In this article we will discuss about the transient in R-L series circuit.
The selection of the circuit breaker does not depend only on the current carrying capacity under normal conditions; but also depends upon the momentary maximum current it may have to carry under abnormal or short circuit conditions. Thus it becomes necessary to determine the initial maximum value of current at the time of short circuit. To approach the problem of determining the initial value when a short circuit occurs, there is a need of studying the transient in R-L series circuit which is the most fundamental type of ac circuit.
Consider an R-L series circuit, which is connected, at the instant t = 0, to a source of alternating voltage v = Vmax sin (ωt/ + α) where α is the phase displacement between the voltage v and the reference wave which passes through zero at the time t = 0.
The equation relating the applied voltage and the current in the given circuit is given as –
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v = iR + L di/dt
or L di/dt + i R = Vmax sin (ωt + α) …(4.1)
The complete solution of the above equation consists of two parts which are called the particular integral and complementary function. The particular integral is the solution corresponding to the steady-state conditions, namely –
Where
= Z, the circuit impedance, ɸ is the phase angle between the current and voltage determined by ɸ = Tan-1 ωL/R.
The above solution is obtained without any constant of integration and cannot, therefore, represent initial conditions. Complementary function is, therefore, necessary to represent the initial conditions.
This is the solution of the equation with zero substituted for Vmax sin (ω t + α), namely the solution of [L (di/dt) + Ri] = 0 which is i = A e [(–R/L) t] and complete solution of Eq. (4.1) is given as –
Now when t = 0; i = 0 hence substituting these values in above Eq. (4.2) we get –
Hence Eq. (4.2) may be written as –
The first term in the above Eq. (4.3) for i is current corresponding to the steady-state condition and the second term is a transient which vanishes theoretically after infinite time. But practically, it vanishes very quickly after two or three cycles.
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The total current (i) curve is shown in Fig. 4.1 (b). As obvious from the figure with decreasing transient current it, the total current i tend towards the forced current value. However, in the time interval between T/4 and 3 T/4 after switching, depending upon the phase angle α, the current value may exceed the peak value of the forced current.
The total (or resultant) current can take the maximum possible value when at the closure of the switch the forced current is at its peak and the time constant is large (i.e., when R is approximately zero, the time constant tends to infinity, and the phase angle ɸ is about 90°), or when the transient decay very slowly.
Under such conditions α is about equal to 180°, and the applied voltage should pass through zero at switch closure. The current waveform for α – ɸ = 90° and for sufficiently large value of time constant is shown in Fig.. 4.1 (c). In about half a cycle after switch closure, the resultant current is twice the peak value of forced current. This is known as doubling effect.
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Thus when a sinusoidal voltage is applied to an R-L network, the resultant current cannot exceed twice the peak value of the forced current under any circumstances.
The initial value of the free current is equal and opposite to that of the forced current. Therefore, if at the closure of the switch the forced current passes through zero, the initial value of free component is also zero. Thus there will be no transient current flowing and the circuit will attain a forced state at once.
According to Eq. (4.3) this occurs at –
α – ɸ = 0 or α – ɸ = π.
Thus if the switch happens to be closed when v is maximum, a current wave which is symmetrical and has a maximum instantaneous value equal to Imax results. It may also be noted that closing of switch at other instants will give asymmetrical current waves with small transients.
If some simplifying assumptions are made, then a transmission line can be represented by the circuit model shown in Fig. 4.2 (a), which is equivalent to an R-L series circuit shown in Fig. 4.1 (a).
Simplifying assumptions made are:
1. The line is supplied from a constant voltage source.
2. Short circuit takes place when the line is unloaded.
3. Line capacitance is negligible and the total resistance and inductance of the line are assumed to be lumped at one place.
Short-circuit current is given by Eq. (4.3):
In power system terminology, the first term of the above equation, steady-state current is called the symmetrical short circuit current and the second term, unidirectional transient component is called the dc offset-current, which causes the total short circuit to be unsymmetrical till the transient decays.
The maximum momentary short circuit current Imm corresponds to the first peak, as obvious from Fig. 4.2 (b).
Neglecting the decay of transient current in this short period we have –
Imm = Imax +Imax sin (ɸ – α)
Since transmission line resistance is small as compared to its reactance, ɸ ≃ 90
... Imm = Imax + Imax cos α
This has the maximum possible value for α = 0 i.e. short circuit occurring when the voltage wave is going through zero.
Thus,
Imm (maximum possible) = 2 Imax
For a selection of circuit breakers momentary short circuit current is taken corresponding to its maximum possible value (a safe choice).
Example:
A transmission line of inductance 0.1 H and resistance 5 Ω is suddenly short circuited at the far end, as shown in the figure. Write the expression for the short circuit current I(t). Find approximately the value of the maximum momentary short circuit current.
Solution:
Applied voltage,
v = 100 sin (100t + 15°).
So Vmax = Coefficient of the sine of the time angle = 100 V
ω = Coefficient to time, t = 100
Z = √R2 + (ωL)2 = √52 + (100 x 0.1)2 = 31.81 Ω
ɸ = Tan-1 ωL/R = Tan-1 10/5 = Tan-1 2 = 80.96°
Imax = Vmax/Z = 100/31.81 = 3.1435 A
Expression for short circuit current is given as –
i = Imax sin (ωt + α – ɸ) + Imax sin (ɸ – α) e-t/λ
= 3.1435 sin (100 t + 15° – 80.96°) + 3.1435 sin (80.96° – 15°) e-5/t/0.1
= 3.1435 sin (100 t – 65.96°) + 2.871 e-50 t Ans.
Assuming that the first current maximum occurs at the same time as the first current maximum of the symmetrical short circuit current,
Imm = 3.1435 + 2.871 = 6 A Ans.