FACULTY OF ENGINEERING
MECHANICAL AND ENERGY DEPARTMENT
Fundamentals Of Electrical Engineering (EEEN211)
EXPERIMENT 12 LAB REPORT;IMPED
NAME:TSAMODIMO MMUTLE
STUDENT ID:21000394
DATE OF SUBMISSION:07/03/2023
Page | 1
Contents
ABSTRACT...................................................................................................................................................... 3
INTRODUCTION ............................................................................................................................................. 4
OBJECTIVES/AIMS ......................................................................................................................................... 5
THEORY ......................................................................................................................................................... 6
APPARATUS ................................................................................................................................................. 10
PROCEDURE ................................................................................................................................................ 11
DISCUSSION................................................................................................................................................. 25
CONCLUSION............................................................................................................................................... 26
Bibliography ................................................................................................................................................ 27
Page | 2
ABSTRACT
With three different parallel circuit types—RL, RC, and RLC—this experiment intends to
determine the admittance and the phase between the AC voltage and current. To ascertain the
correctness of the measurements, the measured values will be contrasted with the calculated
values. The parallel RL circuit's admittance and phase must be measured first, and the parallel
RC circuit must have the same values measured second. The parallel RLC circuit's admittance
and phase must be measured, and the final goal is to contrast the measured and calculated
values. The experimenters will have a better understanding of parallel circuit behaviour under
AC voltage and current circumstances after achieving these objectives.
Page | 3
INTRODUCTION
The reciprocal of impedance is commonly called Admittance, symbol ( Y ).In parallel AC circuits
it is generally more convenient to use admittance to solve complex branch impedance’s
especially when two or more parallel branch impedance’s are involved (helps with the math’s).
The total admittance of the circuit can simply be found by the addition of the parallel
admittances. Then the total impedance, ZT of the circuit will therefore be 1/YT Siemens. .But if
we can have a reciprocal of impedance, we can also have a reciprocal of resistance and
reactance as impedance consists of two components, R and X. Then the reciprocal of resistance
is called Conductance and the reciprocal of reactance is called Susceptance.
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OBJECTIVES/AIMS



Page | 5
To measure the admittance and the phase between the ac voltage and current for a
parallel RL circuit and compare your measured values with the calculated values.
To measure the admittance and the phase between the ac voltage and current for a
parallel RC circuit and compare your measured values with the calculated values.
To measure the admittance and the phase between with; the ac voltage and current
for a parallel RLC circuit and your measured values with the calculated values
THEORY
When an ac sinusoidal voltage is applied across a parallel RL, RC or RLC circuit, as shown in Fig.
13.0, 13.1 and 13.2, there is an opposition to the ac current flow called impedance (Z), and its
unit of measurement is the ohm (Ω). The voltage across each element is sinusoidal. The current
in each element is also sinusoidal and has the same frequency as the applied voltage. However,
the ac current in the inductor will lag the applied ac voltage by 90 degrees, the ac current in the
capacitor will lead the voltage by 90 degrees, and the ac current in resistor will be in-phase with
the applied ac voltage. This will cause a phase difference (θ) between the ac voltage applied to
the parallel circuit and the ac current entering the parallel circuit. This difference can be
between 0 degrees and 90 degrees, depending on the relationship between the reactance and
resistance in the circuit.
