KYAMBOGO UNIVERSITY
FACULTY OF SCIENCE
PHYSICS DEPARTMENT
BACHELOR OF SCIENCE WITH EDUCATION
YEAR: ONE (1)
SEMESTER: TWO(2)
ACADEMIC YEAR: 2024/2025
COURSE NAME: ELECTRICITY AND MAGNETISM
COURSE CODE: SPH 1202
LECTURER'S NAME: Dr. NSAMBA BENARD
TASK: GROUP WORK
GROUP MEMBERS
NAME REG. NUMBER SIGN
AKANDWANAHO JOHNBAPTIST 24/U/ESP/013/GV
MUGOBE HENRY 24/U/ESP/16889/PD
SSETIMBA CHARLES 24/U/ESP/361/GV
EDEKU MARTIN 24/U/ESP/16796/PD
CANOGURA KELLY MICHEAL 24/U/ESP/16786/PD
ODUR AUGUSTINE 23/U/ESP/11658/PD
3(a) Define the following terms
(i) Inductive reactance
(ii) Capacitive reactance
(iii) Impedance
(b) Derive the expression for the impedance of an RLC series circuit.
(C) Given that the series RLC circuit in (b) above, resistance R = 50Ω, an inductance L =
0.1H and a capacitance C = 10µF are connected to an a.c Voltage source V = 100V with
a frequency f = 50Hz.
RESPONSE TO THE TASK
(i) Inductive reactance (𝑋 ) is the non resistive opposition to the flow of
Alternating Current (AC) offered by an inductor.
It's essentially the "resistance" due to the inductor's magnetic field.
It’s directly proportional to the frequency (f) of the A.C and the inductance (L) of the
inductor.
This means that as the frequency increases, the inductive reactance also increases.
XL (Ω)
f (Hz)
Inductors resist changes in current flow. When AC current flows through an inductor, the
changing current creates a changing magnetic field. This changing magnetic field
opposes the change in current, causing the inductor to act like a resistance.
Real-life application: Inductors are used in filters to block certain frequencies of AC
signals. For example, a choke coil in a radio circuit uses inductance to block high-
frequency signals while allowing low-frequency signals to pass through.
Choke coil
Its S.I Units is Ohms (Ω)
𝑋 = 2πfL
f is the frequency of the AC current (in Hertz)
L is the inductance of the inductor (in Henries)
(ii) Capacitive reactance (𝑋 ) is the non resistive opposition to the flow of
Alternating Current (AC) offered by a Capacitor.
It's the "resistance" due to the capacitor's ability to store electrical energy.
It’s inversely proportional to the frequency and capacitance (C) of the
capacitor.
XC (Ω)
f (Hz)
Its S.I Units is Ohms (Ω)
𝑋 =
f is the frequency of the AC current (in Hertz)
C is the capacitance of the capacitor (in Farads)
Capacitive reactance is inversely proportional to both frequency and capacitance. This
means that as the frequency or capacitance increases, the capacitive reactance decreases.
Capacitors resist changes in voltage. When AC current flows through a capacitor, the
changing voltage across the capacitor causes the capacitor to store and release electrical
energy. This process opposes the change in voltage, causing the capacitor to act like a
resistance.
Real-life application: Capacitors are used in filters to block low frequencies and allow
high frequencies to pass through. For example, a capacitor in a power supply circuit can
smooth out the AC voltage from the power grid.
(iii) Impedance (Z) is the total opposition to the flow of current through an A.C
circuit containing resistive and reactive components (inductive and capacitive)
It's the combined effect of resistance and reactance.
Formula; 𝑍 = 𝑅 + (𝑋 − 𝑋 )
Its S.I Units is Ohms (Ω)
Impedance is a complex quantity, meaning it has both magnitude and phase. The
magnitude of impedance represents the total opposition to current flow, while the
phase represents the phase difference between the voltage and current in the
circuit.
The impedance of an RLC series circuit is a function of frequency. At low
frequencies, the capacitive reactance is high, and the impedance is dominated by
the capacitive reactance. At high frequencies, the inductive reactance is high, and
the impedance is dominated by the inductive reactance.
Real-life application: In a speaker system, the impedance of the speaker
determines how much power it can handle. A higher impedance speaker requires
more power to drive it.
(b) R L C
𝑉 𝑉 𝑉
V=𝑉 sin𝜔𝑡
(b) where
𝑉 = 𝐼𝑅 (Ohms law)
𝑉 = 𝐼𝑋 = 𝐼(2𝜋𝑓𝐿)
𝑉 = 𝐼𝑋 = 𝐼( )
The voltage across the resistor (𝑉 ) is in phase with the current (I), the voltage across the
inductor (𝑉 ) leads the current by 90 degrees, and the voltage across the capacitor (𝑉 )
lags the current by 90 degrees.
