RL and RC Circuit Transient Response Analysis
1.Derive the expression for Transient Response of RL Circuits
A
1. Circuit and assumptions
Resistance = R
Inductance = L
DC source = V applied at t=0
Initial current in inductor: i(0)=0
2. Apply Kirchhoff’s Voltage Law (KVL)
For t>0:
V=Ri(t)+L di(t)/dt
Rearranging, L di(t)/dt+Ri(t)=V
3. Solve the differential equation
Step 1: Standard form: di(t)/dt+R/T i(T)=V/L
This is a first-order linear differential equation
Step 2: Complementary (homogeneous) solution
Di/dt+R/L i=0
Ic(t)=Ce _RL t
Step 3: Particular solution:
Or steady state (t→∞),di/dt=0 ip=V/R
Step 4: General solution:
I(t)=ip+ic
I(t)=V/R+Ce _RLt
4. Apply initial condition
At t=0,i(0)=0:
0=V/R+C
C=-V/R
5. Final expression for transient current
I(t)=V/R(1-e_RLt)
6. Time constant of RL circuit
R=L/R
I(t)=I∞(1-e_ t/T)
I∞ =V/R
7. Important transient voltages
Voltage across inductor:
VL(t)=L di/dt=Ve -t/r
Voltage across resistor:
VR(t)=Ri(t)=V(1-e-t/r)
8. Key observations (very important for exams)
Current starts from zero and rises exponentially
At t=τt = \taut=τ, current reaches 63.2% of final value
Inductor behaves like:
Open circuit at t=0
Short circuit at steady state…………………………………………..
2.Derive the expression for Transient Response of RC Circuits
A.1. Circuit description
Consider a series RC circuit with:
Resistance RRR
Capacitance CCC
DC source VVV applied at t=0
Initial capacitor voltage:
VC(0)=0
2. Apply Kirchhoff’s Voltage Law (KVL)
For t>0:
V=vR(t)+vC(t)
VR(t)=Ri(t), i(t)=C dvC(t)/dt
Sub: v=RC dvC(t)/dt+VC(T)
3. Differential equation of RC circuit:
RC dVC(t)/dt+ VC(t)=V
4. Solution of the equation
(a) Complementary (homogeneous) solution
DvC/dt+1/RC vC=0
VC,h(t)=Ke -t/RC
b) Particular solution
At steady state (t→∞), dvc/dt=0
Vc,p=V
(c) General solution
VC(t)=VKe−RCt
5. Apply initial condition
At t=0 vC(0)=0
0=V+K
K=−V
6. Final expression for capacitor voltage (charging)FFB
Vc(t)=V(1-e-t/RC)
7. Current during transient
I(t)=C dvc/dt
I(t)=V/R e_t/RC
8. Time constant of RC circuit
R=RC
So the voltage equation can also be written as:
VC(t)=V(1-e-t/T)
9. Important observations (exam points)
Capacitor voltage rises exponentially
At t=τ, voltage reaches 63.2% of final value
At steady state:
Capacitor behaves as open circuit
Current becomes zero
Initially (t=0):
Capacitor behaves as short circuit
🔹 Discharging case (source removed)
If the capacitor initially has voltage V0:
VC(t)=V0e-t/RC, i(t)=-V0/R e-t/RC
23.Give the V-I relationship of R, L, C elements?
A. 1. Resistor (R)
For a resistor, voltage is directly proportional to current
V(t)=Ri(t)
This is Ohm’s Law
2. Inductor (L)
For an inductor, voltage is proportional to the rate of change of current
V(t)=L di(t)/dt
Inductors oppose change in current
3. Capacitor (C)
For a capacitor, current is proportional to the rate of change of voltage
I(t)=C dv(t)/dt
Capacitors oppose change in voltage
Quick Summary Table
Element | V–I Relationship |
|---|---|
| Resistor | v=Ri |
| Inductor | v=L di/dt |
| Capacitor | i=C dv/dt |
Unit 2
1.Explain the working principle & operation of a single phase transformer?