When dealing with parallel ac circuits, the inverse of impedance, called admittance (Y), is easier
to use: Therefore,
𝑌=
1
𝑍
The unit of measurement for admittance is the siemens (S). As in a dc resistive circuit, Ohm’s
law determines the relationship between the ac rms voltage applied to the parallel circuit (V)
and the ac rms current entering the parallel circuit (I). In an ac circuit, the impedance (Z) of the
circuit replaces the resistance of a resistor in the Ohm’s law equation, therefore,
𝑉 = 𝑍𝐼
Where Z is the circuit impedance measured in ohms (Ω). Because admittance is the inverse of
impedance, Ohm’s law for admittance is
1
𝑉 = 𝑌𝐼
Or
𝐼 = 𝑌𝑉
Because the rms voltage and current are equal to peak value times 0.707, Ohm’s, law can also
be applied to the peak ac voltage (Vp) and peak ac current (Ip) as follows:
𝐼𝑝 = 𝑌𝑉𝑝
Because the ac current in parallel reactance (IX) is 90 degrees out of-phase with the applied ac
voltage (V), and the ac current in the parallel resistance (IR) is in-phase with the applied ac
voltage, they must be treated as phasors. Therefore, IP and IX must be added as if they are
separated by 90 degrees. From the right angle trigonometry (Pythagorean Theorem), Kirchoff’s
current law, and Ohm’s law, the ac rms current entering the parallel circuit (I) can be
represented by
Page | 6
𝑉
𝑉
𝐼 = √(𝐼𝑅)2 + (𝐼𝑋)2 = √( )2 + ( )2 = √(𝐺𝑉)2 + (𝐵𝑉)2 = 𝑉√(𝐺)2 + (𝐵)2 = 𝑉𝑌
𝑅
𝑅
Based on this equation, the parallel circuit admittance (Y) can be determined from
𝑌 = √(𝐺)2 + (𝐵)2
1
1
Where 𝐺 = 𝑅 is the circuit conductance in Siemens, and 𝐵 = 𝑅 the circuit susceptance in
Siemens.
From the right angle trigonometry (Pythagorean Theorem), the phase difference between the
ac rms voltage applied across the parallel circuit and the ac rms current entering the parallel
circuit can be
𝐵𝑉
𝐵
θ = arctan ( ) = arctan( )
𝐺𝑉
𝐺
Where θ is the phase difference in radians or degrees. As can be seen from the equation, the
larger the susceptance (B) compared to the conductance (G), the larger the phase difference (θ)
between the ac rms voltages applied across the parallel circuit and the ac rms current entering
the parallel circuit. The phase difference (θ) between two periodic functions (such as the sine
functions) can be measured by measuring the time difference (t) between the two curve plots
and the time period for one cycle (T), of the curve plots. Because the ratio of the time
difference (t) divided by the number of degrees of phase for one cycle (360 degrees), the phase
difference can be calculated from
θ
𝑡
=
360 𝑇
Therefore,
θ=
𝑡
× 360°
𝑇
The admittance (Y) of the parallel RL ac circuit in Fig. 13.0 is the phasor sum of the conductance
(G), in siemens, the inductive susceptance (BL), in siemens. Therefore, the magnitude of the
admittance (Y) can be found from
𝑌 = √(𝐺)2 + (𝐵𝐿)2
And the phase difference (θ) between the ac rms voltage across the parallel circuit (V) and the
ac rms current entering the parallel circuit (I) can be found from
θ = −arctan(
Page | 7
𝐵𝐿
)
𝐺
Note that for admittances, when the current (I) lags the voltage; (V), the phase difference (θ) is
negative. The conductance (G) is the inverse of the resistance (R). Therefore,
𝐺=
1
𝑅
The inductive susceptance (BL ) is the inverse of the inductive reactance (XL ). Therefore,
𝐵𝐿 =
1
1
=
𝑋𝐿 2πfL
The admittance (Y) of the parallel RC ac circuit in Fig. 13.1 is the phasor sum of the conductance
(G), in siemens, and the capacitive susceptance (BC), in siemens. Therefore, the magnitude of
the admittance (Y) can be found from
𝑌 = √(𝐺)2 + (𝐵𝑐)2
And the phase difference (θ) between the ac rms voltage across the parallel circuit (V) and the
ac rms current entering the parallel circuit (I) can be found from
θ = arctan(
