## Sheet1

Activity 7-1:(b page181) The Resistor in an AC Circuit | |||||||||

Frequency(Hz) | V2 (V) | V1(V) | i (mA) | ||||||

1000 | 5 | 0.169 | |||||||

2000 | 5 | 0.169 | |||||||

3000 | 5 | 0.169 | |||||||

4000 | 5 | 0.164 | |||||||

5000 | 5 | 0.164 | |||||||

c page 181 | |||||||||

V2(V) | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 | 4.5 | 5 |

V1(V) | 0.145 | 0.235 | 0.3 | 0.375 | 0.44 | 0.5 | 0.6 | 0.7 | 0.78 |

I0 | 1.45 | 2.3 | 3 | 3.75 | 4.4 | 5.3 | 6 | 7 | 7.8 |

Activity 7-2 (page 182): The capacitor in an AC Circuit | |||||||||

Frequency(Hz) | V1(V) | io(mA) | |||||||

1000 | 0.03 | ||||||||

2000 | 0.06 | ||||||||

3000 | 0.088 | ||||||||

4000 | 0.116 | ||||||||

5000 | 0.142 | ||||||||

Activity 7-3 (b page 183): The Inductor In the AC Circuit | |||||||||

Frequency(Hz) | V1 (V) | io(mA) | |||||||

1000 | 0.092 | ||||||||

2000 | 0.05 | ||||||||

3000 | 0.034 | ||||||||

4000 | 0.026 | ||||||||

5000 | 0.022 | ||||||||

Activity 7.4 (page 185): RLC in series in an AC circuit | |||||||||

Frequency(Hz) | V2 (V) | V1(V) | i (mA) | ||||||

1000 | 5 | 0.76 | |||||||

2000 | 5 | 1.8 | |||||||

3000 | 5 | 0.82 | |||||||

4000 | 5 | 0.56 | |||||||

5000 | 5 | 0.44 |

0

AC BEHAVIOR OF RESISTORS, CAPACITORS, AND INDUCTORS

Equipment: • Function generator • Resistors, capacitors, and inductors • oscilloscope • electrical probes • circuit board

Introduction

In this unit, you will measure the voltages and currents associated with simple RC, RL and RCL circuits. You will make calculations of the electric current across different components in these electric circuits. By measuring the current across these different components in your circuit you will verify that the equations learned in class actually have physical significance in your own circuits.

Resistors: The resistance of a resistor does not depend on the frequency of an alternating voltage. It is the same for DC as it is for AC. If a current i0sin(2nft) (where f is the frequency) flows through R, then a voltage i0Rsin(2nft) appears across R which is in phase with the current, as shown in Figure 1.

VV V … t

JI t

Figure 1

177

V”‘ Q/C

Figure 2

t t

)o t

~’-~ I\ C,, \JV V

– Vo

Figure 3

Capacitors: The AC analog to resistance for a capacitor is called the capacitive reactance. The capacitive reactance is defined as the ratio of V 0/i0 and it depends on the frequency . The voltage across a capacitor and the capacitor current that is charging the capacitor plates are not in phase.

This can be seen by assuming that Q, the charge on one plate (the other plate has a charge -Q), is Q0cos (2nft). We know that the potential difference across a capacitor is V = Q/C and i = dQ/dt where i is the capacitor current. These quantities are sketched in Figure 2.

We note that at any time t, i (=dQ/dt) is the slope of the Q vs. t graph at that time. V and Q are in phase (V = Q/C), but i and V are not in phase. Also note that the current is shifted one quarter of a period (90°) to the left relative to V, i.e., the capacitor current leads the capacitor voltage by 90° (ICE).

Inductors: An inductor consists of many turns of tightly wrapped wire, symbolically indicated by:

–11>0015oooo”L

Its DC resistance is ideally zero, and its inductive reactance (AC analog to resistance) is frequency dependent. The voltage across the inductor is proportional to the change in inductor current with time, or V = L(di/dt) where Lis called the inductance and has units of henrys:

1 volt = 1 henry x (ampere/sec)

V, i, and di/dt are sketched in Figure 3, where i (the inductor current) is assumed to be i0sin(2nft) . We note that i and V are not in phase and that i lags V by one-quarter of a period or 90° (ELI). Table 1 summarizes the above discussions.

At this point it would be useful to write down the expressions for capacitive reactance and inductive rcactance as below.

Capacitive reactance (in unit of ohm): Xe= 1/(2,ifC)

Inductive reactance (in unit of ohm): Xi= 2,rjL

178

a Resistor Capacitor Inductor

Si n e Wave

J\;- lkH z

Sine Wave

J\;- lkHz

0

TABLE 1

Voci V=Ri i & Vin phase

VocQ V=Q/C i leads V by 90°

V oc di/dt V = -Ldi/dt i lags V by 90°

All three circuit components have the common feature that they impede the flow of electric charge or electric current. The ratio of V 0/i0 is the resistance of the resistor. The ratio of the amplitudes V 0/i0 is the reactance of the capacitor or inductor. Note that since we are using AC we are using amplitude ratios here.

