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1 RMS values Book page cgrahamphysics.com 015
2 Review When angle between normal to loop and field lines is θ = 90 0 max flux, Φ = NAB cos θ min emf, emf = ωnab sin ωt θ = 0 0 min flux, max emf cgrahamphysics.com 015
3 AC alternating current An alternating current can be expressed as I = I 0 sin ωt I 0 = peak current For f = 50 Hz, T = 1 f = 1 50 = 0.0 = 0ms I(A) I 0 I Alternating current is expressed in terms of their root mean square value RMS The instantaneous value of the current is not as important as the average effect over time cgrahamphysics.com 015
4 Consider the heat dissipated in a resistor Current The instantaneous rate of heat energy is E t = P = I R, where I = size of current in that instant The direction of the current is unimportant Only variation in size effects power Consider two identical resistors, one carrying dc and the other ac. Suppose they are dissipating the same power as thermal energy Then the RMS value of ac current that produces the same power, is equal to the dc value of direct current cgrahamphysics.com 015
5 AC maximum power P = IV = I 0 sin ωt V 0 sin ωt = I 0 V 0 sin ωt It would be convenient to find P average Problem: I average or V average would give zero P average is always positive since sine is squared cgrahamphysics.com 015
6 Mathematical manipulation I = I 0 sin ωt always positive But sin θ = I = I 0 1 cos θ (1 cosθ) Over one cycle this part equals zero Hence I = I 0 is I RMS = I 0 and the average current of one cycle cgrahamphysics.com 015
7 Power and voltage We can use the same approach for voltage to get V RMS = V 0 For the maximum value in ac, the power dissipated, is given by P = I 0 V 0 sin ωt This means, the power supplied to the resistor in time by an alternating current is equal to the average value of I R multiplied by time P = I average R = I 0 Rsin ωt, where sin ωt will vary between 0 and 1. The average value will be 0+1 = 1 cgrahamphysics.com 015
8 Power The average power dissipated in the resistor equals P average = 1 I 0 R = I 0 I 0 Conclusion P average = 1 R or V 0 R = V 0 V 0 R The current dissipated in a resistor in an ac circuit that varies between I 0 and - I 0 would be equal to a current I 0 I RMS R = 1 I 0 R in a dc circuit cgrahamphysics.com 015
9 Summary The RMS equivalent current is defined as the dc that will provide the same power in the resistor as the ac does on average Using RMS values of current, voltage drop and emf, ac can be treated as dc provided the circuit contains only components with resistance P average = I RMS R = 1 I 0 R= V RMS R = 1 I 0V 0 = I RMS V RMS cgrahamphysics.com 015
10 Example Domestic ac has a peak value of 35V. What is the RMS voltage? What is the RMS current in a 100W light bulb? Solution V RMS = V 0 = 35 = 30V P average = I RMS V RMS I RMS = P average V RMS = = 0.43A cgrahamphysics.com 015
11 Example A resistor is connected in series with an alternating current supply. The peak value of the supply voltage is 140V and the peak value of current in the resistor is 9.5A. Find the average power dissipated in the resistor Solution P average = 1 I 0V 0 = = 670W cgrahamphysics.com 015
12 Example The value V 0 for a circuit containing a 35Ω resistor is 45V. Calculate the current and the power dissipated in the resistor. Solution V RMS = V 0 = 45 = 31.8V I RMS = V RMS = 31.8 = 0.91A R 35 P average = I RMS R = = 9W cgrahamphysics.com 015
13 Example Calculate a)the resistance and the peak current in a 1000W hair dryer connected to a 10V line. B)What happens if it is brought to China and connected to a 40V line? Solution P average = I RMS V RMS I RMS = P average V RMS = = 8.33A I RMS = I 0 I 0 = I RMS = 8.33 = 11.8A R = V RMS = 10 = 14.4Ω or I RMS 8.33 R = V 0 = 10 = 14.4Ω I b) Now more current would flow T and R Assume the same 14.4 resistor P average = V RMS = 40 = 4000W R 14.4 This is 4 times the dryer s power rating it would melt the heating element cgrahamphysics.com 015
14 Example Each channel in a stereo receiver is capable of an average power output of 100W into an 8Ω loudspeaker. What are the RMS voltage and the RMS current fed to the speaker at maximum power of 100W. Solution P average = V RMS R V RMS = P average R = = 8V P average = I R RMS I RMS = Or I RMS = V RMS R P average R = = 3.5A = 8 8 = 3.5A cgrahamphysics.com 015
15 Topic 11: Electromagnetic induction - AHL 11. Power generation and transmission Root mean square (rms) values of current and voltage Power consumption is given by P = V rms I rms which we just showed to be (1/)V 0 I 0.
16 Topic 11: Electromagnetic induction - AHL 11. Power generation and transmission Root mean square (rms) values of current and voltage This is a straight-forward definition of rms current.
17 Topic 11: Electromagnetic induction - AHL 11. Power generation and transmission Root mean square (rms) values of current and voltage I 0 I 0 T T First we square the values over one period T: We then find the squared area: I 0 T/ + I 0 T/ = I 0 T. Then we find the mean height: A / T = I 0 T / T = I 0. Then we take the square root: I = I 0.
Book Page cgrahamphysics.com Transformers
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