Inductance

10.1.1  Definition of Inductance

We need a method for characterizing how well a device "holds" magnetic fields in much the same way as we needed the capacitance to characterize how well a device holds electric charge. For example, we know that we can study the properties of magnetic fields with a solenoid, just a bunch of coil loops of fixed radius and carrying a current. How do we rate one solenoid as more effective than another? The analog to capacitance for magnetic fields is the inductance. Just as capacitance, C = Q/V, tells us how much charge can be stored on the plates for a given voltage (and hence how much electric field can be produced for a given voltage between the parts of the capacitor), the inductance should tell us the amount of magnetic flux we get for a given current. Hence, the inductance (also referred to as the self-inductance ) of a circuit element is defined as

L = NFB/i

(10.1.1.1)

where N is the number of loops through which the magnetic field produced by current i threads. This means that, for any electrical device with inductance L, we must have

L i = NFB

(10.1.1.2)

Hence, the EMF of this device in any situation in which the current through it is changing with time is

|eL| = N dFB dt = d dt (L·i) = L di dt

(10.1.1.3)

By Faraday's Law, dF_B/dt gives rise to the EMF. We also note that, for any device with inductance L, the EMF is zero if the current through it is steady, i.e. not changing with time.

10.1.2  Qualitative description for RL circuits

We can immediately apply these ideas qualitatively to a circuit containing an inductor, a device with inductance L, a resistor R, and another source of EMF as shown in figure 10.1.

Figure 10.1: A circuit with an open switch, a battery, a resistor, and an inductor. This is an RL circuit.

Now we look at the qualitative behavior when the switch S  is set so that the battery begins to push current through the resistor and into the inductor. As soon as current appears at the first coil of the inductor, a magnetic flux is created. Since there was initially no current in the inductor (we assume), this new flux represents a sudden change of flux and therefore an EMF. This EMF pushes opposite to the EMF causing the flux in the first place, at least according to Lenz's Law.

Figure 10.2: The RL circuit with current flowing through the inductor. Note that the current i is only equal to e/R after a very long time.

This EMF from the inductor has a value which immediately opposes the current coming in. After the first bit of current makes it in, the change in flux as the current hits the next loop goes down, the EMF induced goes down, the current moves to the next loop and so on. Thus, we expect that the current in the inductor, and hence in the entire circuit, must increase over time until it reaches its maximum value of i_max = e/R where e is the voltage provided by the EMF source.

After a long time, we imagine that the circuit reaches equilibrium, i.e. a non-changing state, in which the current is at its maximum value and remains steady. If we then set the switch S so that it disconnects the battery from the circuit, but re-connects the inductor and the resistor, the current would tend to drop to zero. However, Faraday's Law prevents the current from going instantaneously to zero as this would involve an infinite rate of change in magnetic flux through the inductor coils. Instead, the inductor now becomes a source of EMF to push rather than oppose current flow through itself as shown in figure 10.3.

Figure 10.3: The switch S is set so as to disconnect the battery and continue the connection between the inductor and resistor.

The current decreases with time and energy is dissipated in the resistor. Thus the current decreases with time until it eventually reaches zero and the circuit returns to its original state of no current flow anywhere.

10.1.3  Quantitative description for RL circuits

We can add quantitative estimates to the qualitative picture by adding to Kirchoff's rules for circuits. The EMF induced in the inductor is -L(di/dt) as we discussed in the definition of inductance. The direction of this induced EMF is opposite to the change occurring as indicated by the minus sign in front of L. Therefore, for the RL circuit of figure 10.2, we have to have (going clockwise around the circuit starting from the EMF source)

e - iR - L di dt = 0

(10.1.3.4)

We can solve this differential equation using Maple. Note that to get a complete solution we have to use the condition that the current starts with a value of zero. The solution is:

i(t) = e R (1 - e-Rt/L) = imax(1 - e-Rt/L).

(10.1.3.5)

This equation has the qualitative features we expressed earlier, i.e. the current is initially (t = 0) at 0 and builds to a maximum of i_max after a long time (t ®¥).

