Chapter 24 Alternating-Current CircuitChapter 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 24-1An AC Generator Connected to a

LampLamp

)124(sin −= tVV ω )124(sinmax= tVV ω

Where, ω = 2πf, (f = 60 Hz).

According to Ohm’s law,

tItVVI iimax tItRR

I ωω sinsin maxmax ===

Root Mean Square (rms) Value

Both alternating voltage and current have a zero value. So direct average gives no information (or useless).

I d t l t lt ti t i titIn order to evaluate an alternating parameter in quantity, we use root mean square (rms):

We square the alternating current IWe square the alternating current I,

tII ω222 sinmax

=

Now, we can average I2,

22 1)( II 22max2

)( II av =

RMS is square root of the above eq.,

max21 IIrms =

Any quantity x that varies with time as x=xmaxsinωt, or Rms: Square, average, square root

x=xmaxcosωt, obey the relationships:

RMS Value of a Quantity with Sinusoid Time DependenceRMS Value of a Quantity with Sinusoid Time Dependence

21)( 2

max2 = xx av

)424(2

12

max

max

−= xxrms

av

2

So, the rms value of the voltage in a AC circuit is

)524(1−= VV )524(

2 maxVVrms

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

max21 VVrms =Since , we have

The maximum and peak is:

2 2(120 ) 170V V V Vmax 2 2(120 ) 170rmsV V V V= = =

“Average” Power

Since P = I2 R,

Replace I with I we have the average value of PReplace I with Irms, we have the average value of P

RIP rmsav2=

2V

Apply Ohm’s law,

)624( −=R

VP rmsav

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 DetermineFind (a) the rms voltage and (b) the rms current in the circuit. Determine (c) the average and (d) maximum power dissipated in the resistor.

Solution

(a) The rms voltage is

VVVrms 0.17)0.24(11max ===rms )(

22 max

(b) The rms current is

17 0V V17.0 0.0642265

rmsrms

V VI AR

= = =Ω

(c) The average power(c) The average power

WVR

VP rmsav 09.1

265)0.17( 22

=Ω

==R 265Ω

(d) The maximum power

22

WWR

VR

VP rms 18.209.1222

max2

max =×===

Homework Problems: 2, 4

2. In many European homes the rms voltage availablefrom a wall socket is 240 V. What is the maximumvoltage in this case?

4. The rms current in an ac circuit with a resistance of 150 Ω is 0.85 A. What are the (a) average and (b) maximum power consumed by this circuit?

SSummary

The calculate of RMS for a x=xmaxsinωt,

or x=xmaxcosωt, function