Notes for Experiment #4 -- AC Circuits

Notes & Hazards

1. Some theoretical considerations:

If the sine wave is symmetric about the 0-point, then the peak value is simply the value from the 0-point to the top of a crest. If, however, the curve is not symmetric about the 0-point, then there is a DC Offset. In this case, the peak value is given by Peak-to-Peak/2 where the Peak-to-Peak value is measured from the top of one crest to the bottom of the successive trough. The difference between the actual location of the peak of the crest and the 0-point gives you the DC offset. E.g., if the peak value is 2 and the top of the crest is located at the 2-point mark, then the DC offset is 2-2 = 0. If, however, the peak value is 2 and the top of the crest is located at the 2.5 point mark, then the DC offset is 2.5 - 2 = 0.5.

The period, T, is the time (or divisions * time/divisions) between 2 successive peaks (or troughs). The period has units of seconds/cycle. The frequency, n, is the inverse of the period (i.e., 1/T) and has units of cycles/second.

Finally, remember that the plot of Current (I) vs. Time (t) is a straight line for a DC power source (since the current is constant) but varies with time for an AC power source (since current reverses polarity with time). It can be a square wave, a triangle wave, a sine wave, etc. If it happens to be a sine (or a cosine) curve, then we can take advantage of certain mathematical characteristics of sine waves to gain insight/info into our AC current and voltage. For sine (or cos) waves only, the RMS values for current or voltage are given by: Peak value/sqrt. For all other waveforms (e.g., square, triangle, etc.), you're limited to reading the RMS values from your multimeter (e.g., the Goldstar) and cannot use the above formula for RMS. Also, please note that Ohm's Law and Kirchoff's Laws only apply to the RMS values when dealing with AC circuits.
2. Misc. Experimental Notes:

• Right before Question 4.1.9 (on p. 46), remember that Vrms IS NOT EQUAL to Vo/Sqrt. That equation is only valid for sine waves and the square wave isn't a sine wave.
• On p. 51, in Step 1, be sure to set Ch. 2 to DC and make sure that the Volts/division is set to 1.
• On p. 53, Question 4.3.3 wants you to sketch the input (this is on Ch. 1 and is the orginal function that the generator puts out) and the output (this is on Ch. 2 and is the rectified circuit).

For Questions 4.3.4 and 4.3.5, you can consider the capacitor-resistor circuit to be a low pass filter so think about how altering the R in the time constant, RC would affect this behaviour.

Another way to think of this is to consider the 2 extremes; if R goes to infinity, then that wire acts like it wasn't there and you can cut it out of the circuit (think of your ideal voltmeter and how it has infinite voltage so that all the current goes through the rest of the circuit). If you do this, then only the capacitor remains in the circuit; this then means that when the capacitor discharges (during the negative polarization part of the AC cycle), there is no resistor to "bleed" any of the current; end result, it smoothes out the curve. At the other extreme, when R = 0, the current, being inherently lazy, will take the path of least resistance and so will bypass the capacitor altogether and just go down the R=0 wire; end result, you short the circuit.
3. Some derivations (EQs 5-7 in your lab manual):

```EQ5    V = I1R1 + I1Req where Req is the resistance of the
of the parallel resistors.  Since they're in parallel, this is equal to:
1/Req = 1/R2 + 1/R3

Plugging this in above yields,
1
==>  V = I1R1 + I1(----------)
1/R2 + 1/R3

The term in parantheses can be rearranged as:
R2R3
Req =  ----------
R2 + R3

Factoring out the I1 and dividing out yields:
V
==>  I1 = ----------
R1 +   R2R3
-------
R2 + R3

Multiplying and dividing R1 by R2+R3, gives:
V
==>  I1 = ---------------------
R1(R2 + R3) +   R2R3
--------------------
R2 + R3

Which finally yields:
V(R2 + R3)
==>  I1 = ---------------------              (EQ5)
R1R2 + R1R3 +   R2R3

EQ6    I1Req = Veq = I2R2

Using Req from above gives:
I1Req            I1R2R3
I2 = ------   ==>  ------------
R2             R2(R2 + R3)

Cancelling out the R2's, gives:
I1R3
==>  ---------                               (EQ6)
R2 + R3

EQ7    I1Req = Veq = I3R3

Using Req from above gives:
I1Req            I1R2R3
I3 = ------   ==>  ------------
R3             R3(R2 + R3)

Cancelling out the R3's, gives:
I1R2
==>  ---------                               (EQ7)
R2 + R3
```

Corrections

1. Many minor errors.

Required Materials:

1. Laboratory Manual (SGM 407)