applications of differential equations in RL-RC electrical circuit problems

applications of differential equations in RL-RC electrical circuit problems

1. applications of
differential equations in
rl-rc electrical circuit
problems
PRESENTED BY:-
B-TECH (CSE) [Group -5]
PROJECT
-
1
DigbijayaMohapatra-230301120235
AarchiKumari-230301120236
AbhishekDash–230301120237
OmmPrakashMishra–230301120239
AdityaRoshanPradhan-230301120240
SwetalinaPradhan-230301120241
DIFFERENTIAL EQUATION AND LINEAR ALGEBRA

2. Aresistor–inductorcircuit(RLcircuit),orRLfilterorRLnetwork,
isanelectriccircuitcomposedofresistorsandinductorsdriven
byavoltageorcurrentsource.
Afirst-orderRLcircuitiscomposedofoneresistorandone
inductor,eitherinseriesdrivenbyavoltagesourceorinparallel
drivenbyacurrentsource.Itisoneofthesimplestanalogue
infiniteimpulseresponseelectronicfilters.
ALRSeriesCircuitconsistsbasicallyofaninductorof
inductance,Lconnectedinserieswitharesistorofresistance,R.
Applications of differential equations in RL electrical circuit
problems:-

3. The applications of RL circuit include the following :
• RF Amplifiers.
• Communication Systems.
• Filtering Circuits . Processing of Signal.
• Oscillator Circuits.
• Magnification of Current or Voltage.
• Variable Tunes Circuits.
• Radio Wave Transmitters.

4. AnRLcircuitwithaswitchto
turncurrentonandoff.When
inposition1,thebattery,
resistor,andinductorarein
seriesandacurrentis
established.Inposition2,the
batteryisremovedandthe
currenteventuallystops
becauseofenergylossinthe
resistor.
Agraphofcurrentgrowth
versustimewhenthe
switchismovedtoposition
1.ApplicationofDE.
Agraphofcurrentgrowth
versustimewhentheswitch
ismovedtoposition1
RL CIRCUIT

5. Variablevoltageacrosstheresistor:
Vr=Ir
Variablevoltageacrosstheinductor:
VL=Ldi/dt
Kirchhoff'svoltagelaw:
Ri+Ldi/dt=V
Application of DE in a RL circuit

6. The formation of differential equation for an electric
circuit depends upon the following laws :
i)i=dq/dt
ii)Voltagedropacrossresistance(R)=RI
iii)Voltagedropacrossinductance(L)=Ldi/dt
iv)Voltagedropacrosscapacitance(C)=q/c
Kirchhoff’slaw
:Thealgebraicsumofthevoltagedroparound
anyclosedcircuitisequaltoresultantemfinthecircuit.
Currentlaw
:Atajunctioncurrentcomingisequaltocurrent
going.

7. Example:
AnRLcircuithasanemfof5V,aresistanceof50Ω,aninductanceof1H,andno
initialcurrent
.
Findthecurrentinthecircuitatanytimet.Distinguishbetween
thetransientandsteady-statecurrent.
+/=
/+50=5
1st order DE : Y’ + P(x)y = Q(x)
IF: ^∫()= ^∫50
Therefore: ^∫50= ∫5 ^∫50)
^50= ∫5 ^50
^50=1/10 ^50+ C

8. Applications of differential equations in RC electrical circuit
problems:-Aresistor–capacitorcircuit(RCcircuit),orRCfilterorRCnetwork,is
anelectriccircuitcomposedofresistorsandcapacitorsdrivenbya
voltageorcurrentsource.
Afirst-orderRCcircuitiscomposedofoneresistorandonecapacitor,
eitherinseriesdrivenbyavoltagesourceorinparalleldrivenbya
currentsource.Itisoneofthesimplestanalogueinfiniteimpulse
responseelectronicfilters.
AnRCSeriesCircuitconsistsbasicallyofacapacitorofcapacitance,C
connectedinserieswitharesistorofresistance,R.

9. • Signal Filtering.
• Time Constant Calculations.
• Differentiator and Integrator Circuits.
• Time Circuits.
• Waveform Shaping.
• Low-Pass and High-Pass Filters.
The applications of RC circuit include the following:

10. RC CIRCUT
AnRCcircuitwithaswitchtocharge
anddischargethecapacitor.When
inposition1,thebattery,resistor,
andcapacitorareinseries,anda
chargeaccumulatesonthe
capacitor.Inposition2,thebattery
isremoved,andthecapacitor
eventuallydischargesthroughthe
resistor.
A graph of charge
growth versus
time when the
switch is moved to
position 1.
A graph of charge
decay versus time
when the switch is
moved to position 2.

11. Application of DE in a RC circuit
Variable voltage across the resistor:
V
r = iR
Variable voltage across the inductor:
V
C = (1/C) ∫idt
Kirchhoff's voltage law:
Ri + (1/C) ∫idt = V

12. The formation of differential equation for an
electric circuit depends upon the following laws:-
i) i= C * dV/dt
ii) Voltage drops across resistance (R) = RI
iii)Voltagedropsacrosscapacitance(C)=(1/C)∫idt
iv) Voltage drops across inductance (L) = L di/dt
Kirchhoff’slaw:Thealgebraicsumofthevoltagedropsaroundanyclosedcircuitis
equaltoresultantemfinthecircuit.
Currentlaw:Sumofincomingcurrententeringatajunctionisequaltosumof
currentleavingthejunction.

13. Example:
AnRCcircuithasanemfof5V,aresistanceof50Ω,acapacitanceof1F,andnoinitial
charge.Findthevoltageacrossthecapacitoratanytimet.Distinguishbetweenthe
transientandsteady-statevoltage.
Vemf−VR−VC=0
→Since, I=C⋅dVc/dt
→Vemf−(C⋅dVc/dt)⋅R−Vc=0
→dVc /(Vemf−Vc) =dt/RC
→−ln ∣Vemf −VC∣=t/RC+C1
Where C1 is the constant of integration
→∣Vemf −VC∣=e^(−t/RC−C1)
Since the initial charge is 0 (no initial charge), C1=0, and we can drop the absolute value.
→ Vemf –Vc = e^(-t/RC)
→ Vc(t)=Vemf−e^(−t/RC)
→ VC(t)=5 –e^(-t/50 ⋅1)
→ VC(t)=5 –e^(-t/50)