The document discusses an RLC circuit problem involving a resistor, inductor, and capacitor, detailing their definitions and functions within an electronic circuit. It presents a specific circuit problem with given parameters and derives the charge and current at any time t>0 using differential equations and MATLAB code. The document includes references and figures to support the content.

1. RLC CIRCUIT
PROBLEM
Presented By:
Md. Nazmus Saqib Khan
Roll no. 12075768
Email: ratulkhan.jhenidah@gmail.com
Class of Masters,
Department of Mathematics,
University of Rajshahi,
Rajshahi 6205.

2. TOPICS
Definitions
 Resistor (R)
 Inductor (L)
 Capacitor (C)
RLC Circuit
RLC Circuit Problem
MATLAB Code
Figure
References
1

3. RESISTOR (R)
 A resistor is an electrical
component that implements
electrical resistance as a
circuit element. In electronic
circuits, resistors are used to
reduce current flow, adjust
signal levels, to divide
voltages, bias active
elements, and terminate
transmission lines, among
other uses.[1]
Fig.2 Electronic Symbol
(Source: wikipedia)
Fig.1 A typical axial-lead
resistor (Source:
wikipedia)
2

4. INDUCTOR (L)
 An inductor, also called a coil
or reactor, is an electrical
component that stores
electrical energy in a
magnetic field when electric
current is flowing through it.
An inductor typically consists
of an electric conductor, such
as a wire, that is wound into a
coil.[2]
Fig.3 A selection of low-
value inductors
(Source: wikipedia)
Fig.4 Electronic Symbol
(Source: wikipedia)
3

5. CAPACITOR (C)
 A capacitor is an electrical
component that stores
electrical energy in an
electric field. The effect of a
capacitor is known as
capacitance. Capacitors are
widely used in electronic
circuits for blocking direct
current while allowing
alternating current to pass. In
analog filter networks, they
smooth the output of power
supplies.[3]
Fig.5 Different types of
capacitors
(Source: wikipedia)
Fig.6 Electronic
Symbol(Source:
wikipedia) 4

6. RLC CIRCUIT
 An RLC circuit is an electrical circuit consisting of a
resistor (R), an inductor (L), and a capacitor (C),
connected in series or in parallel. The name of the
circuit is derived from the letters that are used to
denote the constituent components of this circuit,
where the sequence of the components may vary from
RLC. [4]
Fig.5 A series RLC circuit: a resistor, inductor, and a
capacitor (Source: wikipedia)
5

7. RLC CIRCUIT PROBLEM
Problem: A circuit has in an electromotive force of 40sin30t volts,
a resistor of 10 ohms, a capacitor of 0.0004 F and an inductor of
0.1 Henry. If the initial current and charge assumed to be zero,
then find current at any time t>0.
Solution: Let i denote the current and q denote
the charge on the capacitor at any time t>0.
Now the voltage drop across the resistor is
ER = i R = 10i
The voltage drop across the inductor is,
EL = L di/dt = 0.1 di/dt
And the voltage drop across the capacitor is
EC = q/C = q/0.0004 = 2500q
Now applying the Kirchhoff’s law, we get
EC + EL + ER = E(t)
Fig. 6 Circuit
diagram for the
given problem 6

8. or, 2500q + 0.1di/dt + 10i = 40sin30t
or, d2q/dt2 + 100dq/dt + 25000q = 400sin30t, (1)
where i= dq/dt.
Subject to initial conditions : q(0) = 0 (2)
And i(0) = dq/dt = 0 (3)
Let q = emt be the trial solution of
d2q/dt2 + 100dq/dt +25000q = 0
Therefore, auxiliary equation is
m2 +100 m +25000 = 0
m = -50 ± 150i
RLC CIRCUIT PROBLEM

7

9. Now, the complementary solution is
qc = e-50t (Asin150t + Bcos150t)
Now, the particular solution is
qp =
1
D2
+100D +25000
(400sin30t)
=
400
−900+100D +25000
(sin30t)
=
4
D +241
(sin30t)
=
4(𝐷−241)
D2
−58081
(sin30t)
=
120𝑐𝑜𝑠30𝑡−964𝑠𝑖𝑛30𝑡
−900−58081
=
964
58981
sin30t -
120
58981
𝑐𝑜𝑠30t
RLC CIRCUIT PROBLEM
8

10. Therefore , the charge at any time t>0 on the
capacitor is
q =
964
58981
sin30t -
120
58981
𝑐𝑜𝑠30t + e-50t (Asin150t +
Bcos150t)
i(t)=
𝑑𝑞
𝑑𝑡
=
28920
58981
cos30t +
3600
58981
𝑠𝑖𝑛30t + e-50t
(150Acos150t - 150Bsin150t) - 50 e-50t (Asin150t
+Bcos150t)
Imposing the initial conditions (2) & (3) , we get
A=
−22920
58981
×
1
150
and B =
120
58981
RLC CIRCUIT PROBLEM
9

11. Therefore ,
q =
964
58981
sin30t -
120
58981
𝑐𝑜𝑠30t + e-50t (
−22920
58981
×
1
150
sin150t +
120
58981
cos150t)
i(t)=
𝑑𝑞
𝑑𝑡
=
28920
58981
cos30t +
3600
58981
𝑠𝑖𝑛30t - e-50t
(
22920
58981
cos150t +
18000
58981
sin150t) +50 e-50t (
22920
58981
×
1
150
sin150t -
120
58981
cos150t)
Which is the required current at any time t>0.
RLC CIRCUIT PROBLEM
10

12. MATLAB CODE
function charge
clear all
clf
clc
n=1000;
h=0.001;
x(1)=0;
y(1)=0;
yp(1)=0;
11

13. for i=1:n
k1=h*f(x(i),y(i),yp(i));
k2=h*f(x(i)+h/2.0,y(i)+h*yp(i)/2.0+h*k1/8
.0,yp(i)+k1/2.0);
k3=h*f(x(i)+h/2.0,y(i)+h*yp(i)/2.0+h*k2/8
.0,yp(i)+k2/2.0);
k4=h*f(x(i)+h,y(i)+h*yp(i)+h*k3/2.0,yp(i)
+k3);
x(i+1)=x(i)+h;
y(i+1)=y(i)+h*((yp(i)+(k1+k2+k3)/6.0));
yp(i+1)=yp(i)+(k1+2.0*k2+2.0*k3+k4)/6.0;
end
12
MATLAB CODE

14. for i=1:n+1
q=(964/58981)*sin(30*x(i)) -
(120/58981)*cos(30*x(i)) + exp(-
50*x(i))*((-2292/(58981*15))*
sin(150*x(i)) + (120/58981)*
cos(150*x(i)));
end
q'
a=[x' y' yp'];
13
MATLAB CODE

15. plot(x,y,':r')
h=xlabel(' time t');
set(h,'FontSize',15)
h=ylabel(' charge q');
set(h,'FontSize',15)
end
function g1=f(x,y,yp)
g1=-100*yp-25000*y+400*sin(30*x);
end
14
MATLAB CODE

16. FIGURE
15
Fig.7 Graph of the solution curve.

17. REFERENCES
[1] https://en.wikipedia.org/wiki/Resistor
[2] https://en.wikipedia.org/wiki/Inductor
[3] https://en.wikipedia.org/wiki/Capacitor
[4] https://en.wikipedia.org/wiki/RLC_circuit
16

18. END OF THE PRESENTATION
THANK YOU!
17