𝐵𝑐
)
𝐺
The capacitive susceptance (Bc) is the inverse of the capacitive reactance (Xc). Therefore,
𝐵𝐿 =
1
= 2πfC
𝑋𝐶
The admittance (Y) of the parallel RLC ac circuit in Fig. 13.2 is the phasor sum of the
conductance (G), in siemens, and the total susceptance (B), in siemens. The total susceptance
(B) is equal to the phasor sum of the inductive susceptance (BL ) and the capacitive susceptance
(BC). Because the inductive susceptance and capacitive phasors are 180 degrees out-of-phase,
the total susceptance (B) is
B = Bc – BL
Therefore, the magnitude of the admittance (Y) for the parallel RLC circuit can be found from
𝑌 = √(𝐺)2 + (𝐵)2
And the phase difference (θ) between the ac rms voltage across the parallel circuit (V) and the
ac rms current entering the parallel circuit (I) can be found from
𝐵
θ = arctan( )
𝐺
The inductive susceptance (BL) and the capacitive susceptance (BC) are a fond of the ac
sinusoidal frequency. In a parallel RLC ac circuit, there is only one frequency at which they are
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equal. At this frequency, the total susceptance is zero (B = Bc – BL) and the circuit
admittance is Y = G and is at the minimum value, making the impedance (Z) be at its minimum
value. [1]
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APPARATUS







One dual – trace oscilloscope
One function generator
One 0–20mA ac multimeter
One 0–10 Vac voltmeter
One 0.1µF capacitor
One 100mH inductor
Resistors - 1Ω, 1kΩ
Page | 10
PROCEDURE
PROCEDURE AND RESULTS
STEP 1:
After connecting the circuit in Fig. 13.0 in the multism and setting the instruments as shown in
the figure, the voltage across the parallel RL circuit (V) and the current entering the parallel RL
circuit (I) were plotted on the oscilloscope. The voltage (V) rms was measured using the
multimeter, and the current (I) rms was measured using the multimeter connected in series
with the 1Ω resistor. The curve plots for the voltage (V) and current (I) were drawn on the
graph. The ac rms voltage (V) and current (I) readings on the ac voltmeter and ammeter were
recorded.
V= 2.826V rms
I= 5.299 *10-3 A rms
Figure 13.0 shows an RL circuit in a parallel combination (multism)
Page | 11
Fig 3.0 above shows voltage against current curve plots for RL circuit (parallel combination)
STEP 2:
Based on the curve plots in step 1, the phase difference (θ) between the voltage and current
was determined.
𝑡
θ=360 (𝑇)
1
T =𝑓
T =10-3 s
−161.5∗10−6 𝑠
θ=360 (
10−3 𝑠
)
θ = -58.14°
θ = -58.1°
STEP 3:
Based on the AC rms voltage (V) and current (I) readings obtained from the multism, the
magnitude of the admittance (Y) of the parallel RL circuit was calculated using the formula
Y
= I / V, where I is the AC rms current and V is the AC rms voltage and also the impedance was
calculated Z =1/Y.
𝑌=
Page | 12
𝐼
𝑉
5.299 ∗ 10−3 𝐴
𝑌=
2.826 𝑉
𝑌 = 1.8751 ∗ 10−3 𝑆
𝑌 = 1.88 ∗ 10−3 𝑆
𝑍=
𝑍=
1
𝑌
1
1.88 ∗ 10−3 𝑆
Z= 531.9 Ω
Z= 0.532 kΩ
STEP 4:
Based on the known value of the inductance (L) and the sinusoidal frequency (f) used in the
experiment, the inductive susceptance (BL) of the inductor was calculated in Siemens.
𝐵𝐿 =
𝐵𝐿 =
1
1
=
𝑋𝐿 2𝜋𝑓𝐿
1
2𝜋 ∗ 1.00 ∗ 103 𝐻𝑧 ∗ 100 ∗ 10−3 𝐻
𝐵𝐿 =
1
628.3185Ω
𝐵𝐿 = 1.592 ∗ 10−3 𝑆
𝐵𝐿 = 1.59 ∗ 10−3 𝑆
STEP 5:
Page | 13
Based on the known value of the resistance of the resistor used in the experiment, the
conductance (G) of the resistor was calculated.
𝐺=
𝐺=
1
𝑅
1
1000Ω
𝑮 = 𝟏. 𝟎𝟎 ∗ 𝟏𝟎−𝟑 𝑺
STEP 6:
Based on the known values of the conductance (G) of resistor R and the inductive susceptance
(BL) of inductor L, the expected magnitude of the admittance (Y) of the parallel RL circuit was
calculated. Then impedance of the RL circuit was calculated using the value of admittance(Y).
