Tiêu đề Alternating Current 4.0.pdf ✅

Alternating Current 1 4 Alternating Current 1. Direct Current The current which flows only in one direction is called a direct current. The current produced by batteries/cells is a direct current. In ordinary circuits such as in a car, a torch light and toys, this current is normally constant as shown in figure and is called a steady current. Time Current 2. Alternating Current Current is said to be alternating if it changes continuously in magnitude and periodically in direction with time. It can be represented by a sine curve or cosine curve. I = I0 sin ω t or I = I0 cos ω t where I = Instantaneous value of current at time t, I0 = Amplitude or peak value ω = Angular frequency ω = 2πf T = time period f = frequency Ι0 Ι=Ι ω 0 sin t T 4 T 2 3 T 4 T t Ι0 – Ι as a sine function of t Ι0 Ι=Ι ω 0 cos t T 4 T 2 3 T 4 T t Ι0 – Ι as a cosine function of t The alternating current continuously varies in magnitude and periodically reverses its direction. Ι Sinosudial AC Ι triangular AC + + – – t Ι Square wave AC t Ι Saw tooth wave t Ι t Not AC (diection does not change) Ι t Not AC (not periodic)
2 Alternating Current 3. Characteristics of AC 3.1 Amplitude The maximum value of current in either direction is called peak value or the amplitude of current. It is represented by I0. Peak to peak value = 2I0 3.2 Time Period The time taken by alternating current to complete one cycle of variation is called periodic time or time period of the current. 3.3 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 4. Mean Value or Average Value The mean value of AC 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 dt dt Ι ∫ ∫ average value of I = I0 sin ωt over the positive half cycle : I T 2 0 0 av T 2 0 sin t dt dt Ι ω = ∫ ∫ = I T 0 2 0 2 cos t T   − ω   ω = 0 2Ι π average value of function over full cycle: 2 2 sin sin2 0 cos cos2 0 sin cos 0 1 sin cos 2 < θ >=< θ >= < θ >=< θ >= < θ >= < θ >=< θ >= Maximum Value • I = a sinθ ⇒ IMax. = a • I = a + b sinθ ⇒ IMax. = a + b (if a and b > 0) • I = a sinθ + b cosθ ⇒ IMax. = 2 2 a +b • I = a sin2 θ ⇒ IMax. = a (a > 0)

4 Alternating Current Example 3: If the current in an AC circuit is represented by the equation i = 5 sin (300t – π/4) Here, t is in second and i is in ampere. Calculate (a) Peak and rms value of current (b) Frequency of AC (c) Average current Solution: (a) As in case of AC, i = i0sin (ωt ± φ) ∴ The peak value i0 = 5A and irms = 0 i 5 2 2 = = 3.535 A (b) Angular frequency ω = 300 rad/s ∴ f = 2 ω π = 300 2π ≈ 47.75 Hz (c) av 0 2 2 i i (5) 3.18A   = = =     π π Example 4: An electric heater draws 2.5 A current from a 220-V, 60-Hz power supply. Find (a) The average current (b) The average of the square of the current (c) The current amplitude (d) The supply voltage amplitude Solution: In AC circuit, the average value of current over a long time interval is zero but I2 is not zero. The r.m.s. value of current and voltage is given by max max rms rms I V I and V 2 2 = = . (a) The average of sinusoidal AC values over any whole number of cycles is zero. (b) RMS value of current = Irms = 2.5 A so, ( ) ( )2 2 2 rms av I I 6.25A = = (c) 0 rms I I 2 = ; So, current amplitude 0 rms I 2I 2(2.5A) 3.5A = = = (d) Vrms = 220V = mV 2 ; So, supply voltage amplitude V 2 V 2(220V) m rms = = ( ) = 311 V.