Page 1 J E E - P h y s i c s 3 2 E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 ALTERNATING CURRENT ALTERNATING CURRENT AND VOLTAGE Voltage or current is said to be alternating if it is change continously in magnitude and perodically in direction with time. It can be represented by a sine curve or cosine curve I = I 0 sin ? ?t or I = I 0 cos ? ?t where I = Instantaneous value of current at time t, I 0 = Amplitude or peak value ? ?= Angular frequency ? ? = 2 T ? = 2 ?f T = time period f = frequency I 0 –I 0 T 4 T 2 3 4 T T t I I as a sine function of t I 0 –I 0 T 4 T 2 3 4 T T t I I as a cosine function of t AMPLITUDE OF AC The maximum value of current in either direction is called peak value or the amplitude of current. It is represented by I 0 . Peak to peak value = 2I 0 PERIODIC TIME The time taken by alternating current to complete one cycle of variation is called periodic time or time period of the current. FREQUENCY The number of cycle completed by an alternating current in one second is called the frequency of the current. UNIT : cycle/s ; (Hz) In India : f = 50 Hz , supply voltage = 220 volt In USA :f = 60 Hz ,supply voltage = 110 volt CONDITION REQUIRED FOR CURRENT/ VOLTAGE TO BE ALTERNATING • Amplitude is constant • Alternate half cycle is positive and half negative The alternating current continuously varies in magnitude and periodically reverses its direction. I sinosudial AC I t + – triangular AC + – I t square wave AC saw tooth wave I t mixture of AC and DC I 0 t Not AC (direction not change) I 0 I 0 t t Not AC (not periodic) JEEMAIN.GURU Page 2 J E E - P h y s i c s 3 2 E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 ALTERNATING CURRENT ALTERNATING CURRENT AND VOLTAGE Voltage or current is said to be alternating if it is change continously in magnitude and perodically in direction with time. It can be represented by a sine curve or cosine curve I = I 0 sin ? ?t or I = I 0 cos ? ?t where I = Instantaneous value of current at time t, I 0 = Amplitude or peak value ? ?= Angular frequency ? ? = 2 T ? = 2 ?f T = time period f = frequency I 0 –I 0 T 4 T 2 3 4 T T t I I as a sine function of t I 0 –I 0 T 4 T 2 3 4 T T t I I as a cosine function of t AMPLITUDE OF AC The maximum value of current in either direction is called peak value or the amplitude of current. It is represented by I 0 . Peak to peak value = 2I 0 PERIODIC TIME The time taken by alternating current to complete one cycle of variation is called periodic time or time period of the current. FREQUENCY The number of cycle completed by an alternating current in one second is called the frequency of the current. UNIT : cycle/s ; (Hz) In India : f = 50 Hz , supply voltage = 220 volt In USA :f = 60 Hz ,supply voltage = 110 volt CONDITION REQUIRED FOR CURRENT/ VOLTAGE TO BE ALTERNATING • Amplitude is constant • Alternate half cycle is positive and half negative The alternating current continuously varies in magnitude and periodically reverses its direction. I sinosudial AC I t + – triangular AC + – I t square wave AC saw tooth wave I t mixture of AC and DC I 0 t Not AC (direction not change) I 0 I 0 t t Not AC (not periodic) JEEMAIN.GURU