In the activities 1, 2, and 3 below, you will investigate the frequency dependence of resistance (resistor) and reactance (capacitor and inductor). In Activity 4 you will investigate how the phase and frequency characteristics of an inductor and a capacitor and a resistor in series connected circuits.

The circuits to be used in Activities 1, 2, and 3 are shown in Figure 4 . When you make connections, leave a slightly large room between points A and C, so you could re-use most of the circuits except that you have to inset different parts between Points A and C for those three different circuits.

A

lk o hm C

R2

Rl 1 00 o hm

B

Figure 4 (a)

A 0 . 0luF C

Capacitor

Rl 100 o hm

B

Figure 4 (b)

179

Sine Wave

f\_r lkHz

A 1.0 H C

Inductor

Rl 100 ohm

B

Figure 4 (c)

Activity 7-1: The Resistor in an AC Circuit

Connect the circuit in Figure 4(a). Use BNC to banana adaptor to connect function generator output to the breadboard posts with red banana lead going to read post at point A, and black lead going to black post at point B. Use BNC to clips adaptor to bring the signal from the circuit board to the oscilloscope. Connect BNC to oscilloscope channel 1 with the other end of red clip going to point C, and black clip going to point B. Use another BNC to clips for channel 2 with its red clip going to point A and its black clip going to point B. Set the function generator frequency to 500 Hz (using kHz x 1 setting), choose sine waveform and set the output amplitude to about 5V (Peak-to-peak is 10 Von the oscilloscope. See Figure 5). Obtain the wave- form of the voltage across the 100 ohm resistor on channel 1 and voltage across the lk ohm resistor on channel 2 on the oscilloscope. Please note that we are monitoring the current in the circuit by measuring the voltage across the 100 ohm resistor V, since voltage and current are in phase for a resistor and i0 = V1/R, = Vi/100 ohm, where V, is the peak-to-peak voltage read from channel 1 on oscilloscope.

The recommended setting for your oscilloscope should be similar to these: 1. Channel 2 inputs set to AC, scale to 2 Volts/Division, while Channel 1

input set to AC with scale of 0.2 Volts/Division; 2. Vertical Mode on DUAL, ALT mode and channel 2 INV is un-pressed; 3. Timing scale set to 0.2 ms/DIV to start with; 4. Trigger mode set to NORM, and trigger source set to Channel 2(CH2);

Trigger level at medium position

(a)First observe the voltage wave-forms from channel l and channel 2. Are two wave-fonns in phase with each other?

180

•

V2(V)

V1(V)

0 io (mA)

(b) Change the frequency of the oscillator from 500 Hz to 5,000 Hz in 500 Hz steps. Change the oscilloscope Timing Scale accordingly for best observation results. Record your data in the table below. If you cannot set the frequencies to the exact values as indicated, write down the frequency in parenthesis. Do not change the function generator output amplitude. Convert voltage from channel 1 to current using i0 = Vi/R1 = Vi/100 ohm.

Frequency (Hz) V2(V) V1 (V) io(mA) 500( ) 1000( ) 1500( ) 2000( ) 2500( ) 3000( ) 3500( ) 4000( ) 4500( ) 5000( )

Does the resistance depend on the frequency of the voltage?

(c) Now fix the function generator frequency to 1000 Hz and vary the output amplitude from 1 V to 5V in 0.5 V steps. Read voltage V I from channel 1 and V2 from channel 2 for each step. When you read the voltage, it is easier to read the peak-to-peak voltage as shown below. Convert voltage from channel 1 to current using io = V i/R1 = V ,/100 ohm. Plot V 2 vs. i0. Is Ohm’s law satisfied?

Figure 5

181

•

Activity 7-2: The Capacitor in an AC Circuit

Replace the lk oh resistor in the previous circuit with a 0.01 µF capacitor and the rest remains the same. Now the circuit should be like Figure 4(b ).

(a)First observe the voltage wave-forms from channel 1 and channel 2. Are two wave-forms in phase with each other? If not, describe the situation.

(b) Set the function generator output to about 5V. Change the frequency of the oscillator from 500 Hz to 5,000 Hz in 500 Hz steps. Record data from the RC circuit in this table. Convert voltage from channel 1 to current using io = Vi/R1 = Vi/100 ohm.

Frequency (Hz) V1 (V) io(mA) 500( ) 1000( ) 1500( ) 2000( ) 2500( ) 3000( ) 3500( ) 4000( ) 4500( ) 5000( )

Does the current i0 depend on the frequency of the voltage? How does the current i0 change as the frequency increase?

