D.C. Transients in RL Circuit

WORKED EXAMPLES

D.C. Transients in RL Circuit

Example 1 A d.c. voltage of 100 volts is applied to a series RL circuit with R = 252 What will be the current in the circuit at twice the time constant?

Solution: As the voltage source is in the circuit, the expression for rise in current is given by

Example 2 Sketch the current given by i (t) = 5 - 4e-^20t.

Solution:

i = 5 - 4e-^20t

= 4 [ 1.25 - e-^20t]

Let us assume that t take some values like t = 0, 0.1 sec, 0.15 sec, 0.2 sec, 0.25 sec. For these values of t, i is calculated and the result is tabulated for convenience.

The sketch is drawn in fig. 3.8.

Example 3 In the circuit of the fig. 3.8, find the expression for the transient current and the initial rate of growth of the transient current.

p

Solution: By applying KVL, we get

5i + 3 di / dt = 100 u (t) ... (i)

As i (0) is directed opposite to that of i (t), i (0) is treated as negative. That is i (0) = -6. Taking Laplace Transformation of equation (i),

Taking inverse Laplace Transformation, on both sides,

i (t) = 20 - 26 e^{-5t/ 3}

To find the initial rate of growth of i.

Example 4 In the circuit shown in the fig. 3.9, switch S is in position 1 for a long time and brought to position 2 at time t = 0. Determine the circuit current.

Solution: After closing the switch to the position 2 applying KVL, we get

5i + 2 di / dt = 10  ... (i)

Taking Laplace Transformation on both sides,

51 (s) + 2 {s I (s)- i (0+) } = 10 / s … (ii)

i (0+) is the current passing through the circuit just after the switch is at position 2. Since the inductor does not allow sudden changes in currents, this currents is equal to the steady state current when the switch was in position 1 [Because switch was in position 1 for long time.]

 i (0+) = 50 / 5 = 10 A

Substituting this in equation (ii),

Example 5 In the circuit shown in fig. 3.10. The switch was initially in the position 1. (a) for Ra =500 Ω find the voltage across the coil the instant at which the switch is changed to position 2. (b) Calculate the value of the Ra for the voltage across the coil to be reduced to 120 V at the instant of switching. (c) with the value of Rain case. Find the time taken to dissipate 95% of stored energy.

Solution:

(a) Steady current with the switch in position 1 = 120 / 20 + 40 = 2A.

Now, with R_d = 500 and the switch changed to position 2, applying Kirchoff's voltage law, we get the following equation.

2 × 500 + 2 × 20 + voltage across the coil = 0 ... (i)

⇒ Voltage across the coil = - 1040 V

[-sign indicates that the voltage is in opposite direction.]

(b) In this case, the voltage across the coil = 120 V

Therefore, R_d × 2 + 2 × 20 -120 = 0

⇒ R_d = 80 / 2 = 40 Ω

(c) Energy stored = 1/2 LI²

Energy stored = 1/2 × 10 × 2² = 20 joule

Given that, 95% of energy is dissipated. So, residual energy = 5% of energy stored = 0.05 × 20 = 1 joule.

Let i be the current at that instant. Then,

This case is a decay of current in R-L combination.

The equation for i is given by,

Here I = 2A, R = 40 + 40 + 20 = 100 Ω

L = 10 H

Substituting these values in (ii), we get,

Example 6 Find the current in the circuit shown in fig.3.12, at an instant t, after opening the switch if a current of 1 amp. had been passing through the circuit at the instant of opening.

Solution:

After t seconds of opening the switch, applying KVL, we get,

iR + L di / dt = E ... (i)

Here R = 6Ω, L = 5H, E = the total voltage of the circuit after opening the switch = 12 + 24 = 36V

Given that I₀₊ = Initial current = 1A

Putting the values of R, L and E in (i), we get,

6i + 5dt/dt = 36 …. (ii)

Taking Laplace transformation on both sides, we get,