In this lab, we calculated the theoretical phase and gain response of an RL circuit and compared the results to actual measurements taken from a passive RL circuit. The circuit used was very simple and only contains a 47 Ohm resistor in series with a 1 mH inductor.
The cutoff frequency was given as R/L, and we were instructed to perform our calculations and measurements of the circuit for the cutoff frequency, 10 * cutoff frequency, and 1/10 * cutoff frequency. The actual resistance of the 47 Ohm resistor was 46.9 Ohms, and the true inductance of the inductor was 0.963 mH. Using these values, we calculated the cutoff frequency to be w = 48182.8 rad/s, and thus a frequency of f = 7669 Hz. The actual circuit used for this experiment is pictured below.
First, we used the cutoff frequency as our voltage input, with an amplitude of 2V. White board calculations and measurements from the actual circuit are pictured below.
The impedance of the inductor was calculated as jwL = 46.4j Ohms. Since the resistor is 46.4 Ohms, the total impedance was calculated to be sqrt(46.4^2 + 46.4^2) = 65.6 Ohms, with a phase shift of atan(1) = 45°, thus Z_tot = 65.6 Ohms, 45°. The current in the circuit can be calculated as V/Z, which is 2, 0°/65.6, 45° = 30.5, -45° mA. The voltage across the inductor was calculated by multiplying the current by the impedance of the inductor, which yielded V_L = 1.42, 45° V. The results obtained from measuring the actual circuit were nearly spot on to our calculations.
Oscilloscope at f = 7669 Hz (cutoff frequency)
Blue = Input Voltage
Red = Current
Oscilloscope at f = 7669 Hz (cutoff frequency)
Blue = Input Voltage
Yellow = Inductor Voltage
Red = Current
The wrong scale was used in the second set of graphs (it should have shown the yellow scale to show voltage across the inductor); however, looking closely it can be seen that the scale used for both voltages are 0.5V / division. The input voltage was 2 V, and the voltage across the inductor is slightly less than 1 division below the input voltage, thus the inductor voltage we calculated was correct at ~1.4 V. It can also be seen that the inductor voltage is out of phase with the input voltage by ~45°.
Next, we changed the input voltage frequency to be 1/10 the cutoff frequency; thus, w = 4818.3 rad/s, and f = 766.9 Hz. White board calculations as well as measurements taken from the circuit are pictured below.
Changing the input frequency affects the impedance of the inductor (by a factor of 1/10), and thus the total impedance was much smaller than the previous setup at Z_tot = 46.6, 5.71° Ohms. The current and voltage were calculated using the same method as before, yielding values of 43.0, -5.71° mA, and 0.2, 34.4°
Oscilloscope at f = 766.9 Hz (1/10 cutoff frequency)
Blue = Input Voltage
Yellow = Inductor Voltage
Red = Current
Oscilloscope at f = 766.9 Hz (1/10 cutoff frequency)
Blue = Input Voltage
Yellow = Inductor Voltage
Red = Current
The first set of graphs shows the vertical scale in terms of mA, and it can be seen that the peak current measured across the circuit was just over ~42 mA with a very slight phase shift from the input voltage, which was nearly identical to our white board calculation. The second graph shows the same image but with a scale indicating the voltage. The voltage across the inductor was measured to be ~0.25 V with a slightly lower phase shift than the first setup.
Finally, we tested the same RL circuit setup but with the input frequency at 10 * the cutoff frequency, or w = 481,828 rad/s, and f = 76,695 Hz. The white board calculations as well as measurements taken from the actual circuit are pictured below.
Since w was increased by a factor of 10, the impedance of the inductor was also increased by the same factor, making Z_L = 464j Ohms, and thus Z_tot = 466.3, 84.3° Ohms. Using the same calculations as previous setups, the current and voltage were calculated to be 4.29, -84.3° mA
Oscilloscope at f = 76695 Hz (10 * cutoff frequency)
Blue = Input Voltage
Yellow = Inductor Voltage
Red = Current
Oscilloscope at f = 76695 Hz (10 * cutoff frequency)
Blue = Input Voltage
Yellow = Inductor Voltage
Red = Current
In the first series of graphs, the vertical scale indicates the voltage across the inductor. Just as we predicted, the voltage across the inductor is nearly the same as the input voltage, with a very slight phase shift of ~5°. The second set of graphs shows the same thing but with the scale depicting the current through the circuit in mA. We predicted a current with an amplitude of 4.3 mA, and the measurements taken show this result exactly. The phase shift of the current was calculated to be -84.3° from the input voltage, which is precisely what our measurements indicate.
In Class Examples
1. Find i x in the circuit shown using nodal analysis.
2. Determine current Io in the circuit below using mesh analysis.
3. We were asked to use mesh currents again for the previous example, but with only a current source that delivers I = -2.353 + j2.353 A
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