(c) Plot the current vs. the frequency. Since we keep the function generator voltage amplitude unchanged, the current measured is essentially the reciprocal of the capacitive reactance. What can you say about the relation between the capacitive reactance vs. frequency?

182

r

•

0

Activity 7-3: The Inductor in an AC Circuit

Replace the 0.01 µF capacitor in the previous circuit with a 1.0 Henry inductor and the rest remains the same. Now the circuits should be like Figure 4(c).

(a)First observe the voltage wave-forms from channel 1 and channel 2. Are two wave-forms in phase with each other? If not, describe the situation.

(b) Set the function generator output to about 5 V and leave it unchanged. Change the frequency of the oscillator from 500 Hz to 5,000 Hz in 500 Hz steps. Record data from the RL circuit in this table. Convert voltage from channel 1 to current using i0 = V i/R 1 = Vi/100 ohm.

Frequency (Hz) V,(V) io(mA) 500( ) 1000( ) 1500( ) 2000( ) 2500( ) 3000( ) 3500( ) 4000( ) 4500( ) 5000( )

Does the current depend on the frequency of the voltage? How does the current change as the frequency increase?

(c) Since we keep the function generator voltage amplitude unchanged, the current measured is essentially the reciprocal of the inductive reactance. In another word, (I/ i0 ) is propo11ional to the inductive reactance. Now plot (I/ io) vs. frequency. What can you say about the relation between the inductive reactance vs. frequency?

183

Activity 7-4: RLC in series in an AC circuit

Connect the RLC circuit as below. Note that we use I 000 ohm resistor in this circuit.

A

0 .0 l uF 1. 0 H C

Sine Wave Capacitor I nductor

~ Rl 1 000 ohm l kHz

B

From Activities 2 and 3, we already observed that phase shifts phenomenon in capacitive and inductive devices. In this activity, we will study the phase shift in more details. We also study the resonance phenomenon. Before we do both, we write down several formulas that we learn in AC circuit.

The total impedance Z of this circuit:

Where XL and Xe are in the same as before:

Capacitive reactance: Xe= 1/(2,ifC)

Inductive reactance: XL = 2 7ifL

. V The current in the circuit is then: z=-z Where the V is the voltage read from channel 2.

The phase between the current and the voltage is:

X -X tan¢= L c

R

Since the XL and Xe go in different direction as frequency changes, there is a certain frequency at which XL = Xe. The frequency is called resonance frequency and is given by:

/’= 1 2JrAf

184

•

At the resonance frequency, the impednace Z is minimum, therefore the current is maximum for a given voltage. Also at resonance frequency, the current and voltage are in phase again. At other frequencies, this phase shift can be positiive or negative.

(a) Set the function generator output to about 5V and leave it unchanged. Change the frequency from 500 Hz to 5000 Hz in 500 Hz steps. Record V2 and V1 for each step. Record data from the RLC circuit in this table. Convert voltage from channel 1 to current using i0 = Vi/R1 = Vi/100 ohm.

Frequency (Hz) V2(V) V1 (V) io(mA) 500( ) 1000( ) 1500( ) 2000( ) 2500( ) 3000( ) 3500( ) 4000( ) 4500( ) 5000( )

Plot current vs. frequency to see how current varies as frequency changes.

Now set frequency =2000 Hz and observe the phase <j> between V2 and V1 for this frequency. To calculate the phase q>, refer to the following graph and follow the procedure.

s ~ D ~ ——-

185

The separation between peaks P and Qare the period, which represents 360° in phase. However, this separation can also be measured as the distance D. On the oscilloscope, one large division is 1.0 cm, and one small division is 2 mm. Record the Din unit ofmm. Now consider the phase difference between channel 1 and channel 2. The phase difference is represented by distance between peaks Sand P. Record the distanced between peaks Sand P. The phase difference then casn be calculated as <I> = (d/O)*360°. The peak Smay be in front of P or may be following P. It always is the closest peak near P. Find this measured phase for frequency = 2000 Hz.

Now use the formulas on page (7-8) to calculate the theoretical phase and compare your results with measured phase. How is your calculation compard with what you just measured?.

Finally let us find the resonance frequency . Slowly change the frequency on the function generator and do not change the output amplitude. Watch the channel 1 for maximum voltage, which means maximum current in the circuit. Fine tune the frequency until you find the maximum channel 1 voltage (maximum current). Record the resonance frequency and compare your result with the theoretical vale on page (7-8). Also verify that, at this moment, channel 1 and channel 2 are in phase with each, since according to our prediction, the phase difference <I> should be zero. Does this resonance frequency agree with the current vs. frequency plot?

186

•