# Inductive load switching (voltage transients)

#### Miguelec

Joined Mar 8, 2006
6
Please refer to the simple series RL circuit in Volume 1 (DC), Chapter 16 (RC and L/R time constants), section "Voltage and current calculations".

I would be grateful if someone could explain to me how the voltage varies across the inductor when the switch is opened (starting from the closed position, steady state condition).

I think I can see what the current will do: decrease exponentially from 15A to 0. But in what direction will the current flow? Towards the resistor or towards the battery? (Please state what convention for current direction you are using).

The voltage drop across the inductor is initially zero. When the switch is opened, how does the voltage vary with time at the following four different points:

a) the point between the battery and the switch
b) the point between the switch and the resistor
c) the point between the resistor and the inductor
d) the point between the inductor and the battery

Any help will be very much appreciated. Thank you!

Miguel

#### paultwang

Joined Mar 8, 2006
80
Originally posted by Miguelec@Mar 9 2006, 06:14 AM
Please refer to the simple series RL circuit in Volume 1 (DC), Chapter 16 (RC and L/R time constants), section "Voltage and current calculations".

I would be grateful if someone could explain to me how the voltage varies across the inductor when the switch is opened (starting from the closed position, steady state condition).
If you open the switch, it ceases to be a complete circuit - no current flow.

Also note that voltage is potential - You must have two different points to have a voltage.

You can't have voltage at a single point, unless you have a common reference point to compare to.

#### Papabravo

Joined Feb 24, 2006
18,408
Originally posted by Miguelec@Mar 9 2006, 08:14 AM
Please refer to the simple series RL circuit in Volume 1 (DC), Chapter 16 (RC and L/R time constants), section "Voltage and current calculations".

I would be grateful if someone could explain to me how the voltage varies across the inductor when the switch is opened (starting from the closed position, steady state condition).

I think I can see what the current will do: decrease exponentially from 15A to 0. But in what direction will the current flow? Towards the resistor or towards the battery? (Please state what convention for current direction you are using).

The voltage drop across the inductor is initially zero. When the switch is opened, how does the voltage vary with time at the following four different points:

a. the point between the battery and the switch
b. the point between the switch and the resistor
c. the point between the resistor and the inductor
d. the point between the inductor and the battery

Any help will be very much appreciated. Thank you!

Miguel
[post=14807]Quoted post[/post]​
With respect to the negative terminal of the battery assuming the inductor had no dynamic behavior
a. The battery and the switch are always at the same potential because the R between them is zero and any current times 0 ohms is 0 volts of drop. No drop -- No change.
b. The point between the switch and the resistor wants to decay from the battery voltage toward zero volts since there is no path for current. No current -- no voltage drop.
c. The point between the resistor and the inductor also wants to decay toward zero for the same reason as point b.
d. The negeative terminal of the battery stays at ground with respect to itself. There is no resistance from ground to ground, and any current times 0 ohms is 0 volts for no drop -- no change.

The problem with all of this is that it ignores they dynamic behavior of an inductor. Understanding inductors and capacitors involves a bit of calculus. As you have observed the voltage drop across an inductor in a steady state(nothing is changing) is zero. The expression for the voltage drop across an inductor is

Rich (BB code):
      di
V = L * --
dt
If our conventions is that current is positive when it flows from the positive terminal of the battery toward the negative terminal of the battery, and
L is the inductance in Henries, and di/dt is the time rate of change of the current. Then this time rate of change of the current will be negative since there is an exponential decay. The switch will open in say 50 milliseconds.
Lets say the current goes from 15 A to 0 A in this time, then with a 1H inductor the voltage across the inductor will be - 300V with respect to ground. To be really precise this average decay over the 50 millisecond interval is not really what is happening. di/dt will start at zero increase rapidly to some maximum value and head for zero.

What is happening is that all the energy stored in the inductor's magnetic field is converted to a big negative voltage(potential) when the switch is opened. This big negative potential is then disippated because there is still a DC path to ground. This is a transient condition and it will persist for only a short time.

With this in mind points a. and d. still behave as described above.
Points b. and c. briefly go to large negative values and then decay toward zero because the collapsing magnetic field of the inductor cannot support sutained current flow.

Hope this helps

#### Miguelec

Joined Mar 8, 2006
6
Thank you for both replies. It does help. I think I understand.

So, using the same convention for current (positive when it flows from positive to negative), it is interesting that, albeit for a short time, you actually have a positive current flowing from a negative potential (upstream of the inductor) to a more positive potential (DC ground o negative battery terminal). Right?

Thank you once again.

#### Papabravo

Joined Feb 24, 2006
18,408
Originally posted by Miguelec@Mar 10 2006, 06:54 AM
Thank you for both replies. It does help. I think I understand.

So, using the same convention for current (positive when it flows from positive to negative), it is interesting that, albeit for a short time, you actually have a positive current flowing from a negative potential (upstream of the inductor) to a more positive potential (DC ground o negative battery terminal). Right?

Thank you once again.
[post=14838]Quoted post[/post]​
Right. It can also happen in AC circuits when the sinusoidal voltage and current are out of phase. At any point in the circuit with respect to ground you can have all four relationships of arithmetic sign. (+,+), (+,-), (-,+), and (-,-).

The signs of voltages and currents tend to confuse people. The equations of KVL and KCL do not uniquely determine the artihmetic signs of voltages and currents they only require that they be consistent. Once a convention is chosen the remainder are determined.

Another way to think about this is that there are two kinds of things which are carriers of current. One of them flows in one direction and the other flows in the opposite direction. The flows are equal and of opposite sign. When we measure current with an ammeter we are only measuring one type of flow and not both. That does not matter because the other one is just equal and opposite. What does matter is consistency. Once we decide which one we are working with we have to stick with that decision.

#### Miguelec

Joined Mar 8, 2006
6
Thanks! I have been reading more about this subject, including arc phenomena in relays and arc suppression methods (including a series resistor and capacitor across the switch or a shunt diode across the load).

I am interested in the case where no arc suppression is used. Here the energy stored in the inductor's magnetic field is dissipated in the arc across the switch contacts. My guess would be that the current still flows from the battery positive terminal to the negative terminal.

However, in some of the literature I have read it seems like the current suddenly changes direction (see for example figure 4 of the attachment) and then decays exponentially to zero.

This is consistent with the idea that the energy is transferred from the inductor back to the switch (which is then dissipated in the arc) but is not consistent with the idea that the inductor will try to keep the current flowing through it constant, and therefore in the same direction.

I would be very grateful for your thoughts.