University:
California State University, Northridge
Course:
PHYS 100B | General Physics II
Academic year:

2008
Views:

473

Pages:

13
Author:

Alexia Cummings

Chapter 24 Alternating-Current Circuit Outline 24-1 Alternating Voltage and Circuit 24-1 Alternating Voltage and Circuit In an alternating circuit, the magnitude and direction of the voltage and current change periodically, and they are a function of time: The current and voltage is the so-called alternating current (AC) and voltage, respectively. Figure 24-1 An AC Generator Connected to a Lamp V = Vmax sin ωt Where, ω = 2πf, (f = 60 Hz). According to Ohm’s law, V Vmax I= = sin i ωt = I max sin i ωt R R (24 − 1) Root Mean Square (rms) Value Both alternating voltage and current have a zero value. So direct average gives no information (or useless). IIn order d to t evaluate l t an alternating lt ti parameter t in i quantity, tit we use root mean square (rms): We square the alternating current II, 2 I 2 = I max sin 2 ωt Now, we can average I2, 1 2 ( I ) av = I max 2 2 RMS is square root of the above eq., I rms 1 = I max 2 Rms: Square, average, square root Any quantity x that varies with time as x=xmaxsinωt, or x=xmaxcosωt, obey the relationships: RMS Value of a Quantity with Sinusoid Time Dependence 1 2 ( x ) av = xmax 2 1 xrms = xmax 2 2 (24 − 4) So, the rms value of the voltage in a AC circuit is Vrms 1 = Vmax 2 (24 − 5) Also suitable for current ! Exercise 24-1 Typical household circuit operates with an rms voltage of 120V. What is the maximum, or peak value of the voltage in the circuit? Solution Since Vrms = 1 Vmax , we have 2 The maximum and peak is: Vmax = 2Vrms = 2(120V ) = 170 V “Average” Power Since P = I2 R, Replace I with Irms, we have the average value of P 2 Pav = I rms R Apply Ohm’s law, 2 Vrms Pav = R (24 − 6) Rms can operate directly for Ohm law ! A Resistor Circuit An AC generator with a maximum voltage of 24.0 V and a frequency of 60.0 Hz is connected to a resistor with a resistance R= 265 Ω. Find (a) the rms voltage and (b) the rms current in the circuit circuit. Determine (c) the average and (d) maximum power dissipated in the resistor. Solution (a) The rms voltage is Vrms = 1 1 Vmax = (24.0) = 17.0 V 2 2 (b) The rms current is I rms = Vrms 17.0 17 0V = = 0.0642 A R 265Ω (c) The average power V 2 rms (17.0V ) 2 = = 1.09 W Pav = R 265Ω 265 Ω (d) The maximum power Pmax V 2 max V 2 rms = =2 = 2 ×1.09 W = 2.18 W R R Summary The calculate of RMS for a x=xmaxsinωt, or x=xmaxcosωt, function