ECE2120

Electrical Engineering Laboratory II

Lab 4: Inductors and Series RL Circuits

Name:

Date:
Objective
The objective of the lab is to

 Understand the series and parallel combination of the inductor and

measurement.
 Calculate the reactance of the inductor at given frequency.
 Observe and verify the behavior of series RL circuit.

Equipment Used
1. Analog Discovery 2
2. Resistor
3. Inductor
4. DMM
5. Breadboard

Part (0)
Calculate theoretical voltage and current values in Figure 3.3 and record
them in Table 4.1. Calculate all voltage and current values as peak-to-peak.

Table 4.1 Calculated Values of RL Circuit

Freq XL ZTotal i 1 (p−¿− p) V R ( p−¿− p) V L ( p−¿− p)

(Hz) (Ω) R(Ω) X (Ω) mag(mA ) angle mag (V ) angle mag(V ) angle

1000 314
5k 3.14 k 0.6775 −32.13 3.387 −32.13 2.13 57.87
0 0

Calculations

Following are the calculations performed for the above table.

Experimental Work

Part (1)

Use the digital multimeter (DMM) to complete the following steps. Record
your values in Table 3.2.

In this part of experiment, nominal values of the capacitor and resistor are
noted as mentioned on the component and color code. Using DMM, value of each
component is measured and noted in the Table 3.2.

Table 4.2 Calculated and Measured Values

Component Nominal∨Calculated Value Measured Value

R 5 kΩ 4.96 kΩ
L 50 mH 48 mH
RW 0 13 0 Ω
Form the above table, it can be observed that the measured values are different
form the nominal values because of the percentage tolerance of the component.

Part (2)

In this part of the experiment, RL series circuit was analyzed. Series RL circuit shown
in Fig. 1 is implemented on the breadboard. Input signal is applied using Wavegen of
Analog Discovery 2, CH1 is connected across the input. To measure voltage across
resistor, CH2 is connected across resistor, and the current is calculated using
following formula.

VR
I p− p=
R

Figure 1 Series RL Circuit

To measure the phase angle, cursor is used to measure the time difference in the
zero crossing of wave, and phase angle is determined using following formula.

∆T
hase angle= ×360o
Total Time Period of wave

Voltage across inductor is measured by connecting CH2 across capacitor, and

reactance is determined using the following formula.

VL
X L=
I p− p
 Table 4.3 Recorded Values for Part 2

Vs VL VR I PP φ XL ZTotal φ
Freq
(V ) (V ) (V ) (mA ) Meas (k Ω) (kΩ) Cal
10000 4 1.9 3 3. 4 0.68 5 −2 9 . 2 2.8 1 4.96 + j 2.8 1 29. 58

From the above results mentioned in Table 4.3, it can be observed that there is a
negligible difference in the values of Table 4.1 and 4.3, the difference is due to the
difference observed in Table 4.2.

1. Using the values listed in the table, draw a diagram of the impedance phasors
and a second diagram of the voltage phasors, as illustrated in Figure 4.2 of
the Background section of this laboratory experiment.

Figure 2 Impedance and voltage phasor diagram

2. Using the data in the table, convert the current I and source voltage VS to
RMS values. Then draw plots of the power phasor diagrams (power triangle)
at the frequency of 10 kHz. Determine the real power, the reactive power,
and the apparent power in the RL circuit at that frequency.

For conversion of peak to peak values to RMS using the following formula:

V RMS= ( )
V pp
2
4
×0.707= ×0.707=1.414 V
2

I RMS= ( )
I pp
2
×0.707=
0.68 5
2
× 0.707=0.24 2mA

S RMS=V RMS I RMS =1.414 ×0.24 2 m∠2 9.2=0.3 42 mVA ∠ 2 9 .2

 Figure 3 Power phasor diagram

Conclusion

From this experiment we concluded that, measured values are different from the
calculated due to the not availability of precise components i-e resistor and inductor
of exact value. From the tables obtained after calculations and measurement it is
observed that the reactance of the inductor decreases with the increase in the
frequency. And hence due to increase in the reactance the current in the circuit
decreases.

Probing Further Questions

What you think would happen to the current in this RL series circuit if the
frequency were decreased, say, to 2000 Hz? Why?
When the frequency of the input signal is increased, the value of reactance is
increased and resultantly the total impedance is also increased. So, due to the
increase in impedance, circuit current is decreased.

Simulate the circuit in LT SPICE and graph VL and VR versus frequency from
1000 Hz to 20,000Hz. Does the simulation support your prediction in question
1?
Figure 4 VR vs Freq

Figure 5 VL vs Freq
 Appendix

Reference figures.

Figure 6 Circuit implementation

Figure 7 output and input