Next: R-C Circuits Up: Introduction to Electronics Previous: Using Norton's equivalent   Contents

# Transient response of RL circuit

For DC circuit analysis, the voltage and current source excitation is constant, so C and L are neglected 6.1. The circuit is assumed to be as it is since time= to . In practice, no excitation is constant from to . A more realistic circuit would include a switch, as shown in Fig.6.2. Also, inductance and capacitances of wires and components cannot be neglected as shown in Fig.6.3, and in Fig.6.4 (for ). Using KVL:

Multiplying both sides by

to get

Therefore,

Integrating both sides, we get

Note that, at , and at , .
Now, , where, , and .

As at , ,

The plot of vs. is shown in the Fig.6.5. Note that when , A.

Voltage across the inductor is given by . Therefore,

The plot of vs is shown in Fig.6.6.
From the above equation, we notice that in time seconds, the voltage across the inductor would reduce to of its original value and would go on decreasing by a further factor of every seconds thereafter. Therefore, summing it up, we have for an inductor-resistor pair with a constant voltage applied at ,
 and

Now, consider the circuit shown in Fig.6.7.
Before , we have the circuit looking as in Fig.. Therefore we have the initial current (at ) through the inductor as A.

At , the circuit looks as in Fig. and therefore, we have the following equations for .

At , A. Hence,

The plot for vs would therefore be as in Fig 5.10

Hence, and, as shown in 6.11, discharge will be immediate. We write equations for across the inductor.

Sign of is as shown in 6.12. As ,

Switching off causes a discharge in the tube or spark at switch 6.13.

At , 6.14

In generic form,

Now, have a look at the circuit shown in the figure 5.15. As the resistance is 0, the equations are indeterminate and are of the form

So, we solve the circuit directly

At , , therefore,

We will use the above circuit to analyse the circuit shown in Figure 5.16. As the resistance of is in parallel with the voltage source and also the rest of the circuit, the current drawn by it will be constant and will not affect the analysis of the rest of the circuit. So, for , we can consider the circuit to be as in Figure 5.17. Analysing it as in the previous example, we get

Further, for , the circuit can be equivalently considered as in Figure 5.19. Notice that still, Amps. as the inductor is in parallel with the voltage source. The plot for vs. would therefore be linear as in Figure 5.18. Therefore,

After , the circuit can be considered equivalently to be that in Fig 5.20. Now, there is no constant voltage source across the resistance of . This, the current flowing through it also comes into the analysis.

The solution thus is: , where, given the initial conditions, we can solve for and .

Next: R-C Circuits Up: Introduction to Electronics Previous: Using Norton's equivalent   Contents
ynsingh 2007-07-25