2
𝑌 = √𝐵𝐿2 + 𝐺 2
2
𝑌 = √(1.59 ∗ 10−3 𝑆)2 + (1.0 ∗ 10−3 𝑆)2
𝒀 = 𝟏. 𝟖𝟕𝟖𝟑 ∗ 𝟏𝟎−𝟑 𝑺
𝒀 = 𝟏. 𝟖𝟖 ∗ 𝟏𝟎−𝟑 𝑺
𝑍=
𝑍=
1
𝑌
1
1.88 ∗ 10−3 S
𝑍 = 531.9Ω
𝒁 = 𝟎. 𝟓𝟑𝟐Ω
.
QUESTION 13.0
How did your calculated admittance and impedance magnitudes in step 6 compare
with the admittance and impedance calculated from the measured ac rms voltage
and current in step 3?
Page | 14
STEP 7:
Based in on the inductive susceptance (BL) and the conductance (G), the expected phase difference
between the current and voltage sinusoidal functions was calculated.
𝑩𝑳
θ = -arctan ( 𝑮 )
𝟏.𝟓𝟗∗𝟏𝟎−𝟑 𝑺
θ = -arctan (𝟏.𝟎𝟎∗𝟏𝟎−𝟑 𝑺)
θ = -57.8°
QUESTION 13.1
a) How did the calculated value for the phase difference in step 7 compare with the
measured phase difference between the current and voltage curve plots in steps 1 and
2?
Same
b) Is the voltage leading or lagging the current?
Lagging
c) Is this what you expected?
Yes
STEP 8:
The circuit in Fig. 13.1 was connected in the multism and the instruments was set as shown in
the figure below, the voltage across the parallel RC circuit (V) and the current entering the
parallel RC circuit (I) were plotted on the oscilloscope. The voltage (V) rms was measured using
the multimeter, and the current (I) rms was measured using the multimeter connected in series
with the 1Ω resistor. The curve plots for the voltage (V) and current (I) were drawn on the
graph. The ac rms voltage (V) and current (I) readings on the ac voltmeter and ammeter were
recorded.
V= 2.826V rms
Page | 15
I =3.34*10-3 A rms
Figure 13.1 shows an RC circuit in a parallel combination (multism)
Fig 4.0 above shows voltage against current curve plots for RC circuit (parallel combination)
STEP 9:
Based on the curve plots in step 6, the phase difference (θ) between the voltage and current
was determined.
Page | 16
𝑡
θ=360 (𝑇)
1
T =𝑓
T = 10-3 s
90.10∗10−6 𝑠
θ=360 (
10−3 𝑠
)
θ = 32.436°
θ = 32.4°
STEP 10:
Based on the AC rms voltage (V) and current (I) readings obtained from the multism, the
magnitude of the admittance (Y) of the parallel RC circuit was calculated using the formula
Y
= I / V, where I is the AC rms current and V is the AC rms voltage and also the impedance was
calculated Z =1/Y.
𝑌=
𝐼
𝑉
3.34 ∗ 10−3 𝐴
𝑌=
2.826 𝑉
𝑌 = 1.182 ∗ 10−3 𝑆
𝑌 = 1.18 ∗ 10−3 𝑆
𝑍=
𝑍=
1
𝑌
1
1.18 ∗ 10−3 𝑆
Z= 847.46 Ω
Z= 0.847 kΩ
STEP 11:
Page | 17
Based on the known value of the capacitive (C) and the sinusoidal frequency (f) used in the
experiment, the capacitive susceptance (BC) of the capacitor was calculated in Siemens.
𝐵𝐶 =
1
= 2𝜋𝑓𝐶
𝑋𝐿
𝐵𝐶 = 2𝜋 ∗ 1.00 ∗ 103 𝐻𝑧 ∗ 1.00 ∗ 10−7 𝐹
𝑩𝑪 =6.283*10-4S
𝑩𝑪 =6.28*10-4S
STEP 12:
Based on the known value of the resistance of the resistor used in the RC circuit, the
conductance (G) of the resistor was calculated.