J E E - P h y s i c s E 3 3 Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 AVERAGE VALUE OR MEAN VALUE The mean value of A.C over any half cycle (either positive or negative) is that value of DC which would send same amount of charge through a circuit as is sent by the AC through same circuit in the same time. average value of current for half cycle < I > = T / 2 0 T / 2 0 Idt dt ? ? Average value of I = I 0 sin ?t over the positive half cycle : < sin > = < sin 2 >=0 < cos >= < cos 2 >= 0 < sin cos > = 0 < sin > = < cos >= ? ? ? ? ? ? ? ? 2 2 1 2 T 2 0 0 av T 2 0 I sin t dt I dt ? ? ? ? = ? ? T 0 2 0 2 I cos t T ? ? ? 0 2 I ? ? • For symmetric AC, average value over full cycle = 0, Average value of sinusoidal AC Full cycle (+ve) half cycle (–ve) half cycle 0 2I 0 ? –2I 0 ? As the average value of AC over a com plete cycle is zero, it is always defined over a half cycle which must be either positive or negative MAXIMUM VALUE • I = a sin ? ? I Max. = a • I = a + b sin ? ? ? I Max. = a + b ( if a and b > 0) • I = a sin ? + b cos ? ? ? I Max. = 2 2 a b ? • I = a sin 2 ? ? I Max. = a (a > 0) ROOT MEAN SQUARE (rms) VALUE It is value of DC which would produce same heat in given resistance in given time as is done by the alternating current when passed through the same resistance for the same time. T 2 0 rms T 0 I dt I dt ? ? ? rms value = Virtual value = Apparent value rms value of I = I 0 sin ?t : T 2 0 0 rms T 0 (I sin t) dt I dt ? ? ? ? = 2 T 2 0 0 I sin t dt T ? ? = 0 T 0 1 1 cos2 t I dt T 2 ? ? ? ? ? ? ? ? ? T 0 0 1 t sin 2 t I T 2 2 2 ? ? ? ? ? ? ? ? ? ? ? = 0 I 2 • If nothing is mentioned then values printed in a.c circuit on electrical appliances, any given or unknown values, reading of AC meters are assumed to be RMS. Current Average P e a k R M S Angular fequency I 1 = I 0 sin ?t 0 I 0 0 I 2 ? I 2 = I 0 sin ?t cos ?t = 0 I sin 2 t 2 ? 0 0 I 2 0 I 2 2 2 ? I 3 = I 0 sin ?t + I 0 cos ?t 0 0 2 I I 0 ? • For above varieties of current rms = Peak value 2 JEEMAIN.GURU Page 3 J E E - P h y s i c s 3 2 E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 ALTERNATING CURRENT ALTERNATING CURRENT AND VOLTAGE Voltage or current is said to be alternating if it is change continously in magnitude and perodically in direction with time. It can be represented by a sine curve or cosine curve I = I 0 sin ? ?t or I = I 0 cos ? ?t where I = Instantaneous value of current at time t, I 0 = Amplitude or peak value ? ?= Angular frequency ? ? = 2 T ? = 2 ?f T = time period f = frequency I 0 –I 0 T 4 T 2 3 4 T T t I I as a sine function of t I 0 –I 0 T 4 T 2 3 4 T T t I I as a cosine function of t AMPLITUDE OF AC The maximum value of current in either direction is called peak value or the amplitude of current. It is represented by I 0 . Peak to peak value = 2I 0 PERIODIC TIME The time taken by alternating current to complete one cycle of variation is called periodic time or time period of the current. FREQUENCY The number of cycle completed by an alternating current in one second is called the frequency of the current. UNIT : cycle/s ; (Hz) In India : f = 50 Hz , supply voltage = 220 volt In USA :f = 60 Hz ,supply voltage = 110 volt CONDITION REQUIRED FOR CURRENT/ VOLTAGE TO BE ALTERNATING • Amplitude is constant • Alternate half cycle is positive and half negative The alternating current continuously varies in magnitude and periodically reverses its direction. I sinosudial AC I t + – triangular AC + – I t square wave AC saw tooth wave I t mixture of AC and DC I 0 t Not AC (direction not change) I 0 I 0 t t Not AC (not periodic) JEEMAIN.GURU J E E - P h y s i c s E 3 3 Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 AVERAGE VALUE OR MEAN VALUE The mean value of A.C over any half cycle (either positive or negative) is that value of DC which would send same amount of charge through a circuit as is sent by the AC through same circuit in the same time. average value of current for half cycle < I > = T / 2 0 T / 2 0 Idt dt ? ? Average value of I = I 0 sin ?t over the positive half cycle : < sin > = < sin 2 >=0 < cos >= < cos 2 >= 0 < sin cos > = 0 < sin > = < cos >= ? ? ? ? ? ? ? ? 2 2 1 2 T 2 0 0 av T 2 0 I sin t dt I dt ? ? ? ? = ? ? T 0 2 0 2 I cos t T ? ? ? 0 2 I ? ? • For symmetric AC, average value over full cycle = 0, Average value of sinusoidal AC Full cycle (+ve) half cycle (–ve) half cycle 0 2I 0 ? –2I 0 ? As the average value of AC over a com plete cycle is zero, it is always defined over a half cycle which must be either positive or negative MAXIMUM VALUE • I = a sin ? ? I Max. = a • I = a + b sin ? ? ? I Max. = a + b ( if a and b > 0) • I = a sin ? + b cos ? ? ? I Max. = 2 2 a b ? • I = a sin 2 ? ? I Max. = a (a > 0) ROOT MEAN SQUARE (rms) VALUE It is value of DC which would produce same heat in given resistance in given time as is done by the alternating current when passed through the same resistance for the same time. T 2 0 rms T 0 I dt I dt ? ? ? rms value = Virtual value = Apparent value rms value of I = I 0 sin ?t : T 2 0 0 rms T 0 (I sin t) dt I dt ? ? ? ? = 2 T 2 0 0 I sin t dt T ? ? = 0 T 0 1 1 cos2 t I dt T 2 ? ? ? ? ? ? ? ? ? T 0 0 1 t sin 2 t I T 2 2 2 ? ? ? ? ? ? ? ? ? ? ? = 0 I 2 • If nothing is mentioned then values printed in a.c circuit on electrical appliances, any given or unknown values, reading of AC meters are assumed to be RMS. Current Average P e a k R M S Angular fequency I 1 = I 0 sin ?t 0 I 0 0 I 2 ? I 2 = I 0 sin ?t cos ?t = 0 I sin 2 t 2 ? 0 0 I 2 0 I 2 2 2 ? I 3 = I 0 sin ?t + I 0 cos ?t 0 0 2 I I 0 ? • For above varieties of current rms = Peak value 2 JEEMAIN.GURU J E E - P h y s i c s 3 4 E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 Example If I = 2 t ampere then calculate average and rms values over t = 2 to 4 s Solution < I > = 4 3 4 2 2 2 4 4 2 2 2 t.dt (t ) 4 2 8 2 2 3 3 (t) dt ? ? ? ? ? ? ? ? ? and I rms = 4 4 2 2 2 4 2 (2 t ) dt 4t dt 2 dt ? ? ? ? = 4 2 2 t 2 2 3 A 2 ? ? ? ? ? ? ? Example Find the time required for 50Hz alternating current to change its value from zero to rms value. Solution ? I = I 0 sin ? ?t ? 0 0 I I sin t 2 ? ? ? 1 sin t 2 ? ? t 4 ? ? ? ? 2 t T 4 ? ? ? ? ? ? ? ? ? ? T t 8 ? 