A.A single-phase transformer is a static electrical device that transfers AC power from
One circuit to another at the same frequency, but usually at a different voltage level,
Using electromagnetic induction
1. Working Principle:
The working principle of a single-phase transformer is Mutual Induction
When an alternating voltage (AC) is applied to the primary winding, an alternating current flows
This current produces an alternating magnetic flux (ϕ) in the transformer core
The magnetic flux links both the primary and secondary windings
If a load is connected to the secondary, the induced EMF causes a secondary current to flow
2. Construction (Brief);
A single-phase transformer mainly consists of:
Primary winding – connected to the AC supply
Secondary winding – connected to the load
Magnetic core – made of laminated silicon steel to reduce losses
3. Operation of a Single-Phase Transformer:
(a) No-Load Condition:
Secondary is open-circuited
A small current called no-load current (I₀) flows in the primary
This current produces the required magnetic flux in the core
(b) Load Condition:
When a load is connected to the secondary:
Secondary current (I₂) flows
This current produces its own magnetic flux opposing the main flux
To maintain constant flux, the primary draws additional current
4. EMF Equation:
The RMS value of induced EMF is given by:
E=4.44fNΦm
Where:
F = supply frequency (Hz)
N = number of turns
Φm= maximum flux (Wb)
5. Voltage & Current Relationship:
Voltage Ratio:
V1/V2=N1/N2
Current ratio:
I1/I2=N2/N1
Step-up transformer: N2>N1⇒V2>V1
Step-down transformer: N2
6. Power Transfer:
Ideally: Pinput=Poutput
In practice, some losses occur:
Core losses (hysteresis & eddy current)
Copper losses (I²R losses)
7. Key Points to Remember:
Works only with AC supply
Frequency remains constant
No direct electrical connection between primary & secondary
Energy transfer is through magnetic field
2. Derive the EMF equation of Transformer?
A.Assumptions:
1.The transformer is supplied with a sinusoidal AC voltage
2.The magnetic flux in the core is sinusoidal
3.Leakage flux and winding resistance are neglected (ideal case)
Step 1: Expression for Magnetic Flux
Let the alternating flux in the core be sinusoidal:
ϕ=ϕmsinωt
Where:
ϕm = maximum flux (Wb)
ω=2πf
F = supply frequency (Hz)
Step 2: Induced EMF in a Coil
According to Faraday’s Law of Electromagnetic Induction:
e=−N dϕ/dt
Substitute the flux equation:
e=−N d/dt (ϕmsinωt)
e=−Nϕmωcosωt
Step 3: Maximum Value of Induced EMF
The maximum value of EMF occurs when:
cosωt=1
emax=Nϕmω
Substitute ω=2πf: emax=2πfNϕm
Step 4: RMS Value of Induced EMF
Erms= emax/underoot2
E= 2πfNϕm/underroot2
E=4.44fNϕm
EMF Equation of Transformer
Primary Winding:E1=4.44fN1ϕm
Secondary Winding:E2=4.44fN2ϕm
Where:
E1,E2= RMS induced EMF in primary and secondary (Volts)
f = frequency (Hz)
N1,N2 = number of turns in primary and secondary
ϕm = maximum magnetic flux (Wb)
Voltage Ratio
Dividing the two EMF equations:
E1/E2=N1/N2
This shows that voltage ratio equals turns ratio.
3.A single-phase transformer has 400 primary and 1000 secondary turns. The net cross sectional
area of the core is 60 cm². If the primary winding be connected to a 50-Hz supply at
520 V, calculate (i) the peak value of flux density in the core (ii) the voltage induced in the secondary winding.
A.Given
Primary turns, N1=400
Secondary turns, N2=1000
Supply voltage, V1=520 V (rms)
Frequency, f=50 Hz
Core cross-sectional area,
A=60 cm2=60×10−4=0.006 m2
(i) Peak value of flux density in the core:
For a transformer, the rms induced emf equation is
E1=4.44fN1Φmax
Since the primary is connected to the supply
E1=V1=520 V
So, Φmax=V1/4.44fN1
Φmax= 520/4.44×50×400
Φmax=5.86×10−3 Wb
Now, peak flux density:
Bmax= Φmax/A
Bmax= 5.86×10−3 /0.006
Bmax≈0.98 T
Peak flux density in the core:
(ii) Voltage induced in the secondary winding
Using the turns ratio: V2/V1=N2/N1
V2=V1× N2/N1
V2=520×1000/400
V2=1300 V (rms)
Final Answers:
Peak flux density in the core:Bmax≈0.98 T
Secondary induced voltage: V2=1300 V (rms) …………………………….