TUTORIAL 2 - Inductor

Inductor 

 

An inductor is an element which stores a magnetic field. An inductor is
a wire coiled around a material called a core. The core is typically made of 
a magnetic material however the core can be anything from a toilet paper roll to a piece of wood.
(The behavior of the inductor IS effected by the core material though.)

Recall from Physics that static charge sets up an electrostatic force. Also recall that
moving charge sets up a magnetic field called flux around the current-carrying wire.

If the wire is then coiled as shown below, then the flux increases linearly with every coil turn.




The flux is proportional to the current flowing through the inductor. 
We can use the following equation where L is the inductance of the inductor.(L is constant)

Flux = Li

Through experimentation it was found that if the flux changes then a voltage is induced.
If the flux is constant then there is no voltage across the inductor.

V = d(Flux)/dt
V = d(Li)/dt


V = L di/dt
When the current is changing a voltage is induced across the inductor.
This voltage opposes the change in the current.
When the current stops changing and stays constant, then the voltage collapses.
Therefore when conditions are constant (unchanging) the inductor appears like a
wire or short circuit.

Current is continuous in an inductor. In other words, current can not
change instantaneously. If current could change instantaneously then an infinite
voltage would be induced across the capacitor. Infinite voltage is a practical impossibility. 
Therefore current can not change instantaneously through an inductor.
Voltage across an inductor can change instantaneously.

The energy stored in an inductor is:

E = 1/2 Li²

Using the above concepts, let's analyze the following circuit:

This circuit has both a switch and an inductor: The switch closes at t=0 The switch is open for t<0 and is closed for t>0. This can be seen by inspecting the switch's arrow direction. We will not try to analyze the circuit at t=0 since the circuit's state at t=0 is unkonwn. Instead we will look at the circuit at t=0- (the time right before the switch moves) and t=0+ (the time right after the switch moves). In this problem it is given that i6 at t=0- is 2A With this information find the following things at t=0- and t=0+: The current through the inductor iL The current through the 6 ohm, i6 diL/dt for the inductor

Here are snapshots of the circuit at t=0- and t=0+: _______________________________________________________ t = 0-: At t=0- the switch is open. Since we know that i6 is given to be 2A, we can apply KCL to the marked node to find iL. Σ  ientering = Σ  ileaving 5A = i6 + iL 5A = 2A + iL iL = 3A at t=0-. Now to find diL/dt: diL/dt = VL/L We need to find VL first: Apply KVL to the right-most loop: VL + 4iL - (2A)(6 Ohms) = 0 VL + 4(3A) - 12 = 0 VL = 12 - 12 VL = 0 diL/dt = VL/L = 0/2 = 0 _______________________________________________________ t = 0+: iL(0-) = iL(0+)  This is always true! Therefore iL(0+) = 3A Current through an inductor is continuous  We also need to fine i6: Note that the 6 Ohm resistor has been shorted out by the closed switch. A resistor in parallel with a wire has zero voltage across is. If the 6 Ohm resistor has zero volts across it then the current through the 6 Ohm has to be zero due to Ohm's Law, V=IR. Thus i6(0+) = 0 Now to find diL/dt: diL(0+)/dt = VL(0+)/L We need to find VL(0+) first: VL + 4iL - 0 = 0 VL + 4(3A) = 0 VL = -12 diL/dt = VL/L = -12/2 = -6A/s