𝐺=
𝐺=
1
𝑅
1
1000Ω
𝑮 = 𝟏. 𝟎𝟎 ∗ 𝟏𝟎−𝟑 𝑺
STEP 13:
Based on the known values of the conductance (G) of resistor R and the capacitive susceptance
(BC) of capacitor C, the expected magnitude of the admittance (Y) of the parallel RC circuit was
calculated. Then impedance of the RC circuit was calculated using the value of admittance(Y).
2
𝑌 = √𝐵𝐶2 + 𝐺 2
2
𝑌 = √(6.28 ∗ 10−4 S)2 + (1.0 ∗ 10−3 𝑆)2
𝒀 = 𝟏. 𝟏𝟖𝟏 ∗ 𝟏𝟎−𝟑 𝑺
𝒀 = 𝟏. 𝟏𝟖 ∗ 𝟏𝟎−𝟑 𝑺
𝑍=
𝑍=
Page | 18
1
𝑌
1
1.18 ∗ 10−3 S
𝑍 = 847.46Ω
𝒁 = 𝟎. 𝟖𝟒𝟕Ω
QUESTION 13.2
How did your calculated admittance and impedance magnitudes in step 13 compare with the
admittance and impedance calculated from the measured ac rms voltage and current in step
10?
Same
STEP 14:
Based in on the capacitive susceptance (BC) and the conductance (G), the expected phase
difference between the current and voltage sinusoidal functions was calculated.
𝑩𝑪
θ = arctan ( 𝑮 )
𝟔.𝟖𝟒∗𝟏𝟎−𝟒 𝑺
θ = arctan (𝟏.𝟎𝟎∗𝟏𝟎−𝟑 𝑺)
θ = 34.4°
QUESTIONS:
1. Is the voltage leading or lagging the current?
Leading
2. Is this what you expected?
Yes
STEP 15:
The circuit in Fig. 13.2 was connected in the multism and the instruments was set as shown in
the figure below, the voltage across the parallel RLC circuit (V) and the current entering the
parallel RLC circuit (I) were plotted on the oscilloscope. The voltage (V) rms was measured using
the multimeter, and the current (I) rms was measured using the multimeter connected in series
with the 1Ω resistor. The curve plots for the voltage (V) and current (I) were drawn on the
graph. The ac rms voltage (V) and current (I) readings on the ac voltmeter and ammeter were
recorded.
Page | 19
V= 2.826V rms
I =3.909*10-3 A rms
Figure 13.2 shows an RLC circuit in a parallel combination (multism)
Fig 5.0 above shows voltage against current curve plots for RLC circuit (parallel combination)
STEP 16:
Based on the curve plots in step 15, the phase difference (θ) between the voltage and current
was determined.
Page | 20
𝑡
θ=360 (𝑇)
1
T =𝑓
T = 10-3 s
−122∗10−6 𝑠
θ=360 (
10−3 𝑠
)
θ = -43.92°
θ = -43.9°
STEP 17:
Based on the AC rms voltage (V) and current (I) readings obtained from the multism, the
magnitude of the admittance (Y) of the parallel RLC circuit was calculated using the formula
Y = I / V, where I is the AC rms current and V is the AC rms voltage and also the impedance was
calculated Z =1/Y.
𝑌=
𝐼
𝑉
3.909 ∗ 10−3 𝐴
𝑌=
2.826 𝑉
𝑌 = 1.383 ∗ 10−3 𝑆
𝑌 = 1.38 ∗ 10−3 𝑆
𝑍=
𝑍=
1
𝑌
1
1.38 ∗ 10−3 𝑆
Z= 722.947 Ω
Z= 0.723 kΩ
STEP 18:
Based on the known value of the capacitive (C) and the sinusoidal frequency (f) used in the
experiment, the capacitive susceptance (BC) of the capacitor was calculated in Siemens.
