value. Sketch the waveform for on complete cycle. Solution: i = 7.071 Sin (157.08t – π/4). Effective or RMS value I = 0.707 Im JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 15 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ I = 0.707 x 7.071 = 5 A. Periodic time w = 2πf 157.08 = 2πf f = 157.08/2π T = 1/f = 2π/157.08 T = 0.04 sec. Time for positive maximum At positive maximum I = 7.071 A 7.071 = 7.071 Sin (157.08t – π/4). Sin (157.08t – π/4) =1 157.08t – π/4 = 1.5707 157.08t = 2.3561 t = 0.015 sec. Waveform Ex. 5) An alternating current is represented by expression i = 10 Sin ( 120πt – π/6). Find (a) Its periodic time (b) Instantaneous value at t=0 (c) Time at which it reaches negative maximum for the first time after t =0 (d) Draw neat waveform for one complete cycle. Solution: i = 10 Sin ( 120πt – π/6) Periodic time w = 2πf 120π = 2πf f = 120/2 = 60 Hz T = 1/f = 1/60 T = 0.01666 sec. Instantaneous value at t=0 i = 10 Sin (120π x 0 – π/6) i = – 5 A. Time at which it reaches zero first time after t =0 0 = 10 Sin (120πt – π/6) JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 16 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ 120πt = π/6 t =0.001388 sec. Time at which it reaches negative maximum for the first time after t =0 After t =0 it reaches negative maximum at ¾ th cycle t = 0.001388 + ¾ x 0.01666 t = 0.01388 sec. Waveform Ex. 6) A 50 Hz sinusoidal current has peak factor 1.4 and form factor 1.1. Its average value is 20 A. The instantaneous value of the current is 15 A at t =0 sec. Write equation of current and draw waveform. Solution: Form factor = 1.1, Peak factor = 1.4, Iavg =20A, f = 50 Hz. Form factor = Irms/Iavg Peak factor = Im/Irms Form factor x Peak factor = Im/ Iavg 1.1 x 1.4 = Im/20 Im = 30.8 A. At t =0, i = 15A, the alternating quantity is leading alternating quantity i = Im Sin (100πt + Ø) 15 = 30.8 Sin (100π x 0 + Ø) Ø = 29.1440 = 0.508 rad. Current equation i = 30.8 Sin (100πt + 0.508) Waveform JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 17 Introduction to Electrical Engineering F.Y. B. TECH. ______________________________________________________________________________________ Ex. 7) An alternating voltage v = 200 sin (100πt) is individually applied across resistance of 25 Ω, inductance of 32 mH and capacitance of 80µF. Find instantaneous expression for current in each element and power consumed by each element. Solution: v = 200 sin (100πt), R =25Ω, L= 32 mH, C=80µF. Current in resistor Voltage across the resistance and current through resistance are always in phase. Therefore, i = Im sin (wt) Im = Vm/R = 200/25 =8 A i = 8 sin (100πt) Power consumed by resistor P=VI P = 0.707 x 200 x 0.707 x 8 P = 800 W. Current in inductor Current through inductor always lags behind the voltage across the inductor by π/2. Therefore, i = Im sin (wt – π/2) Im = Vm/R = 200/25 =8 A i = 8 sin (100πt – π/2) Power consumed by inductor P =0 Current in capacitor Current through capacitor always leads the voltage across the capacitor by π/2. Therefore, i = Im sin (wt + π/2) Im = Vm/R = 200/25 =8 A i = 8 sin (100πt + π/2) Power consumed by capacitor P =0 JSPM’s RAJARSHI SHAHU COLLEGE OF ENGINEERING, PUNE 18