What about for the second case in which the battery is removed from the circuit and only the connection between the inductor and resistor remains? We restart the clock so that t = 0  now corresponds to the switch S being changed to disconnect the battery. In this case, the direction of the inductor's EMF reverses so that it is now in the same direction as the current flow. Kirchoff's rules applied in the clockwise direction now states that

-L di dt - iR = 0.

(10.1.3.6)

The minus sign for the first term probably appears incorrect. We have to remember though that this is a differential equation. If we wish to treat the current as always positive (i.e. always going in the direction for which we evaluated Kirchoff's rules around the circuit), then we need to note that energy conservation demands that di/dt < 0 for all times t. Since di/dt is negative, -L di/dt yields a positive EMF as we expect. The solution to equation 6, again acquired through Maple with the initial condition being that i(0) = i_max, is

i(t) = imaxe-Rt/L.

(10.1.3.7)

We notice the same factor of Rt/L as for the charging case. We therefore describe the circuit as having a time constant, t_L, which is

tL = L R

(10.1.3.8)

We should take the lessons above for setting down some rules of thumb:

When charging: An inductor acts like a break in the circuit, i.e. it allows no current to pass at the instant an external EMF begins to push current through it. The inductor therefore immediately creates an EMF that opposes completely the external EMF attempting to push current through it. The inductor EMF can provide this canceling EMF for only an instant after current flow starts entering. The EMF of the inductor reduces to zero over time. After a long time of action by an external EMF, the inductor acts like a perfect conductor, offering no opposition to the flow of current. When discharging: An inductor allows no immediate change in the current flowing through its coils. It initially maintains the current going through it at the instant before the external EMF responsible for the current in the first place is disconnected. The inductor does this by becoming a source of EMF in the direction of the current flow through its coils. If energy is being dissipated or changed into a different form as the current flows, the inductor EMF will decrease with time until the current flow goes to zero.

Note also that these rules are the reverse of the behavior for a capacitor, initially uncharged, exposed to an external EMF source that attempts to push current onto its plates. It's important to understand why: capacitor plates accumulate charge which builds a potential difference that opposes current flow from depositing more charge. Inductors oppose any change in current. If you want, you can think of the capacitor as responding to the integral, with respect to time, of the current while the inductor responds to the derivative of the current with respect to time.

Energy in an Inductor

While the inductor is "discharging" it is sending current through the resistor which dissipates energy in the form of heat. Where does this energy come from? The only answer possible is: the inductor. So, during the time that current was being established in the inductor, we must also have been storing energy as well. The amount of energy stored can be determined from the definition of power input into the inductor.

PL = |eL|iL = L diL dt iL

(10.1.3.9)

where P_L is the power input to the inductor and i_L is the current through the inductor at any instant of time. To get the total energy stored in an inductor that has current i going through it, we perform an integration of the power over time.

UL = ó õ PL dt = ó õ t 0  Li¢ di¢ dt¢  dt¢ = L ó õ i 0  i¢ di¢ UL = 1 2 Li2 (10.1.3.10)

It is reasonable to suppose that this energy is stored in the magnetic field in analogy to our supposition that gravitational potential energy is "stored" in the gravitational field of the earth when an object is lifted vertically upward. Note that we use the same analogy in stating that energy can be stored in the electric field between the plates of a capacitor as the charge on the capacitor plates is increased. To find the relationship between the magnetic field magnitude and the energy stored, we use a solenoid, but the formula derived is quite general. For a solenoid with N coils and length l, we have

Bsolenoid = m0 N l i

(10.1.3.11)

The energy density inside the solenoid is

uB = UL volume = UL Area·l = Li2 2(Area)l = NFB i · i2 2(Area)l = N 2(Area)l (B·Area)i = N l B 2 Bl Nm0 uB = B2 2m0 (10.1.3.12)

This is an important result! We have shown that both the electric and magnetic fields can hold energy. We also now see clearly a mechanistic view of the fundamental constants e₀ and m₀. They figure into the amount of work necessary to establish an electric or magnetic field in a volume of space.