1 2.5 8 50 ? ? ? ms Example If E = 20 sin (100 ? ?t) volt then calculate value of E at t = 1 600 s Solution At t = 1 600 s E = 20 Sin 1 100 600 ? ? ? ? ? ? ? ? = 20 sin 6 ? ? ? ? ? ? ? = 20 × 1 2 = 10V Example A periodic voltage wave form has been shown in fig. Determine. (a) Frequency of the wave form. (b) Average value. Solution (a) After 100 ms wave is repeated so time period is T = 100 ms. ? f = 1 T = 10 Hz (b) Average value = Area/time period = ( / ) ( ) 1 2 100 10 100 ? ? = 5 volt Example Explain why A.C. is more dangerous than D.C. ? Solution There are two reasons for it : ? ? A.C. attracts while D.C. repels. ? A.C. gives a huge and sudden shock. O -E E t +E = 311.08 0 -E = -311.08 0 0.01 sec for 220 V ac V rms = 220 V Hence, V 0 = rms 2.V = 12.414 × 220 = 311.08 V Voltage change from +V 0 (positive peak) to –V 0 (negative peak) = 311.08 – (–311.08) = 622.16 V This change takes place in half cycle i.e., in 1 100 s (for a 50 Hz A.C.) A shock of 622.16 within 0.01 s is huge and sudden, hence fatal. JEEMAIN.GURU Page 4 J E E - P h y s i c s 3 2 E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 ALTERNATING CURRENT ALTERNATING CURRENT AND VOLTAGE Voltage or current is said to be alternating if it is change continously in magnitude and perodically in direction with time. It can be represented by a sine curve or cosine curve I = I 0 sin ? ?t or I = I 0 cos ? ?t where I = Instantaneous value of current at time t, I 0 = Amplitude or peak value ? ?= Angular frequency ? ? = 2 T ? = 2 ?f T = time period f = frequency I 0 –I 0 T 4 T 2 3 4 T T t I I as a sine function of t I 0 –I 0 T 4 T 2 3 4 T T t I I as a cosine function of t AMPLITUDE OF AC The maximum value of current in either direction is called peak value or the amplitude of current. It is represented by I 0 . Peak to peak value = 2I 0 PERIODIC TIME The time taken by alternating current to complete one cycle of variation is called periodic time or time period of the current. FREQUENCY The number of cycle completed by an alternating current in one second is called the frequency of the current. UNIT : cycle/s ; (Hz) In India : f = 50 Hz , supply voltage = 220 volt In USA :f = 60 Hz ,supply voltage = 110 volt CONDITION REQUIRED FOR CURRENT/ VOLTAGE TO BE ALTERNATING • Amplitude is constant • Alternate half cycle is positive and half negative The alternating current continuously varies in magnitude and periodically reverses its direction. I sinosudial AC I t + – triangular AC + – I t square wave AC saw tooth wave I t mixture of AC and DC I 0 t Not AC (direction not change) I 0 I 0 t t Not AC (not periodic) JEEMAIN.GURU J E E - P h y s i c s E 3 3 Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 AVERAGE VALUE OR MEAN VALUE The mean value of A.C over any half cycle (either positive or negative) is that value of DC which would send same amount of charge through a circuit as is sent by the AC through same circuit in the same time. average value of current for half cycle < I > = T / 2 0 T / 2 0 Idt dt ? ? Average value of I = I 0 sin ?t over the positive half cycle : < sin > = < sin 2 >=0 < cos >= < cos 2 >= 0 < sin cos > = 0 < sin > = < cos >= ? ? ? ? ? ? ? ? 2 2 1 2 T 2 0 0 av T 2 0 I sin t dt I dt ? ? ? ? = ? ? T 0 2 0 2 I cos t T ? ? ? 