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Chapter 7, , Alternating Current, , Chapter Con, , , , ts, Introduction, , A.C. Voltage Applied to a, Resistor, , Representation of AC, Current and Voltage by, , Rotating Vectors Phasors, , AC Voltage Applied to an, Inductor, , AC Voltage Applied to a, Capacitor, , AC Voltage Applied to a, Series LCR Circuit, , Power in AC Circuits : The, Power Factor, , LC Oscillations, Transformers, Quick Recap, , Introduction, , Currents whose direction does not change with time through a load,, are known as direct current (D.C.), whereas currents whose direction, changes periodically through a load are known as alternating, currents and the voltage is known as alternating voltage (ac voltage)., Most of the electrical devices require ac voltage because electrical, energy in a.c. form can be easily transmitted over long distances, without much loss. A.C. voltage can be easily converted to other, voltages by step up/step down transformers., , AC. VOLTAGE APPLIED TO A RESISTOR, , Let a.c. voltage applied is, , v=v,, Sinwt -.- (i), !, , , , R, , where v,,, is the maximum value of applied voltage (or amplitude of oscillating, potential difference) and is the angular frequency. From Kirchhoff's, loop rule, v,, sinwt = IR, , = 1-(%2)sin oot = jp Sinot (il), , , , where |, -(%2) is the amplitude of current. From Eq. (i) & (ii), itis clear that, , voltage across the resistor and current through the resistor are in same phase., , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456

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204 Alternating Current Board & Competitive Exams., , The sum of the instantaneous current values over one complete cycle is ZERO and thus the average of the current, over one complete cycle is ZERO., , The instantaneous power dissipated in the resistor is, , p=PR=1,?Rsin2ot we (ii), The average value of ‘p’ over a cycle is, P=<PR>=</?Rsin? wt > - (Iv), T, jane wtdt 7=28 r, se Ta ! (1-cos2ut)dt = >, o, , «++ (¥), , The effective value of a.c. current for heat dissipation is equal to d.c. current which produces same heat as due to a.c., Thus,, , , , , , .-. (vi), , , , Similarly Vins = *g = 0.707 vo -. (wil), , The house-hold line voltage of 220 V is the rms value with a peak voltage of, , Vm = V2 V =(1.414)x 220 = 311V, , Calculation of mean and mms values for some specific cases of a.c., , 1. For acurrent varying as shown in the graph, the mathematical expression may be given as i = (22 )e, , Now let us calculate mean value for positive half cycle, , hy a, , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Deihi-110005 Ph. 011-47623456

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Board & Competitive Exams. Alternating Current 205, 2. For sinusoidal a.c., ‘ean = 0 for one complete cycle, , 2h, Inaan = —_- for half cycle, , = /o, ae, , 3. For the out put of a half wave rectifier, , , , I, , meen = for one complete cycle, 2Ig, , mean = “2 for t= Oto t= Ti2, , A, ime = 3 for one complete cycle, , 4. For the out-put of a full wave rectifier, , , , I, ‘ime = pe for one complete cycle, , 5. For the square wave shown in figure, , , , , , , , , , , , , , , , , , Imean = 0 for one complete cycle lk Square wave, le, yp for t= Oto t= T2, o t, J, for one complete cycle 4 Wa T, Ima = 'p for t= Ti2 to t= T, , , , Phasor, , Any quantity that varies sinusoidally, can be represented as the projection of a uniform circular motion on a, diameter of the circular path., , To represent an alternating emf, the peak value Ey is taken as radius vector of the circle. When this radius, vector Eo rotates with angular velocity m as shown in figure, the projection on the diameter along y-axis gives, instantaneous value of emf. In this scheme of things, we have, , 1. The vector rotating in anticlockwise direction with angular velocity ‘w' is called a phasor., , , , 2. Its length is equal to amplitude of the alternating quantity., 3. Projection of the vector on y-axis gives the instantaneous value of alternating quantity., , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Dethi-110005 Ph. 011-47623456

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206 Altemating Current Board & Competitive Exams., , , , Content Builder, , , , Example 1:, , Solution :, , , , Mean Value of Alternating Current, , Let an alternating current be represented by / = /, sin wt, where /, is the peak value. Clearly, the, mean value of the current over a complete cycle is zero. It has no significance. Hence, the, mean value of alternating current is defined as its average over half a cycle. For positive half cycle,, , 4 pT, , , , Imeon = 775 ot, Putting / = /, sin wt and T = (27/w), we get, mde a | Io A, =2 f, sin wt dt = > 2 [- cos wf =-2 [cos x-cos ]=- 2 - 1-1), 2, Thus, Irnean = =o .-.[Mean current over positive half cycie], , , , , , , , Similarly, the mean value of alternating EMF is 2E,/x for positive half cycle. For negative half cycie, , 2h,, the mean value of alternating current is — and the average value of EMF for negative half cycle, 2E,, is — ae So, the average value of EMF of current for complete cycle is zero., , Note : Whenever you are asked to calculate charge flown in a circuit then mean value of altemating, current comes into picture., , , , , , , , A light bulb has the rating 200 W 220 V. Find (i) resistance of the bulb filament (ii) rms value, of current flowing through the filament., , 2, @ r=o - 220xa20 - 2x2? | 2422, , (ii) The rms value of current, , , , A light bulb has the rating 200 W 220 V., 1. Find the peak value of current., 2. If the supply voltage is 200 V, then find power consumed by the bulb., , , , 1. The voltages of domestic ac is 220 V. What does this represent?, (1) Mean voltage (2) Peak voltage, (3) Root mean voltage (4) Root mean square voltage, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456

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Board & Competitive Exams. Alternating Current 207, , , , 2., , The equation of an alternating voltage is V= 4100/2 sin 100nt volt. The RMS value of voltage and frequency, will be respectively, , (1) 100 V, 50 Hz (2) 50V, 100 Hz, (3) 150 V, 50 Hz (4) 200V, 50 Hz, , The average value of current for the current shown for time period 0 to = is, L, , 2, , , , ly, o, I 2h, , The phase difference between current and voltage in an AC circuit is 3 radian. If the frequency of AC is 50 Hz,, then the phase difference is equivalent to the time difference of, , (4) 075s (2) 10.5 ms, , (3) 2.5 ms (4) 0.25 ms, , Equation of alternating current is given by / = 1042 sin( 1000 +z). The time taken by current to reach the, root mean square value from f= 0 is f then the value of f is, , 1) tos zis, , © 3200 ®) 250, , 1 1, ®) 2008 008, , In the AC circuit, the current is expressed as / = 100 sin 200nt. In this circuit the current rises from zero to, peak value in time, , 1 1, 4005 2 3005, , @) z58 (4), , 700 200°, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Dethi-110005 Ph. 011-47623456