Experiement 10

• Objectives: 1. To know the behavior of the resistors, coils, & capacitors in the AC circuits. 2. To measure the reactance in indirect method for the capacitors, & inductors. 3. To be familiar with the concept of the reactance of the coil & the capacitor. 4. To measure the phase angle in the coil & the capacitors. • Equipment: 1. 2. 3. 4. 5. 6. 7.

Bread board. Computer software PSpice. Oscilloscope. Function generator. Wires. Digital multimeter. Resistors, capacitors, inductors.

• Contents: 1. The resistor in an AC circuit. 2. The coil in an AC circuit. 3. The capacitor in an AC circuit. • Theory:  The behavior of the resistor which converts electrical energy

into heat is independent of the direction of current flow and independent of the frequency. So we see that the AC impedance of a resistor is the same as its DC resistance.  The reactance of the capacitor which store electrical energy

is dependent on the frequency, so we see that the impedance of the capacitor is Хc = 1 /2π fC. Where: ХL: The reactance of the capacitor, in ohms. f: The frequency, in Hz. C: The capacitance, in farads.

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 The reactance of the inductor is dependent on the frequency,

so we see that the impedance of the inductor is ХL = 2π f L. Where: ХL: The reactance of the inductor, in ohms. f: The frequency, in Hz. L: The inductance of the inductor, in henries.

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• Calculation & data analysis: • The resistor in an AC circuit.

Frequency (Hz)

We connect the circuit shown in figure 10.1 & we set the generator to generate a sinusoidal wave with effective value measured by the DMM equals 5v & various frequencies & fill the table 10.1 Measured quantity VA-C measure d (v)

200 4.98 400 4.95 600 4.97 800 4.93 1000 4.98 Table 10.1

VB-C measure d (v)

VA-B calculate d (v)

IR1 calculate d (mA)

IR2 calculate d (mA)

IR3 calculated (mA)

1.62 1.60 1.57 1.67 1.64

3.36 3.35 3.36 3.30 3.34

3.36 3.35 3.36 3.30 3.34

1.62 1.60 1.57 1.67 1.64

1.62 1.60 1.57 1.67 1.64

VA-B = VA-C - VB-C VA-B = 4.98 – 1.62 = 3.36 v. VA-B = 4.95 – 1.60 = 3.35 v. VA-B = 4.97 – 1.57 = 3.36 v. VA-B = 4.93 – 1.67 = 3.30 v. VA-B = 4.98 – 1.64 = 3.34 v.

IR1 = VA-B/R1 IR1 = 3.36/1000 = 3.36 mA. IR1 = 3.35/1000 = 3.35 mA. IR1 = 3.36/1000 = 3.36 mA. IR1 = 3.30/1000 = 3.30 mA. IR1 = 3.34/1000 = 3.34 mA.

IR2 = IR3 = VB-C/R IR2 = 1.62/1000 = 1.62 mA. IR2 = 1.60/1000 = 1.60 mA. IR2 = 1.57/1000 = 1.57 mA. IR2 = 1.67/1000 = 1.67 mA. IR2 = 1.64/1000 = 1.64 mA.

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•

The coil in an AC circuit.

We connect the circuit shown in figure 10.2 & we set the generator to generate a sinusoidal wave with peak value measured by the oscilloscope equals 3v & various frequencies & fill the table 10.2 Measured quantity Reactance Frequency Measured Calculated Calculated according to (kHz) voltage across current reactance Measurement Resistor (mV) (mA) (Ω) s (Ω) 1.6 1050 10.5 102 100.5 3.2 735 7.35 188 201 4.8 530 5.3 300 301.6 6.4 410 4.1 417 402 8 330 3.3 542 502.6 9.6 310 3.1 584 603 11.2 260 2.6 716 703.7 12.8 230 2.3 822 804 14.4 200 2 960 904.8 16 180 1.8 1078 1005 Table 10.2 Φ = 90o = sin-1(b/a) = sin -1 1 = 90o at f = 16 KHz ХL = 2π f L ХL = 2π *1600*0.01 = 100 Ω. ХL = 2π *3200*0.01 = 201 Ω. ХL = 2π *4800*0.01 = 301 Ω. ХL = 2π *6400*0.01 = 402 Ω. ХL = 2π *8000*0.01 = 502 Ω. ХL = 2π *9.600*0.01 = 603 Ω. ХL = 2π *11200*0.01 = 703 Ω. ХL = 2π *12800*0.01 = 804 Ω. ХL = 2π *14400*0.01 = 905 Ω. ХL = 2π *16000*0.01 = 1005 Ω. I = VR/R I = 1050/100 = 10.5 mA. I = 735/100 = 7.35 mA. I = 530/100 = 5.3 mA. I = 410/100 = 4.1 mA. I = 330/100 = 3.3 mA. I = 310/100 = 3.1 mA. I = 260/100 = 2.6 mA. I = 230/100 = 2.3 mA. I = 200/100 = 2.0 mA. I = 180/100 = 1.8 mA.