0 2 I ? ? • For symmetric AC, average value over full cycle = 0, Average value of sinusoidal AC Full cycle (+ve) half cycle (–ve) half cycle 0 2I 0 ? –2I 0 ? As the average value of AC over a com plete cycle is zero, it is always defined over a half cycle which must be either positive or negative MAXIMUM VALUE • I = a sin ? ? I Max. = a • I = a + b sin ? ? ? I Max. = a + b ( if a and b > 0) • I = a sin ? + b cos ? ? ? I Max. = 2 2 a b ? • I = a sin 2 ? ? I Max. = a (a > 0) ROOT MEAN SQUARE (rms) VALUE It is value of DC which would produce same heat in given resistance in given time as is done by the alternating current when passed through the same resistance for the same time. T 2 0 rms T 0 I dt I dt ? ? ? rms value = Virtual value = Apparent value rms value of I = I 0 sin ?t : T 2 0 0 rms T 0 (I sin t) dt I dt ? ? ? ? = 2 T 2 0 0 I sin t dt T ? ? = 0 T 0 1 1 cos2 t I dt T 2 ? ? ? ? ? ? ? ? ? T 0 0 1 t sin 2 t I T 2 2 2 ? ? ? ? ? ? ? ? ? ? ? = 0 I 2 • If nothing is mentioned then values printed in a.c circuit on electrical appliances, any given or unknown values, reading of AC meters are assumed to be RMS. Current Average P e a k R M S Angular fequency I 1 = I 0 sin ?t 0 I 0 0 I 2 ? I 2 = I 0 sin ?t cos ?t = 0 I sin 2 t 2 ? 0 0 I 2 0 I 2 2 2 ? I 3 = I 0 sin ?t + I 0 cos ?t 0 0 2 I I 0 ? • For above varieties of current rms = Peak value 2 JEEMAIN.GURU J E E - P h y s i c s 3 4 E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 Example If I = 2 t ampere then calculate average and rms values over t = 2 to 4 s Solution < I > = 4 3 4 2 2 2 4 4 2 2 2 t.dt (t ) 4 2 8 2 2 3 3 (t) dt ? ? ? ? ? ? ? ? ? and I rms = 4 4 2 2 2 4 2 (2 t ) dt 4t dt 2 dt ? ? ? ? = 4 2 2 t 2 2 3 A 2 ? ? ? ? ? ? ? Example Find the time required for 50Hz alternating current to change its value from zero to rms value. Solution ? I = I 0 sin ? ?t ? 0 0 I I sin t 2 ? ? ? 1 sin t 2 ? ? t 4 ? ? ? ? 2 t T 4 ? ? ? ? ? ? ? ? ? ? T t 8 ? 1 2.5 8 50 ? ? ? ms Example If E = 20 sin (100 ? ?t) volt then calculate value of E at t = 1 600 s Solution At t = 1 600 s E = 20 Sin 1 100 600 ? ? ? ? ? ? ? ? = 20 sin 6 ? ? ? ? ? ? ? = 20 × 1 2 = 10V Example A periodic voltage wave form has been shown in fig. Determine. (a) Frequency of the wave form. (b) Average value. Solution (a) After 100 ms wave is repeated so time period is T = 100 ms. ? f = 1 T = 10 Hz (b) Average value = Area/time period = ( / ) ( ) 1 2 100 10 100 ? ? = 5 volt Example Explain why A.C. is more dangerous than D.C. ? Solution There are two reasons for it : ? ? A.C. attracts while D.C. repels. ? A.C. gives a huge and sudden shock. O -E E t +E = 311.08 0 -E = -311.08 0 0.01 sec for 220 V ac V rms = 220 V Hence, V 0 = rms 2.V = 12.414 × 220 = 311.08 V Voltage change from +V 0 (positive peak) to –V 0 (negative peak) = 311.08 – (–311.08) = 622.16 V This change takes place in half cycle i.e., in 1 100 s (for a 50 Hz A.C.) A shock of 622.16 within 0.01 s is huge and sudden, hence fatal. JEEMAIN.GURU J E E - P h y s i c s E 3 5 Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 Example If a direct current of value a ampere is superimposed on an alternating current I = b sin ?t flowing through a wire, what is the effective value of the resulting current in the circuit ? I t AC b =? I t DC a + Solution As current at any instant in the circuit will be, I = I DC + I AC = a + b sin ?t ? T T 2 2 eff 0 0 1 1 I I dt (a b sin t) dt T T ? ? ? ? ? ? T 2 2 2 0 1 (a 2ab sin t b sin t)dt T ? ? ? ? ? ? but as T 0 1 sin tdt 0 T ? ? ? and T 2 0 1 1 sin tdt T 2 ? ? ? ? 