Revision of Circuits and Electronics - Inductors and First-Order Circuits (Lecture 13)

Video and lecture notes:- edX  MITx: 6.002.2x Circuits and Electronics 2:  week 3



Capacitors and inductors are both energy storage devices.  They are duals of each other. The structure of the inductor is a core  with a magnetic permeability  μ and a coil of wire  wrapped around it.

 Let's say a current i is flowing through the wire, there is a voltage between the two terminals of the wire, there are N turns  of wire , the area of the core is A and the circumference of the core is l.

Let's say that it has a total magnetic flux linked  to it called λ.  So when  a current is passed through this coil of wire, there is a magnetic flux that links with each of these turns of the wire. And add them all up,  and  a total flux linked called λ is obtained

So the inductance of this is given by (μ N^2 A)/ l.    This is quite similar to the equation of capacitance C=eA/d.  

For the inductor, there is a relationship between the magnetic flux λ,  inductance and current  where  λ=Li.  (Units for  λ is Webers and units for L is Henries). This is quite similar to the equation of q=cv for the capacitor.  

From Maxwell's equations, v=dλ/dt = L di/dt.  This is similar to C dv/dt for capacitors.


i can be   expressed in terms of v.  i is   the integral of v dt multiplied by 1/L with  the limits between t and  negative infinity.


Power can also be computed. P=vi= iLdi/dt). From integration, E=(L*i^2)/2. This is similar to
 E=(c*v^2)/2 for the capacitor. This is an important equation as it proves that the inductor is an energy storage device.  This also makes it a memory device. It stores energy and that energy it stores, remembers what had happened before.


Inductor L is connected to a  the voltage source v(t). The current flowing into the inductor is i(t). The initial current flowing in the inductor is i(t_o).

We need to pick some voltage-a pulse  which has a height of V for some amount of time T.  We need to find i(t) for this input voltage.

To do that, we need to use the integral ofv dt multiplied by 1/L with  the limits between t and  negative infinity. By splitting the integral into several parts,  the graph  of i(t) versus t can  be drawn.  The height during the time T  is VT/L.  This is similar to what was obtained  for the capacitor   IT/C.

Inductors  like to  hold on to  the same current.  If  a large current is pumped into the inductor, the inductor looks  like an instantaneous open  and it could breakdown.


Analyzing RL circuits, a Norton equivalent type of circuit is used where a current source, a resistor and an inductor  are connected in parallel. Applying the node method and rearranging, we will get a first order differential equation  L/R (di_L/dt) +i_L=i_I.

 For the  next step, a solution to this equation, can be obtained by inspection based on the solution obtained for the capacitor circuit. The initial conditions of the inductor is i_L(0)=I_0  and the  input source i_I(t)= I_I. So now we will have a differential equation   L/R (di_L/dt) +i_L=I_I.  The time constant is L/R.

If we base the solution of the inductor with that of the capacitor, the equation obtained will be  i_L=I_I+(I_0 - I_I)e^(-t/τ)   where time constant  τ = L/R