𝐵𝐶 =
Page | 21
1
= 2𝜋𝑓𝐶
𝑋𝐿
𝐵𝐶 = 2𝜋 ∗ 1.00 ∗ 103 𝐻𝑧 ∗ 1.00 ∗ 10−7 𝐹
𝑩𝑪 =6.283*10-4S
𝑩𝑪 =6.28*10-4S
STEP 19:
Based on the known value of the inductance (L) and the sinusoidal frequency (f) used in the
experiment, the inductive susceptance (BL) of the inductor was calculated in Siemens.
𝐵𝐿 =
𝐵𝐿 =
1
1
=
𝑋𝐿 2𝜋𝑓𝐿
1
2𝜋 ∗ 1.00 ∗ 103 𝐻𝑧 ∗ 100 ∗ 10−3 𝐻
𝐵𝐿 =
1
628.3185Ω
𝐵𝐿 = 1.592 ∗ 10−3 𝑆
𝑩𝑳 = 𝟏. 𝟓𝟗 ∗ 𝟏𝟎−𝟑 𝑺
STEP 20:
Based on the known value of the resistance of the resistor used in the RC circuit, the
conductance (G) of the resistor was calculated.
𝐺=
𝐺=
1
𝑅
1
1000Ω
𝑮 = 𝟏. 𝟎𝟎 ∗ 𝟏𝟎−𝟑 𝑺
STEP 21:
Page | 22
Based on the known values of the conductance (G) of resistor R and the capacitive susceptance
(BC) of capacitor C, the expected magnitude of the admittance (Y) of the parallel RLC circuit was
calculated. Then impedance of the RLC circuit was calculated using the value of admittance(Y).
2
𝑌 = √𝐺 2 + (𝐵𝐶 − 𝐵𝐿)2
2
2
𝑌 = √(1.00 ∗ 10−3 𝑆)2 + (6.28 ∗ 10−4 S − 1.59 ∗ 10−3 S)
𝒀 = 𝟏. 𝟑𝟖𝟖 ∗ 𝟏𝟎−𝟑 𝑺
𝒀 = 𝟏. 𝟑𝟗 ∗ 𝟏𝟎−𝟑 𝑺
𝑍=
𝑍=
1
𝑌
1
1.39 ∗ 10−3 S
𝑍 = 719.4Ω
𝒁 = 𝟎. 𝟕𝟏𝟗Ω
QUESTION 13.4
How did your calculated admittance and impedance magnitudes in step 21 compare with the
admittance and impedance calculated from the measured ac rms voltage and current in step
17?
Same
STEP 22:
Based in on the capacitive susceptance (BC) and the conductance (G), the expected phase
difference between the current and voltage sinusoidal functions was calculated.
𝑩𝑪−𝑩𝑳
θ = arctan (
𝑮
)
𝟔.𝟖𝟒∗𝟏𝟎−𝟒 𝑺−𝟏.𝟓𝟗∗𝟏𝟎−𝟑 𝑺
θ = arctan (
θ = -42.2°
Page | 23
𝟏.𝟎𝟎∗𝟏𝟎−𝟑 𝑺
)
QUESTION 13.5
a) How did the calculated value for the phase difference in step 22 compare with the
measured phase difference between the current and voltage curve plots in steps 15 and
16?
Same
b) Is the voltage leading or lagging the current?
Lagging
c) Is this what you expected?
Yes
Page | 24
DISCUSSION
The measured values of admittance and phase difference were compared with the calculated
values. The values for all circuits (RL, RC AND RLC circuit) were close to the calculated values,
with only a small discrepancy in the admittance.to improve the results ensure that multisim
software was up to date and also the signs of components are observed correctly to avoid
errors.
Page | 25
CONCLUSION
In this lab, we successfully measured the admittance and phase difference between the AC
voltage and current for parallel RC, RL, and RLC circuits which was the main objective of the
experiment. Our measured values were in good agreement with the calculated values,
indicating that our experimental setup was accurate and reliable. This lab helped us to
understand the concepts of admittance and phase difference in AC circuits and how to measure
them experimentally.
Page | 26
Bibliography
[1] “ELECTRONICS TUTORIAL,” [Online]. Available: www.electronics-tutorials.ws.
Page | 27