2 2 eff 1 I a b 2 ? ? SOME IMPORTANT WAVE FORMS AND THEIR RMS AND AVER AGE VALUE Nature of Wave–for m RMS Value Average or mean wave form Val ue Sinusoidal 0 + – ? 2 ? 0 I 2 0 2I ? = 0.707 I 0 = 0.637 I 0 Half wave 0 ? 2 ? 0 I 2 0 I ? rectifired = 0.5 I 0 = 0.318 I 0 Full wave 0 ? 2 ? 0 I 2 0 2I ? rectifired = 0.707 I 0 0.637 I 0 Square or + – I 0 I 0 Rectangular Saw Tooth wave 2 ? ? 0 0 I 3 0 I 2 JEEMAIN.GURU Page 5 J E E - P h y s i c s 3 2 E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 ALTERNATING CURRENT ALTERNATING CURRENT AND VOLTAGE Voltage or current is said to be alternating if it is change continously in magnitude and perodically in direction with time. It can be represented by a sine curve or cosine curve I = I 0 sin ? ?t or I = I 0 cos ? ?t where I = Instantaneous value of current at time t, I 0 = Amplitude or peak value ? ?= Angular frequency ? ? = 2 T ? = 2 ?f T = time period f = frequency I 0 –I 0 T 4 T 2 3 4 T T t I I as a sine function of t I 0 –I 0 T 4 T 2 3 4 T T t I I as a cosine function of t AMPLITUDE OF AC The maximum value of current in either direction is called peak value or the amplitude of current. It is represented by I 0 . Peak to peak value = 2I 0 PERIODIC TIME The time taken by alternating current to complete one cycle of variation is called periodic time or time period of the current. FREQUENCY The number of cycle completed by an alternating current in one second is called the frequency of the current. UNIT : cycle/s ; (Hz) In India : f = 50 Hz , supply voltage = 220 volt In USA :f = 60 Hz ,supply voltage = 110 volt CONDITION REQUIRED FOR CURRENT/ VOLTAGE TO BE ALTERNATING • Amplitude is constant • Alternate half cycle is positive and half negative The alternating current continuously varies in magnitude and periodically reverses its direction. I sinosudial AC I t + – triangular AC + – I t square wave AC saw tooth wave I t mixture of AC and DC I 0 t Not AC (direction not change) I 0 I 0 t t Not AC (not periodic) JEEMAIN.GURU J E E - P h y s i c s E 3 3 Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 AVERAGE VALUE OR MEAN VALUE The mean value of A.C over any half cycle (either positive or negative) is that value of DC which would send same amount of charge through a circuit as is sent by the AC through same circuit in the same time. average value of current for half cycle < I > = T / 2 0 T / 2 0 Idt dt ? ? Average value of I = I 0 sin ?t over the positive half cycle : < sin > = < sin 2 >=0 < cos >= < cos 2 >= 0 < sin cos > = 0 < sin > = < cos >= ? ? ? ? ? ? ? ? 2 2 1 2 T 2 0 0 av T 2 0 I sin t dt I dt ? ? ? ? = ? ? T 0 2 0 2 I cos t T ? ? ? 