An-Najah University
ELECTRICAL CIRCUITS LAB
Lab(10641215) Experiment #8
Created by
:
Adel Bani Jaber
2020
)5-2&2-11(تعويض الدكتورة خديجة ميالة
Instructor Name: Dr.Omar Khaled Tamimi Course Name: ELECTRICAL CIRCUITS LAB
Time: Wednesday (2-5) . Date: 3-6-2020.
Section: Electrical Engineering. Semster :second Semester 2020-2021.
Group Members: Adel Bni Jaber .
RLC series & parallel circuits
*Introduction and Objectives *
In this experiment we will investigate on RLC circuit, and how we can connected this
.circuit
RLC Circuit is a circuit that have resistors, Capacitors and inductors on parallel or
series form, however we can use this type of circuits like a switching circuit or a
.storage circuit
Moreover, we will investigate on the impedance of RLC circuit to determine the
distribution of the applied voltage V and current in the R, L and C, after that we will
study the phase angle shifting between the voltage and the current for each element
.in the RLC circuit
**Tools and Equipment's
: the tools we used in this Exp are
.Training Electronic Board (TEB)♦
.Resistors (1 KΩ), Capacitors (0.1 μF, 0.22 μF, 1 μF), inductors (100 MH)♦
♦Different Function Generator ♦
.Wires
.oscolisscope♦
.Digital Multimeter (DMM)♦
:Series RLC circuits ***
As we said the current which pass through every element is the same and
the voltage across the resistor V Rin phase, however the voltages across the
capacitor and the inductor experience a phase shift with their respective
:currents
𝑉
𝐼
𝑡𝑜𝑡 = , = ට 𝑉 𝑅 2+ (𝑉 𝐿− 𝑉 𝐶 )2
𝑉
𝑍
𝑢𝑞𝑒
𝑤𝑒𝑟𝑒
ℎ 𝑣𝑞𝑒 = ඥ 𝑅 + (𝑋 𝐿+ 𝑋 𝐶 ) 𝑋
∶𝑍 2
𝐿 , 𝐶= 1 𝛚 𝐶
2 , = 𝛚 𝐿ൗ
𝑋
.“Where Z equis called the apparent resistance “Impedance
.The phase angle φ is the angle between the applied voltage and the current for the circuit
V−
L V C X−
L X C
nat φ= ¿ = ¿
VR R
***Parallel Circuits:
In parallel RLC circuit, the voltages across all the elements “R, C and L”
are the same, and the current is divided into active current 𝐈𝐑 ,capacitor
current 𝐈𝐂 and the inductor current 𝐈𝐋 . A phase shift occurs between each
of the current 𝐈𝐂 , 𝐈𝐋 and the total current 𝐈 due to the reactance’s 𝐗 𝐋 of the
inductor and 𝐗 𝐂 of the capacitor “ It’s clear that there is no phase shift
exist between 𝐈𝐑 and 𝐈”
The current 𝐈𝐂 leads 𝐈𝐑 (or I ) constantly by 90°and by assuming that the
capacitor contains no resistance ,While the current 𝐈𝐋 lags the active
current 𝐈𝐑 constantly 90° ,also assuming that the inductor is pure
inductive. The currents 𝐈𝐂 opposes 𝐈𝐋 (180° angle phase) and thus tends to
equalize each other depending on their magnitude.
So, the apparent current I (The total current supplied to the circuit) can
be calculated as the following equation:
2
I = I R2 + ( I C −I L )
√
Also the apparent conductance “Admittance” Y of the circuit may be
obtained from the following equation:
1 1
Y = G 2 + ( BC −B L )2 , G=
√ , BL = , B =ω C
R ωL C
Tan of the phase angle :
I C −I L B C −BL
tan φ=¿ = ¿
IR G
**************************************************
*Procedures*
A. Series circuit:
1)We connected the series RLC circuit shown in figure 1:
2)After we connected the previous circuit, we measured the voltages and the
current across R and L and C using the DMM and the results were as shown in
table 1:
C VL(AB) VC(BC) VR(CD) I.tot)mA) Vtotcal φ cal φmeasure
0.1μF 1.429 3.19 2.08 1.94 2.72 40.25 -38.46
0.22μF …… …… …….. ……. …… …… ……
1μF 1.588 0.415 2.441 2.28 2.71 25.66 +21.77
In previous table we calculate the total voltage by using the equation which we mentioned
2 2
√
before a little time,V = V R +(V L −V C ) and we calculate the phase angle by this
:equation
−1 V−
L V C −1 X−
L X C
φ=nat =nat
VR R
3(
Then we connected the oscilloscope’s Ch1 to point C, Ch2 to point A and the point
D on the ground ,and the displayed voltage was as the following figures ”for 0.1
.” µF , 1 µF respectively
B. Parallel Circuits:
a. We connected Parallel RLC circuit shown in Figure 8:
Figure 1
b. After we connected the previous circuit, we measured the
voltages and the current across R and L and C using the
DMM and the results were as shown in table 1:
I φ cal
C I )mA) I R )mA) I C)mA) I L)mA)
totcal)mA)
0.1μF 4.6 2.91 1.92 4.23 3.71 -38.4
0.22μF …. …. …. ….. ….. …..
1μF 15.99 3.01 19.5 4.55 15.2 78.61
In previous table we calculate the total current by using the equation which
2 2
√
we mentioned before a little time, I = I R + ( I C −I L ) , and we calculate the phase
:angle by this equation
I− I L
−1 C −1 B C− B L
φ=
. nat =nat
IR G
:As
)Bc =1/Xc = wc (
)Bl = 1/Xl = 1/wl (
)G = 1/R (
****:***VECTOR DIAGRAM
***********************************************************
*Conclusion*
In this experiment we learnt that in the resistor the current
through is in the phase with the voltage across the resistor. Due
to this current and this voltage reach to their maximum value at
the same time.
However the inductor, the current through it lags the voltage
across it by 90̊ .Because the voltage reaches its maximum value
one quarter of a period before the current reaches its maximum
value.
In other hand, in the capacitor, the current through it leads the
voltage across it by 90̊. That’s because the current reaches its
maximum value one quarter of a period before the voltage
reaches its maximum value.
In series combination in RLC circuit the current through all the
elements are the same and the voltage source is divided on three
elements. However in parallel combination in RLC circuit the
voltage across each element is the same but the total current is
divided on each element.