Example Problem for R-L circuit

An example is probably useful to lock in the quantitative description of R-L circuits.

Problem 1:

In the circuit in figure 10.4, the switch S  is closed at time t = 0. Find the power being delivered by the battery and the current through the resistors at the instant just after the switch S is closed and a long time after the switch is closed.

Figure 10.4: An inductor, EMF source, and two resistors. The switch closes at time t = 0.

Solution:

Just after the switch is closed, current flows through R₁. The same current must flow through R₂ since the inductor initially allows no current to flow through it. A long time after the switch is closed, the circuit reaches a steady state and the inductor gives no opposition to current flow, hence there is no current through resistor R₂ since the inductor acts like a short. The circuit appears, for these two times, as shown in figure 10.5.

Figure 10.5: The circuit shown in figure 10.4 just after and a long time after the switch S is closed.

The current through the resistors is therefore identical just after the switch closes and

i0 = e Req = e R1 + R2

(10.1.3.13)

Therefore, just after the switch closes, the battery outputs a power

Pe = i0·e = e Req e Pe = e2 R1 + R2 (10.1.3.14)

The current through the resistors a long time after the switch is closed finds the inductor acting like a perfect conductor and providing a short across R₂. Therefore, no current flows through R₂ and the current through R₁ is

it = e R1 .

(10.1.3.15)

The power output of the battery is

Pe = it e = e2 R1 .

(10.1.3.16)

Suppose you were crazy enough to want to know what the current through the inductor was between t = 0 and t = ¥? There is a general approach which will work. Apply Kirchoff's rules to the circuit for any time t. We get, in this case, two equations for the two obvious loops (shown in brown and magenta traces) in figure 10.6.

Figure 10.6: Finding the current through the resistors for any time t after the switch is closed.

e - i1R1 - L diL dt = 0 e - i1R1 - i2R2 = 0 (10.1.3.17)

We also have the current continuity equation.

i1 = iL + i2.

(10.1.3.18)

Now we have to do a little algebra

i1 = e - i2R2 R1 Þ iL + i2 = e - i2R2 R1 Þ iLR1 + i2R1 = e - i2R2 Þ i2(R1 + R2) = e - iLR1 Þ i2 = e - iLR1 R1 + R2 (10.1.3.19)

Now we can plug this back into the current continuity equation to get the relationship between i_L  and i₁.

i1 = iL + i2 = iL(R1 + R2) + e - iLR1 R1 + R2 i1 = e + iLR2 R1 + R2 (10.1.3.20)

Now we can go back to our first loop equation and transform it into a differential equation which we then solve.

e - i1R1 - L diL dt = 0 Þ e = i1R1 + L diL dt Þ L diL dt + eR1 + iLR1 R2 (R1 + R2) - e = 0 (10.1.3.21)

This equation depends only on i_L, the current through the inductor. We can appeal to Maple to solve it if we note that the initial condition is for the current through the inductor to be zero. The condition after a long time is for the current to be i_max, which, as we have already determined previously, must be e/R₁. Maple's dsolve command returns the solution as

iL(t) = e R1 [1 - e-R1R2t/(L(R1 + R2))]

(10.1.3.22)

Thus, the time constant looks like that of an equivalent circuit which has R₁ and R₂ in parallel. This may be a surprising result to you! We should check that the above result makes sense. Does it match our expectation? First, at t = 0, the current through the inductor from the above formula gives

iL(t = 0) = 0

(10.1.3.23)

which is what we expect since the inductor should initially carry no current. As t approaches infinity, the current in the inductor goes to

iL(t®¥) = e R1

(10.1.3.24)

again as we expect because now there is no current through R₂.

Send comments to larryg@upenn5.hep.upenn.edu.
This page was last modified on 03/23/2003 at 18:03:39 (EST).
Current date/time is Monday, 23-Nov-2009 15:00:15 EST
07088 hits since