ХL = (V-VR)/I ХL = (2.12-1.05)/(10.5*10^-3) = 102 Ω. ХL = (2.12-0.735)/(7.35*10^-3) = 188 Ω. ХL = (2.12-0.530)/(5.3*10^-3) = 300 Ω. ХL = (2.12-0.410)/(4.1*10^-3) = 417 Ω. ХL = (2.12-0.330)/(3.3*10^-3) = 542Ω. ХL = (2.12-0.310)/(3.1*10^-3) = 584 Ω. ХL = (2.12-0.260)/(2.6*10^-3) = 716 Ω. ХL = (2.12-0.230)/(2.3*10^-3) = 822 Ω. ХL = (2.12-0.200)/(2.0*10^-3) = 960 Ω. ХL = (2.12-0.180)/(1.8*10^-3) = 1078 Ω. 4

•

The capacitor in an AC circuit.

We connect the circuit shown in figure 10.3 & we set the generator to generate a sinusoidal wave with peak value measured by the oscilloscope equals 3v & various frequencies & fill the table 10.3 Measured quantity Reactance Frequency Measured Calculated Calculated according to (Hz) voltage across current reactance Measurement Resistor (mV) (mA) (kΩ) s (kΩ) 1600 185 1.850 1.05 1 800 105 1.050 1.90 2 530 70 0.700 2.93 3 400 52.2 0.522 3.96 4 320 40 0.400 5.20 5 265 35.5 0.355 5.87 6 225 31 0.310 6.75 7 200 25 0.250 8.38 8 180 23.5 0.235 8.92 8.8 160 22 0.220 9.50 10 Table 10.3 Φ = 90o = sin-1(b/a) = sin -1 1 = 90o at f = 160Hz Хc = 1 /2π fC Хc = 1 /(2π *1600*0.1*10^-6) = 1 kΩ Хc = 1 /(2π *800*0.1*10^-6) = 2 kΩ Хc = 1 /(2π *530*0.1*10^-6) = 3 kΩ Хc = 1 /(2π *400*0.1*10^-6) = 4 kΩ Хc = 1 /(2π *320*0.1*10^-6) = 5 kΩ Хc = 1 /(2π *265*0.1*10^-6) = 6 kΩ Хc = 1 /(2π *225*0.1*10^-6) = 7 kΩ Хc = 1 /(2π *200*0.1*10^-6) = 8 kΩ Хc = 1 /(2π *180*0.1*10^-6) = 8.8 kΩ Хc = 1 /(2π *160*0.1*10^-6) = 10 kΩ I = VR/R I = 185/100 = 1.85 mA. I = 105/100 = 1.05 mA. I = 70/100 = 0.70 mA. I = 52.2/100 = 0.522 mA. I = 40/100 = 0.40 mA. I = 35.5/100 = 0.355 mA. I = 31/100 = 0.31 mA. I = 25/100 = 0.25 mA. I = 23.5/100 = 0.235 mA. I = 22/100 = 0.22 mA.

Хc = (V-VR)/I Хc = (2.12-0.185)/(1.85*10^-3) = 102 Ω. Хc = (2.12-0.105)/(1.05*10^-3) = 188 Ω. Хc = (2.12-0.070)/(0.70*10^-3) = 300 Ω. Хc = (2.12-0.0522)/(0.52*10^-3) = 417 Ω. Хc = (2.12-0.040)/(0.40*10^-3) = 542Ω. Хc = (2.12-0.0355)/(0.35*10^-3) = 584 Ω. Хc = (2.12-0.031)/(0.31*10^-3) = 716 Ω. Хc = (2.12-0.025)/(0.25*10^-3) = 822 Ω. Хc = (2.12-0.0235)/(0.23*10^-3) = 960 Ω. Хc = (2.12-0.022)/(0.22*10^-3) = 1078 Ω. 5

• Conclusions: 1. we see that the reactance of the capacitor goes to infinity in DC circuit & it can represented by open circuit, & by short circuit at infinite frequency. 2. we see that the reactance of the coil goes to zero in DC circuit & it can represented by short circuit, & by open circuit at infinite frequency 3. We see that the behavior of the resistor, which converts electrical energy into heat, is independent of the direction of current flow and independent of the frequency. So we see that the AC impedance of a resistor is the same as its DC resistance. 4. we see that the dissipated power to the capacitor & inductor is zero because the load phase is 90o & cos 90 is zero.

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• circuit connection:

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XL (Ω)

reactance Vs frequency for the coil 1200 1000 800 600 400 200 0 0

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10

15

20

f KHz

reactance Vs frequency for the capacitor

Xc (KΩ)

10 8 6 4 2 0 0

500

1000

1500

f Hz

8

2000