Circuits with inductors

We will begin to analyze circuits with inductors in addition to resistors, capacitors and batteries. The same procedure developed in a previous chapter will be used, together with the new information we just learned about inductors. Roughly, this is the procedure again:
• Indicate current arrows, and draw (or just imagine) implied voltmeters consistant with your current arrows for the elements and consistent with the battery polarity for the batteries.
• Write node equations (KCL) and loop rules (KVL).
• Write constitutive laws.
• Specify the state variable(s).
• Find the state equation(s).
Wait! What about state variables??? Before we had one state variable for each capacitor. What about inductors? Going back to the analogy with mechanical systems, just like the velocity of the mass is a state variable, the current in the inductor is also a state variable. So, for electrical systems we will have one state variable for each capacitor and each inductor.

Here is an example of a simple circuit that we will analyze:

In the circuit above, C = 5 mF, L = 20 mH (milli-henries), and Vb = 9 V (volts).

Initially, the capacitor is uncharged: qc(t=0) = 0. The switch is open. At this point, the current in the inductor must be zero and the voltages across each element is zero, except for the battery of course.

Now close the switch:

 KCL: iB= iC= iL= i  KVL: -VB+ VC+ VL= 0  Const. Laws: VC= qC /C  VL= L i'L

• In the first moment, what do the voltmeters read ( VC, VL)? At the first moment, t=0, thus looking at the constitutive law of the capacitor and rewriting as
we see that VC(t=0) =0. Thus by KVL, VL(t=0) =VB(t=0) .
• What are the values of iB, iC, iL in the first moment after closing the switch? Again, since t=0, we rewrite the constitutive law of the inductor as
and we see that iL(t=0) =0. Thus by KCL, i(t=0) = 0
• State Variables: We will choose VC and iL to be the state variables.

• Thus, we will find equations of motion, of the form
• i'L = ...
V'C = ...
where the right hand side involves only no dynamic variables except the state variables iL and VC. To do this, we use the constitutive laws, continuity equations, loop sum rules, and of course the definition q' = i.

We will obtain:

• We can now put the state equations in matrix form by making a state two-vector x = ( iL, vC)T and a one-vector of inputs, B: x' = A x + B
• Of course, we can solve this system numerically via the same Euler method developed for two-state systems in previous domains. This is done by dividing time up into small time steps, Dt, and assuming that during each time step the currents and voltages persist at the values of the previous timestep. Using the notation that x with an arrow underneath represents the state vector and that the subscript i indicates time step number i, we obtain the update equation:

• So, we start from the initial conditions that VC(t=0) =0 and i(t=0) = 0 , calculate V'C and i' at time zero via the state equations, then use the update equations to obtain the values of VC and i at the first time step. Then we proceed to use the state equations and update equation in succession to move forward in time and determine the values of the state variables at each time step. . . . Easy!

What do you think the solution for this circuit will look like as a function of time? That is, how will the voltage in the capacitor and the current in the inductor vary with time? Remember that for a RC circuit, the capacitor charges up to the voltage of the battery in an exponential fashion. Once it reaches the voltage of the battery, the current in the system stops and there is a steady state. What about for this LC circuit?

Remember that in a steady state (V'C=0, i'L=0), the capacitor wants to have a constant voltage and zero current through it, while the inductor wants to have a zero voltage and constant current. These conflicting demands cause oscillatory behavior! This should not be too surprising if we make the analogy to mechanical systems: an LC circuit is analgous to a spring-mass system, which indeed will oscillate. Here is the solution for the following parameters: C=5*10^(-6); Vb=9; L=20*10^(-3).

So, what is happening?

• the switch is closed and current starts flowing from the battery into the system
• the inductor allows current to begin flowing slowly and to slowly increase (remember the current in the inductor cannot have discontinuous jumps)
• the capacitor begins charging and so voltage increases on the capacitor
• at time 0.5 seconds, note that the capacitor has attained the voltage of the battery (9v). The inductor has also achieved a constant current and zero voltage state, which is what it desires for steady state. So why does the system not stop here, like for RC circuits? At this point, the current is still flowing in the inductor and cannot stop instantly, which is what the capacitor desires for steady state. So the current in the inductor begins to decrease slowly, but the voltage in the capacitor continues to increase since current is still flowing through it.
• The current in the inductor reaches zero at 1 second, so the capacitor is happy with the zero current, but since di/dt is not zero current reverses direction and the capacitor begins discharging through the battery.
• The cycle contines in this fashion such that every time the steady state desire of one element is satisfied, that of other element is not. Hence oscillatory behavior.