2 Series RL Circuits
Thursday, 12 July 2018
8:11 AM
When a sine wave at some frequency drives a circuit that contains only resistors and inductors, the waveforms throughout the circuit are also sine waves at that same frequency. 
Recall the following two key concepts of phasors:
A sinusoid given by:
can be represented by the phasor:
Phasors add, subtract and multiply just like vectors. 
Figure 7-3 shows a series  RL circuit with voltage magnitudes labelled, as measured by a DMM.
Figure 7-3
Recall that we are dealing with sinusoids, and we have to perform a phasor (i.e. a vector) addition when applying KVL. The voltage phasor diagram for this circuit is shown below:
Figure 7-4
Now we can see, using phasor (i.e. vector) addition that 3∠0°+4∠90°=5∠53.1°. All the DMM has measured is the magnitudes of the sinusoids, not their phases. To see the phase relationships between the voltages, we have to use a DSO and display the sinusoidal voltages with respect to time.
The voltage across the resistor is given by Ohm's Law:
𝑉𝑅=𝑅𝐼
We saw in the previous section that for an inductor subjected to a sinusoidal excitation, we have a similar v-i relationship to Ohm's Law:
𝑉𝐿=𝑋𝐿𝐼
Recall the general relationship:
𝑉=𝑍𝐼
where Z is the impedance of a circuit element. It plays a similar role to resistance and reactance, except it has to be represented as a quantity with a magnitude and also a direction, i.e. like a vector. The voltage V and the current I in this equation are phasors, i.e. they also carry information about both magnitude and direction (i.e. amplitude and phase).
From Ohm's Law, since R is just a number (not a vector), we see that the sinusoidal current and voltage for a resistor are in phase. Therefore, in Figure 7-4, we can draw in the current phasor I in the same direction as VR.
We also saw in the previous section that current lags the voltage in an inductor by 90º. Therefore, if we assign an angle of 0º for the current (as we have done in the phasor diagram), then the inductor voltage must have an angle of +90º. Hence, we have drawn the inductor voltage phasor "pointing up" (the angle it makes with the horizontal axis is +90º).
Using KVL, the sinusoidal voltage source is the phasor sum of the voltage across the resistor and the voltage across the inductor. It has a magnitude of 5 V, and a phase angle of +53.1º. From the perspective of the voltage source, it is delivering a current that lags the voltage VS by 53.1º.
The phasor diagram illustrated by Figure 7-4 is correct at one frequency only. This is because the reactance of an inductor is frequency dependent as given by the equation:
As the frequency is increased, the reactance (XL) of the inductor increases. This changes the phase angle and voltage magnitudes across the components.
Series RL Circuits
Construct the series RL circuit shown in Figure 7-5. Set the FG for a 20 kHz sine wave at 3.0 Vpp. Connect the DSO channels as shown. Check the frequency and voltage with the DSO.
Note: Record all voltages and currents throughout this section as peak-to-peak values.
Figure 7-5
Use the Math function of the DSO to subtract Channel 2 from Channel 1, i.e. set up a Math waveform that represents the voltage across the resistor R1.
Measure the peak-to-peak voltage across the resistor (VR). Then measure the peak-to-peak voltage
across the inductor (VL ). Record the voltage readings on the first line of Table 7-4.
Table 7-4
Frequency
VR
(measured)
Step 4
VL
(measured)
Step 4
I
(computed)
Step 5
XL
(computed)
Step 6
Z
(computed)
Step 7
Measured
Phase
Angle
𝜃
Step 8
Computed
Phase
Angle
𝜃
Step 9
20 kHz
30 kHz
50 kHz
80 kHz
130 kHz
200 kHz
Compute the peak-to-peak current in the circuit by applying Ohm's Law to the measured value of the resistor:
𝐼=𝑉𝑅𝑅
Since the current is the same throughout a series circuit, this is a simple method for finding the current in both the resistor and the inductor. Enter this computed current in Table 7-4.
Compute the inductive reactance, XL. The reactance is found by dividing the voltage across the inductor (step 4) by the current in the circuit (step 5):
𝑋𝐿=𝑉𝐿𝐼
 Enter the inductive reactance in Table 7-4.
Compute the total impedance of the circuit by applying Ohm's Law to the entire circuit. Use the generator voltage set in step 2 and the current determined in step 5:
𝑍=𝑉𝑆𝐼
Enter the computed impedance in Table 7-4.
Measure the phase of VR with respect to VS . Enter the measured phase angle in Table 7-4.
Note: Use VS as the reference. Therefore, measure the phase of the Math channel (VR) with respect to Channel 1 (VS).
Visually confirm that your phase measurement is sensible.
Compute the phase angle of VR with respect to VS using the trigonometric relation:
Enter the computed phase angle in Table 7-4.
Change the frequency of the generator to 30 kHz. Repeat steps 4 through 9, entering the data in Table 7-4. Continue in this manner for each frequency listed in Table 7-4.
From the data in Table 7-4, draw the voltage phasors for the circuit at a frequency of 20 kHz on Plot 7-1.
Plot 7-1
From the data in Table 7-4, draw the voltage phasors for the circuit at a frequency of 80 kHz on Plot 7-2.
Plot 7-2
The phasor drawings reveal how the voltage phasors change with frequency. Investigate the frequency effect further by graphing both the voltage magnitude across the inductor and the voltage magnitude across the resistor as a function of frequency. Label each curve. Either use Plot 7-3 or use Excel for your graph.
Note: The horizontal frequency axis is logarithmic to the base 10, i.e. log10𝑓
Plot 7-3
Graph the phase of the voltage across the resistor (VR ), with respect to the source voltage (VS), as a function of frequency.
Either use Plot 7-4 or use Excel for your graph.
Note: The horizontal frequency axis is logarithmic to the base 10, i.e. log10𝑓
Plot 7-4