3 Parallel RL Circuits
Thursday, 12 July 2018
8:11 AM
The circuit phasor diagram for a parallel circuit is drawn with current phasors, since voltage is the same across all components. The direction of the current phasor in a resistor is always in the direction of the voltage. Since current lags the voltage in an inductor, the current phasor is drawn at an angle of -90º from the voltage reference. Figure 7-6 shows a parallel  RL circuit with current magnitudes labelled, as measured by a DMM.
Figure 7-6
The current phasor diagram for this circuit is shown below:
Figure 7-7
In parallel circuits, the phase angle of interest is usually between the total current, IS, and one of the branch currents. To use the DSO to measure the phase angle in a parallel circuit, we must convert the current to a voltage. If a branch contains a resistor, then the voltage across the resistor is directly proportional to the current (Ohm's Law) and we can simply scale the voltage waveform to obtain the current waveform. In branches which do not contain resistors, we can insert a "current-sense" resistor and measure the voltage across it - but the resistor must not to have a major effect on the circuit.
Parallel RL Circuits
From your kit, retrieve the components given in Table 7-5 below. To measure the total current in the circuit, we will be using a
1 kW "current-sense" resistor in series with the supply voltage and a 220 W "current-sense" resistor in series with the inductor. You will not be able to measure the inductance with the DMM - only measure and record the value of the resistors in Table 7-5.
Table 7-5
Component

Listed Value
Measured Value
L1
4.7 mH
R1
5.6 kΩ
RS1
1.0 kΩ
RS2
220 Ω
Construct the parallel RL circuit shown in Figure 7-8. Set the FG for a 200 kHz sine wave at 6.0 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 7-8
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 RS1.
Measure the peak-to-peak voltages across R1 (Ch 2), RS1 (Math)and RS2 (move Ch 2). Record the voltage readings in Table 7-6.
Table 7-6
Branch -->   
Resistor
R1
(Ch 2)
Inductor
RS2
(Ch 2)
Source
RS1
(Math)
Voltage
Step 4
VR1 =
VRS2 =
VRS1 =
Current
Step 5
IR =
IL =
IS =
Phase Angle
Between
Step 7
Compute the peak-to-peak currents in the branches of the circuit by applying Ohm's Law to the measured values of the resistors:
𝐼=𝑉𝑅
Enter the computed currents in Table 7-6.
The currents measured indirectly in step 5 are phasors - all we have measured is their magnitudes. Assume that the effect of the sense resistor RS2 can be ignored - then the current in the inductor should lag the current in R1 by 90º (because they have the same voltage across them). Using the computed peak-to-peak currents from Table 7-6, draw the current phasors for the circuit on Plot 7-5.
Plot 7-5
The phasor diagram illustrates the relationship between the total current and the current in each branch. Using the measured currents, compute the phase angle between the total current (IS) and the current in R1 (IR). Then compute the phase angle between the total current (IS) and the current in L1 (IL). Enter the computed phase angles in Table 7-7.
Table 7-7
Phase Angle
Between  
Computed
Step 7
Measured
Step 8
IS and IR 
IR and IL
IS and IL
Measure the phase difference between the 3 branch currents. Enter the measured phase angles in Table 7-7.
Hint: You will need to get each voltage that is directly proportional to the desired current waveform on the DSO screen to do the measurements - this will require moving the DSO probes to different nodes of the circuit, and for some measurements, using the Math waveform (Channel 1 - Channel 2).