Electrical and Electronic Principles: Transients in R-L-C Circuits
TASK 1
Q1. Given: Vs = 10V, R = 29 Ώ, L = 0.45 H and C = 17 µ F all in a series arrangement.
- Current flow in steady domain
To solve current for steady state, the capacitor is replaced with open circuit and an inductor is replaced with a short circuit and then solving the remaining. A DC current cannot flow through a capacitor, only AC current.
Therefore we will remain with 10V source, and a 29 Ώ resistor.
Using ohm’s law, I = V/R, then current I = 10/29 = 0.345 Amperes.
Current flow in a transient (dynamic) state
This is not as easy as solving in steady state since Ohm’s law does not apply to capacitor and inductor loads which vary with time as the current is passed on. Therefore in this case we use Laplace transforms to solve for the current and voltage across each load.
Using Laplace transformation
For impedances of generalized resistors (inductor and capacitor)
Calculating current and voltage using Laplace
Therefore V(s) = 10,
Z1(s) = 0.45,
Z2(s) = 1/(17*10-6 ) = 58823.53,
R = 29
I(s) =
I(s) = 10/(29+0.45+58823.53) = 1.7 * 10-4 Amperes
- The voltage across the inductor in steady state
Time constant τ = L/R
τ = 0.45/29 = 0.0155 S = 15.5 mS
Transient time = 5 * τ = 15.5 * 5 = 77.5 mS
Induced E.M.F (VL ) = V e (-Rt/L)
(VL ) = 10 * e (-29*0.0775/0.45) = 10 * 0.0068 = 0.068 V
Therefore inductive voltage VL = 0.068 V
The voltage across the inductor in transient state
= I(s) * 29Ώ + I(s) *
- The voltage across the capacitor in transient state
Time constant τ = R*C
τ = 29 * 17*10-6 = 493 *10-6 S = 493 µ S
Transient time = 5 * 493 *10-6 = 0.002465 S = 2.465 mS
Induced e.m.f. = V e (-t/RC) = 10* e (-0.002465/29*17*10-6) = 0.067 V
Q2.
Total effective resistance = 5*200/(5+200) = 4.87 Ώ
- Initially the inductor acts as a short circuit and the current flowing through the circuit is given by
I = I1 + I2 where I1 flows through 5Ώ and I2 flows though 200Ώ resistor.
I = 200/4.87 = 41A
I1 current = (200*41)/205 = 40 A
I2 current = (5*41)/205 = 205/205 = 1 A
After some time t, the current through the inductor will increase at a constant rate.
Time constant = L/R = 3/ 4.87 = 0.62 S
Transient time = 5 * 0.62 = 3.1 s
V(t) = R I(t) + Ldi/dt
Ldi/dt = V(t) – R I(t)
di/dt = { V(t) – R I(t)}/L
I (t) = 1/L
Solving this and applying initial conditions you get;
I(t) = 41 + K e (-Rt/L)
At initial state, the current is 41 A at time 0 s and the current through the coil is 1.
Hence, 1 = 41 +K e (-4.87*0/3)
Therefore, K e (0) =40, and K should be a negative integer in this case.
Thus, K = -40
Therefore after t seconds, the current through the coil is given by;
I(t) =41-40e (-4.87t / 3)
- After 0.5 seconds
Current through coil is
I (0.5) = 41 -40 e (-4.87*0.5/3)
= 41 – 40*0.444 = 41-17.76 = 23.23 A
Therefore the current through the shunt component after switching on for 0.5 seconds = 23.23 A
- Peak voltage VAB after switch off in steady state peak conditions.
The current through 5Ώ resistor and the inductor is the same = 23.23 A i.e. through AB
The total resistance is given by;
= = 5.8 Ώ
Therefore, the voltage across AB = I*R = 23.23 A * 5.8 Ώ = 135.45 V
- Time t before the coil to 10% of normal working value
Normal working value is at transient time 3.1 seconds.
Working current I (s) = I (3.1) = 41 -40 e (-4.87*3.1/3) = 41 – 0.26 = 40.74A
Hence 10% of 40.74A = 4.074A
Therefore, 4.074 = 41-40 e (-4.87t/3) = 41-40 e (-1.623t)
36.926/40 = e (-1.623t)
0.923 = e (-1.623t)
-1.623t = ln 0.923= -0.0801
T= 0.0801/1.623 = 0.493 s
References
Bird, J. (2012). Electrical and electronic principles and technology. Routledge.
Nicolaides, A. (2008). Electrical and electronic principles.PASS PUBLICATIONS.
Rizzoni, G., & Hartley, T. T. (2000). Principles and applications of electrical engineering(Vol. 3). McGraw Hill.
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