Ac fundamentals and AC CIRCUITS Q1. Explain and derive an expression for generation of AC quantity. According to Faradays law of electromagnetic induction when a conductor is moving within a magnetic field, an emf is induced in the conductor. such as that shown in Fig. 1, is rotated within a magnetic field the induced emf can be measured between the ends of the conducting loop as a potential difference. The magnitude of the emf induced depends on the strength of the magnetic field, the length of the conductor forming the loop and the speed at which it is rotated. It also depends, however, on the angle at which the conducting loop moves with respect to the direction of the magnetic field. It can be shown that the induced emf is proportional to the sine of this angle. The emf is a maximum when the conductor is moving at right angles to the direction of the magnetic field and a minimum when it is moving in the direction of the field. Due to the circular motion of the armature against the straight lines of force, a variable number of lines of force will be cut even at a constant speed of the motion. 1.At zero degrees, the rectangular arm of the armature does not cut any lines of force, giving zero voltage output. As the armature arm rotates at a constant speed toward the 90 position, more lines are cut. 2.The lines of force are cut at most when the armature is at the 90 position, giving out the most current on one direction. As it turns toward the 180 position, lesser number of lines of force are cut, giving out lesser voltage until it becomes zero again at the 180 position. 3.The voltage starts to increase again as the armature heads to the opposite pole at the 270 position. Toward this position, the current is generated on the opposite direction, giving out the maximum voltage on the opposite side.
4. The voltage decrease again as it completes the full rotation(360 0 ). In one rotation, the AC output is produced with one complete cycle as represented in the sine wave. According to flemings right hand rule we can find the direction of generated emf. Q2. DIFINITIONS :
1. Cycle : One complete set of positive and negative values of alternating quantity is known as cycle. Hence, diagram represents one complete cycle. A cycle may also be sometimes specified in terms of angular measure. In that case, one complete cycle is said to spread over 360º or 2π radians. 2. Time Period The time taken by an alternating quantity to complete one cycle is called its time period T. For example, a 50-Hz alternating current has a time period of 1/50 second. T = 1/ f (sec) where f is frequency in Hz. 3. Frequency The number of cycles/second is called the frequency of the alternating quantity. Its unit is hertz (Hz). It may be noted that the frequency is given by the reciprocal of the time period of the alternating quantity. f = 1/T or T = 1/f (Hz) 4. Amplitude: The maximum value, positive or negative, of an alternating quantity is known as its amplitude. In above diagram Vmax is amplitude. 5. Angular frequency (ω): it is the angular distance covered in one second. It is also called as angular velocity. ω = 2 π f = 2 π / T ( rad/sec) where f is frequency in Hz. And T is time period. 6. Instantaneous value Value at any instance is called instantaneous value. I-Im sinωt Where i is instantaneous value that changes with t. Alternating current and Alternating EMF
An alternating current is one whose magnitude changes sinusoidal with time.thus alternating current is given by i = I m sin (ωt + Ф) Where im=current amplitude or peak value of alternating current The emf or voltage whose magnitude changes sinusoidal with time is known as alternating emf and is represented by v = V m sin (ωt + Ф) where Vm is the peak value of alternating current. Q3. Define and derive the expression for RMS value of AC. Root-Mean-Square (R.M.S.) Value The r.m.s. value of an alternating current is given by that steady (d.c.) current which when flowing through a given circuit for a given time produces the same heat as produced by the alternating current when flowing through the same circuit for the same time. It is also known as the effective or virtual value of the alternating current.
Q4. Define and derive the expression for Average value of AC. The average value Iavg of an alternating current is expressed by that steady current which transfers across any circuit the same charge as is transferred by that alternating current during the same time. In the case of a symmetrical alternating current (i.e. one whose two half-cycles are exactly similar, whether sinusoidal or non-sinusoidal), the average value over a complete cycle is zero. Hence, in their case, the average value is obtained by adding or integrating the instantaneous values of current over one half-cycle only. But in the case of an unsymmetrical alternating current (like half-wave rectified current) the average value must always be taken over the whole cycle.
Phase Difference of a Sinusoidal Waveform The current waveform can also be said to be lagging behind the voltage waveform by the phase angle, Φ if current starts it zero and maximum value after voltage. where, i lags v by angle Ө Likewise, if the current, i has a positive value and crosses the reference axis reaching its maximum peak and zero values at some time before the voltage, v then the current waveform will be leading the voltage by some phase angle. Then the two waveforms are said to have a Leading Phase Difference and the expression for both the voltage and the current will be. where, i leads v by angle Ө
Two Sinusoidal Waveforms in-phase Voltage (vt) = Vm sin ωt Current (it)= Im sin ωt Both voltage and current reaches it zero and maximum value at same time called as in phase quntities. Assignment 1. Explain and derive an expression for generation of AC quantity. 2. Define 1) cycle 2) frequency 3) instantaneous value 4) phase difference 5) periodic waveform 6) Peak factor 7) form factor 3. Define and derive the expression for RMS value of AC. 4. Define and derive the expression for average value of alternating quantity and also prove that average value of an alternating quantity over a complete cycle is zero. Problems for practice---- Q1. The following four emfs act together in a circuit e1=10 sin ωt e2=8 sin (ωt + π /3) e3= 4 sin ( ωt - π /6) e4=6 sin (ωt + 3π /4) calculate 1) the emf represented by e1-e2+e3-e4 2) write the expression for above resultant voltage. 3) Draw phasor diagram. Q2. An alternating current of frequency 50 Hz has a maximum value of 100 A. calculate 1) Its value 1/600 second after the instant the current is zero and its value is decreasing there afterwards. 2) how many seconds after the current is zero and then increasing will attain the value of 86.6 A? Q3. Calculate the maximum value of the e.m.f. generated in a coil which is rotating at 50 rev/s in a uniform
magnetic field of 0.8 Wb/m3. The coil is wound on a square former having sides 5 cm in length and is woundwith 300 turns. [188.5 V] Q4. A sinusoidal alternating voltage has an r.m.s. value of 200 V and a frequency of 50 Hz. It crosses the zero axis in a positive direction when t = 0. Determine (i) the time when voltage first reaches the instantaneous value of 200 V and (ii) the time when voltage after passing through its maximum positive value reaches the value of 141.4 V. [(i) (0.0025 second (ii) 1/300 second)] Q5. The values of the instantaneous currents in the branches of a parallel circuit are as follows : i1 = 5 sin 346 t; i2 = 10 sin (346t + π/4); i3 = 7.5 sin (346 t + π/2); i4 = 8 sin (346 t π/3) Express the resultant line current in the same form as the original expression and determine the r.m.s. value and the frequency of this current. [12.5 A; 55 Hz] Q6. The instantaneous voltage across each of the four coils connected in series is given by v1 = 100 sin 471 t; v2 = 250 cos 471 t; v3 = 150 sin (471 t + π/6); v4 = 200 sin (471 t π/4) Determine the total p.d. expressed in similar form to those given. What will be the resultant p.d. If v2 is reversed in sign? [v = 414 sin (471 t + 26.5 ); v = 486 sin (471 t 40 )] Q7. Three alternating currents are given by i1 = 141 sin (ωt + π/4) i2 = 30 sin (ωt + π/2) i3 = 20 sin (ωt π/6) and are fed into a common conductor. Find graphically or otherwise the equation of the resultant current and its r.m.s. value. [i = 167.4 sin (ωt + 0.797), Irms = 118.4 A]