0 2 I ? ? • For symmetric AC, average value over full cycle = 0, Average value of sinusoidal AC Full cycle (+ve) half cycle (–ve) half cycle 0 2I 0 ? –2I 0 ? As the average value of AC over a com plete cycle is zero, it is always defined over a half cycle which must be either positive or negative MAXIMUM VALUE • I = a sin ? ? I Max. = a • I = a + b sin ? ? ? I Max. = a + b ( if a and b > 0) • I = a sin ? + b cos ? ? ? I Max. = 2 2 a b ? • I = a sin 2 ? ? I Max. = a (a > 0) ROOT MEAN SQUARE (rms) VALUE It is value of DC which would produce same heat in given resistance in given time as is done by the alternating current when passed through the same resistance for the same time. T 2 0 rms T 0 I dt I dt ? ? ? rms value = Virtual value = Apparent value rms value of I = I 0 sin ?t : T 2 0 0 rms T 0 (I sin t) dt I dt ? ? ? ? = 2 T 2 0 0 I sin t dt T ? ? = 0 T 0 1 1 cos2 t I dt T 2 ? ? ? ? ? ? ? ? ? T 0 0 1 t sin 2 t I T 2 2 2 ? ? ? ? ? ? ? ? ? ? ? = 0 I 2 • If nothing is mentioned then values printed in a.c circuit on electrical appliances, any given or unknown values, reading of AC meters are assumed to be RMS. Current Average P e a k R M S Angular fequency I 1 = I 0 sin ?t 0 I 0 0 I 2 ? I 2 = I 0 sin ?t cos ?t = 0 I sin 2 t 2 ? 0 0 I 2 0 I 2 2 2 ? I 3 = I 0 sin ?t + I 0 cos ?t 0 0 2 I I 0 ? • For above varieties of current rms = Peak value 2 JEEMAIN.GURU J E E - P h y s i c s 3 4 E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 Example If I = 2 t ampere then calculate average and rms values over t = 2 to 4 s Solution < I > = 4 3 4 2 2 2 4 4 2 2 2 t.dt (t ) 4 2 8 2 2 3 3 (t) dt ? ? ? ? ? ? ? ? ? and I rms = 4 4 2 2 2 4 2 (2 t ) dt 4t dt 2 dt ? ? ? ? = 4 2 2 t 2 2 3 A 2 ? ? ? ? ? ? ? Example Find the time required for 50Hz alternating current to change its value from zero to rms value. Solution ? I = I 0 sin ? ?t ? 0 0 I I sin t 2 ? ? ? 1 sin t 2 ? ? t 4 ? ? ? ? 2 t T 4 ? ? ? ? ? ? ? ? ? ? T t 8 ? 1 2.5 8 50 ? ? ? ms Example If E = 20 sin (100 ? ?t) volt then calculate value of E at t = 1 600 s Solution At t = 1 600 s E = 20 Sin 1 100 600 ? ? ? ? ? ? ? ? = 20 sin 6 ? ? ? ? ? ? ? = 20 × 1 2 = 10V Example A periodic voltage wave form has been shown in fig. Determine. (a) Frequency of the wave form. (b) Average value. Solution (a) After 100 ms wave is repeated so time period is T = 100 ms. ? f = 1 T = 10 Hz (b) Average value = Area/time period = ( / ) ( ) 1 2 100 10 100 ? ? = 5 volt Example Explain why A.C. is more dangerous than D.C. ? Solution There are two reasons for it : ? ? A.C. attracts while D.C. repels. ? A.C. gives a huge and sudden shock. O -E E t +E = 311.08 0 -E = -311.08 0 0.01 sec for 220 V ac V rms = 220 V Hence, V 0 = rms 2.V = 12.414 × 220 = 311.08 V Voltage change from +V 0 (positive peak) to –V 0 (negative peak) = 311.08 – (–311.08) = 622.16 V This change takes place in half cycle i.e., in 1 100 s (for a 50 Hz A.C.) A shock of 622.16 within 0.01 s is huge and sudden, hence fatal. JEEMAIN.GURU J E E - P h y s i c s E 3 5 Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 Example If a direct current of value a ampere is superimposed on an alternating current I = b sin ?t flowing through a wire, what is the effective value of the resulting current in the circuit ? I t AC b =? I t DC a + Solution As current at any instant in the circuit will be, I = I DC + I AC = a + b sin ?t ? T T 2 2 eff 0 0 1 1 I I dt (a b sin t) dt T T ? ? ? ? ? ? T 2 2 2 0 1 (a 2ab sin t b sin t)dt T ? ? ? ? ? ? but as T 0 1 sin tdt 0 T ? ? ? and T 2 0 1 1 sin tdt T 2 ? ? ? ? 