The above theorem can be easily remembered and grabbed from ‘CIVIL’ which stands
for “for Capacitors, I (current) leads V (voltage) and V (voltage) leads I (current) for L
(inductor)”
The total Voltage V across the circuit is the phasor sum of 𝑉 , 𝑉 𝑎𝑛𝑑 𝑉
Since 𝑉 and 𝑉 are 180 degrees out of phase, their net reactance is 𝑋 − 𝑋 for 𝑋 > 𝑋
𝑉 I
VC-VL V
Where β is the phase angle between current and voltage
From the diagram, using the Pythagorean Theorem, we can find the magnitude of the
total voltage:
𝑉 =𝑉 + (𝑉 − 𝑉 )2………………………………………..(i)
But
𝑉 =IR, 𝑉 =I𝑋 , and 𝑉 = 𝐼𝑋 , we can then substitute these values into the equation (i)
𝑉 = 𝐼 2R2 + (𝐼𝑋 − 𝐼𝑋 )2
V= √[ 𝐼 2R2 + (𝐼𝑋 − 𝐼𝑋 )2]
Factoring out I²:
𝑉= 𝐼 [𝑅 + (𝑋 − 𝑋 ) ]
= √[𝑅 2 + (𝑋 − 𝑋 )2]
But; =𝑍
Z = √[𝑅 2 + (𝑋 − 𝑋 )2]
Thus impedance is independent of the current.
Therefore, the impedance (Z) of an RLC series circuit is calculated using the phasor sum
of the resistance (R), inductive reactance (𝑋 ), and capacitive reactance (𝑋 ).
The impedance of an RLC series circuit can be minimized at a specific frequency called
the resonant frequency (f0). This happens when the inductive reactance and capacitive
reactance are equal (𝑋 = 𝑋 ).
At resonance, the impedance is equal to the resistance (R).
Resonant frequency is the frequency at which the current flowing in the circuit is
maximum for a given value of voltage.
Resonant frequency, 𝑓 =
√
Where
L=inductance in Henries.
C= capacitance in Farads.
R, L and C Components in parallel arrangement
Phasor diagram
𝐼
V 𝐼
𝐼
Supposing 𝐼 > 𝐼 Then;
𝐼 −𝐼
𝜃
V 𝐼
I2 = (𝐼 − 𝐼 )2 + 𝐼 2
( )2 = ( − )2 + ( )2
( )2 = V 2 ( − )2 + ( )2
( )2 = ( − )2 + ( )2
Phase angle;
From the diagram above
tan𝜃=
tan𝜃=
( )
tan𝜃=𝑅 ( − )
𝜃= tan 𝑅( − )
Where 𝜃 is the phase angle.
(c) R= 50Ω, L=0.1H, C=10µF, V=100V and f=50Hz
Tanβ =
Tanβ =
𝛽 = tan ( )
But; 𝑋 =
1
=
2 × 𝜋 × 50 × 10 × 10
= 318.310Ω
Also; 𝑋 = 2𝜋fL
= 2× 𝜋 × 50 × 0.1
= 31.416Ω
. .
β = tan ( )
= 80.1 degrees
The phase angle 80.10 is the phase difference between voltage and current in this circuit.
For voltage and current to be in phase, the phase angle must be 0 0
(ii)
Z = √[𝑅 2 + (𝑋 − 𝑋 )2]
= √[502 + (318.310-31.416)2]
= 291.218Ω
The impedance of the circuit is 291.218Ω.
This means the circuit has a significant reactive component, indicating that it is highly
reactive, dominated by capacitance since 𝑋 > 𝑋 .
BIBLIOGRAPHY
1. Alexander, C. K., & Sadiku, M. N. O. (2017). Fundamentals of electric circuits
(6th ed.). McGraw-Hill Education.
2. Dorf, R. C., & Svoboda, J. A. (2020). Introduction to electric circuits (10th ed.).
Wiley.
3. Floyd, T. L. (2018). Principles of electric circuits: Conventional current version
(10th ed.). Pearson.
4. Hayt, W. H., Kemmerly, J. E., & Durbin, S. M. (2018). Engineering circuit
analysis (9th ed.). McGraw-Hill Education.
5. Nilsson, J. W., & Riedel, S. A. (2019). Electric circuits (11th ed.). Pearson.
6. Sedra, A. S., & Smith, K. C. (2020). Microelectronic circuits (8th ed.). Oxford
University Press.
