CIRCUIT ANALYSIS-II
LAB REPORT 02
S.NO CONTENT MARKS
1. objective
2. equipments
3. introduction
4. procedure
5. observations
6. readings
7. Conclusion
OBJECTIVE:
1. Examine the behavior of a inductor when a sinusoidal source voltage is applied.
2. Describe the relationship between current and voltage in an RL circuit.
3. Understand the phase shift and its measurement.
4. Understand the inductive reactance and its measurement.
EQUIPMENTS:
1. Function generator
2. Oscilloscope
3. Digital multimeter (DMM)
4. breadboard
5. jumpers
6. probes
INTRODUCTION:
An RL circuit is one containing a resistor R and a inductor.
Consider a simple RL circuit in which resistor, R and inductor, L are connected in series with a
voltage supply of V volts. Let us think the current flowing in the circuit is I (amp) and current
through resistor and inductor is IR and IL respectively. Since both resistance and inductor are
connected in series, so the current in both the elements and the circuit remains the same. i.e IR
= IL = I. Let VR and Vl be the voltage drop across resistor and inductor.
Phasor Diagram for RL Circuit
Before drawing the phasor diagram of series RL circuit, one should know the relationship
between voltage and current in case of resistor and inductor.
Resistor
In case of resistor, the voltage and the current are in same phase or we can say that the phase
angle difference between voltage and current is zero.
Inductor
In inductor, the voltage and the current are not in phase. The voltage leads that of current by
90o or in other words, voltage attains its maximum and zero value 90o before the current
attains it.
RL Circuit
In case of series RL circuit, resistor and inductor are connected in series, so current flowing in
both the elements are same i.e IR = IL = I. So, take current phasor as reference and draw it on
horizontal axis as shown in diagram.
In case of resistor, both voltage and current are in same phase. So draw the voltage phasor, VR
along same axis or direction as that of current phasor. i.e VR is in phase with I.
We know that in inductor, voltage leads current by 90o, so draw VL (voltage drop across
inductor) perpendicular to current phasor.
Now we have two voltages VR and VL. Draw the resultant vector(VG) of these two voltages.
Such as, and from right angle triangle we get, phase angle
PROCEDURE:
1. First, let’s select component values for testing. We choose an inductor value somewhere
around 100µH. We selected a frequency between 1.11kHz.first three readings we took a
resister of 1MΩ . But for the last reading we calculate XL and choose
a value for R ≈ 1Ω.
2. Configure the circuit for testing.Insert one of the multimeters
to record the AC current.Voltage supply is given by power generator.
3. Let’s measure the phase shift between the current and applied voltage.First make sure the
the oscilloscope is displaying the correct waveform by switching off the one and then the other.
Make sure the two signal baselines are centered with respect to the horizontal and vertical axes
of the oscilloscope,and adjust the voltage and time scales so that slightly more than one cycle
ofboth waveforms is visible.
4. Now note down the frequency , T and ΔT .
5. Find the time “T” period of the input signal.Find a peak on the channel one sine wave. Follow
that point down to the horizontal time-division marks in the oscilloscope’s display’s center.
Then find the corresponding peak on channel two and its time-division mark. Count the major
divisions between the time-division marks for channel one and two (including any fractions of a
division) it is ΔT.
6. Using the following relation find the phase angle
OBSERVATIONS:
When inductors involved in an AC circuit, the current and voltage do not peak at the same time.
The fraction of a period difference between the peaks expressed in degrees is said to be the
phase difference. The phase difference is <= 90 degrees. It is customary to use the angle by
which the voltage leads the current. This leads to a positive phase for inductive circuits since
current lags the voltage in an inductive circuit. The phase is negative for a capacitive circuit
since the current leads the voltage.
READINGS:
S NO. Time Period of input Frequency ΔT Phase angle
signal
1 0.960ms 1.11 hz 0.16ms 90⁰
2 0.560ms 1.78hz 0.120ms 77.14⁰
3 0.720ms 1.4217hz 0.160ms 80⁰
4 0.00909µs 1.1khz 0.04016ms 15.9⁰
CONCLUSION:
In case of pure resistive circuit, the phase angle between voltage and current is zero and in case of pure
inductive circuit, phase angle is 90⁰ but when we combine both resistance and inductor, the phase angle
of a series RL circuit is between 0 to 90 degree.