2 2 eff 1 I a b 2 ? ? SOME IMPORTANT WAVE FORMS AND THEIR RMS AND AVER AGE VALUE Nature of Wave–for m RMS Value Average or mean wave form Val ue Sinusoidal 0 + – ? 2 ? 0 I 2 0 2I ? = 0.707 I 0 = 0.637 I 0 Half wave 0 ? 2 ? 0 I 2 0 I ? rectifired = 0.5 I 0 = 0.318 I 0 Full wave 0 ? 2 ? 0 I 2 0 2I ? rectifired = 0.707 I 0 0.637 I 0 Square or + – I 0 I 0 Rectangular Saw Tooth wave 2 ? ? 0 0 I 3 0 I 2 JEEMAIN.GURU J E E - P h y s i c s 3 6 E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Phy\Unit No.-10\EMI & AC\English\03 AC.p65 MEASUREMENT OF A.C. Alternating current and voltages are measured by a.c. ammeter and a.c. voltmeter respectively. Working of these instruments is based on heating effect of current, hence they are also called hot wire instruments. Te r m s D.C. meter A.C. meter Name moving coil hot wire Based on magnetic effect of current heating effect of current Reads average value r.m.s. value If used in A.C. circuit then they reads zero A.C. or D.C. then meter works ? average value of A.C. = zero properly as it measures rms value Deflection deflection ? current deflection ? heat ? ? ? (linear) ? ? ? ? rms (non linear) Scale Uniform Seperation Non uniform sepration ? = Number ? - 1 2 3 4 5 ? - 1 2 3 4 5 of divisions ? - 1 2 3 4 5 ? - 1 4 9 16 25 • D.C meter in AC circuit reads zero because < AC > = 0 ( for complete cycle) • AC meter works in both AC and DC PHASE AND PHASE DIFFERENCE ( a ) P h a s e I = I 0 sin ( ?t + ?) Initial phase = ? (it does not change with time) Instantaneous phase = ?t + ? (it changes with time) • Phase decides both value and sign. ? UNIT: radian (b ) Phase difference Voltage V = V 0 sin ( ?t + ? 1 ) Current I ? ? ? ? I 0 sin ( ?t + ? 2 ) • Phase difference of I w.r.t. V ? ?= ? 2 – ? 1 • Phase difference of V w.r.t. I ? = ? ? 1 – ? 2 LAGGING AND LEADING CONCEPT ( a ) V leads I or I lags V ? It means, V reach maximum before I I=I 0 ? sin ( t ) ? ? ? ?t V ,I V=Vsin t 0 ? Let if V = V 0 sin ?t then I = I 0 sin ( ?t – ? ? and if V = V 0 sin ( ?t+ ? ) then I = I 0 sin ?t (b ) V lags I or I leads V ? ?It means V reach maximum after I I=I 0 ? sin ( t + ) ? ? ?t V ,I V=V sin t 0 ? Let if V = V 0 sin ?t then I = I 0 sin ( ?t + ? ? and if V = V 0 sin ( ?t – ? ) then I = I 0 sin ?t PHASOR AND PHASOR DIAGRAM A diagram representing alternating current and voltage (of same frequency) as vectors (phasor) with the phase angle between them is called phasor diagram. ? ?t V V 0 I 0 I ? V 0 I 0 fig (a) fig (b) Y X Y X Let V = V 0 sin ?t and I = I 0 sin ( ?t + ?) In figure (a) two arrows represents phasors. The length of phasors represents the maximum value of quantity. The projection of a phasor on y-axis represents the instantaneous value of quantity ADVANTAGES OF AC • A.C. is cheaper than D.C • It can be easily converted into D.C. (by rectifier) • It can be controlled easily (choke coil) • It can be transmitted over long distance at negligible power loss. • It can be stepped up or stepped down with the help of transformer. JEEMAIN.GURURead More

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