Name: _______________________
Partner(s): ____________________
Desk #: ______________________
Date: ________________________
The RL Series AC Circuit
Purpose
•
To investigate the effects produced by an AC voltage applied to an RL series circuit.
Introduction and Theory
Consider the circuit in Figure 1a. If we connect an inductor to a sinusoidal voltage source, the
current will lag the applied voltage by 90°*. The magnitude of the current depends on the
frequency of the AC source and the inductance.
I
V = Vmax sin ωt
V
L
I
Figure 1a
Figure 1b
Figure 1. Voltage and current of an inductor connected with an AC power source
Basic rules for an inductor in AC circuit
•
I lags V by 90°;
•
Imax =
Vmax
where X L = ωL = 2πfL is called inductive reactance.
XL
Again, it is impossible for us to build an inductive circuit without resistance, especially since
inductors are typically made out of a long piece of wire wind up in a coil with non-zero
resistance. We will investigate RL series circuit.
*
The phrase “ELI the ICE man” is a handy way to remember the relations between the voltages
of RC and RL circuits. The current is I and the voltage is E. For inductance L, current I lags E by
90° so “ELI”, and for capacitance C, current leads E by 90° so “ICE”.
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Phasor Diagram
V
R
V = Vmax sin ωt
I
VL
L
φ
ωt
VR
Figure 2b
Figure 2a
Figure 2. Phasor diagram for a RL series AC circuit
Consider the circuit in Figure 2a. A resistor and an inductor are in series with a sinusoidal
voltage source of amplitude Vmax and angular frequencyω. The phasor diagram is shown in Fig.
2b.
Similar to our last lab, we have:
V (t ) = VR (t ) + VL (t ) or Vmax sin ωt = VR max sin(ωt − φ ) + VL max sin(ωt − φ + 90 o )
From the phasor diagram, we know
Vmax = VR max + VL max
2
Because Vrms =
2
1
2
2
Vmax , it follows that Vrms = VRrms + VLrms .
2
We also get from the phasor diagram that
tan φ =
VL max
VR max
Because VR max = RImax and VL max = X LImax , we have tan φ =
XL
R
The impedance of the whole circuit will depend on both the inductor and the resistor:
VR max + VL max
V
Z = max =
Imax
Imax
2
2
(RImax )2 + ( X LImax )2
2
=
= R 2 + X L = R 2 + (2πfL )2
Imax
We will test these relations in our lab. The uncertainties of calculated results are not required.
But you must record the uncertainties for all measured values and give proper significant digits
for all results, whether they are from direct measurements or from calculations.
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Saved: 2/16/16, printed: 2/16/16
Apparatus
BK Precision 2120 dual-trace oscilloscope, BK Precision 4011 function generator, two Fluke
73III multimeters, BK Precision 875B LCR meter, decade resistance box, 25 mH, and 4 H
inductors.
Record the following:
•
Actual Inductance (and uncertainty) of the 25 mH inductor:
•
Resistance (and uncertainty) of the 25 mH inductor:
•
Actual Inductance (and uncertainty) of the 4 H inductor:
•
Internal resistance of the Fluke multimeter through 400 mA ammeter:
Data
Connect the circuit shown in Figure 2a, using the decade box as the resistor and the 4 H
inductor.
Part A: Testing Vmax = VR max + VL max LLLLLLLL(1)
2
2
Set the output of the function generator to sine wave with f = 1 kHz and Vmax = 2.0 V (using the
oscilloscope). Set the decade resistance box to 50 kΩ.
Measure Vmax, VRmax and VLmax using the multimeter. Then set R = 10 kΩ and repeat the
measurements. In the space below, list the results with a table of these columns: f, R, L, VRmax,
VLmax and Vmax. Verify Eq. (1) for each case.
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Part B: Testing the impedance Z = R 2 + (2πfL )2 LLLLLLLL(2)
Remove the decade resistor box and replace the 4 H inductor with the 25 mH inductor or the
2.5 mH inductor. R is now the internal resistance of the 25 mH inductor. Set the signal
generator output to f = l00 Hz and Vrms = 2.0 V (using the multimeter).
Vary f from 200 Hz to 2000 Hz in 100 Hz increments. For each frequency, determine the
impedance of the circuit by measuring Z = Vrms I rms with the multimeters. Record f, Vrms, Irms and
Z in a table below.
With Excel, graph Z 2 vs. f 2 . If Eq. (2) is valid, your graph should be linear. Sketch the graph and
state whether it supports Eq. (2). From the graph, calculate the values of R and L and compare
them to the values given by the multimeter.
Explain why we cannot test the inductive reactance X L = 2πfL directly like we did for the
capacitive reactance for the RC circuit?
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Part C: The phase angle between the current and the applied voltage: tan φ =
XL
LLL(3)
R
Connect the decade box back and set R = 5 kΩ. Use the 25 mH inductor and set the signal
generator to f = 10 kHz and Vmax = 2.0 V.
Measure φ using first the peak-shift technique and then the Lissajous technique (note:
remember that with RL circuits, the voltage leads the current). Repeat the measurements for
f = 10 kHz, 30 kHz and 60 kHz. Record your results.
Compare the measured φ with the values obtained by Eq. (3), using X L = 2πfL . Remember to
use the total resistance.
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Saved: 2/16/16, printed: 2/16/16