inductive circuit question

S

stephen hallacy

Guest
Hello,

I decided to review some electronics in my spare time. I came across
an inductor circuit in my electric circuit's text book. The circuit
is given below.

-----------R1 (7.5k)-------------
| |
| |
----L1 (1.25H)--------L3 (6H)----
| |
----L2 (10H)---
|
\\\

The initial currents for L1, L2 and L3 are 2A, 2A and 0A respectively,
and the currents are going from right to left in circuit shown above.
I went through the math and came up with

i1(t) = 2*exp(-1500*t)
i2(t) = 5/4+3/4*exp(-1500*t)
i3(t) = -5/4+5/4*exp(-1500*t)}

where i1(t) is the current in L1, i2(t) is the current in L2 and i3(t)
is the current in L3.

I used the following differential equations to come up with the
solution:

d(i1(t))/dt - d(i2(t))/dt - d(i3(t))/dt = 0
10*d(i2(t))/dt - 6*d(i3(t))/dt = 0
10*d(i2(t))/dt + 1.25*d(i1(t))/dt + 7500*i1(t) = 0

The currents should go to zero, however, the equations show that i2(t)
and i3(t) approach 1.25 Amps. Am I doing something wrong?

By the way, OrCAD Pspice showed the same result.

Regards,
Stephen
 
stephen hallacy wrote:
Hello,

I decided to review some electronics in my spare time. I came across
an inductor circuit in my electric circuit's text book. The circuit
is given below.

-----------R1 (7.5k)-------------
| |
| |
----L1 (1.25H)--------L3 (6H)----
| |
----L2 (10H)---
|
\\\

The initial currents for L1, L2 and L3 are 2A, 2A and 0A respectively,
and the currents are going from right to left in circuit shown above.
I went through the math and came up with

i1(t) = 2*exp(-1500*t)
i2(t) = 5/4+3/4*exp(-1500*t)
i3(t) = -5/4+5/4*exp(-1500*t)}

where i1(t) is the current in L1, i2(t) is the current in L2 and i3(t)
is the current in L3.

I used the following differential equations to come up with the
solution:

d(i1(t))/dt - d(i2(t))/dt - d(i3(t))/dt = 0
10*d(i2(t))/dt - 6*d(i3(t))/dt = 0
10*d(i2(t))/dt + 1.25*d(i1(t))/dt + 7500*i1(t) = 0

The currents should go to zero, however, the equations show that i2(t)
and i3(t) approach 1.25 Amps. Am I doing something wrong?
Click to expand...
Put some small resistance in series with each inductor and try again.
That loop composed entirely of two ideal inductors may be giving
you strange results.
 
On 17 Oct 2003 21:49:05 -0700, steven4443@yahoo.com (stephen hallacy)
wrote:

Hello,

I decided to review some electronics in my spare time. I came across
an inductor circuit in my electric circuit's text book. The circuit
is given below.

-----------R1 (7.5k)-------------
| |
| |
----L1 (1.25H)--------L3 (6H)----
| |
----L2 (10H)---
|
\\\

The initial currents for L1, L2 and L3 are 2A, 2A and 0A respectively,
and the currents are going from right to left in circuit shown above.
I went through the math and came up with

i1(t) = 2*exp(-1500*t)
i2(t) = 5/4+3/4*exp(-1500*t)
i3(t) = -5/4+5/4*exp(-1500*t)}

where i1(t) is the current in L1, i2(t) is the current in L2 and i3(t)
is the current in L3.

I used the following differential equations to come up with the
solution:

d(i1(t))/dt - d(i2(t))/dt - d(i3(t))/dt = 0
10*d(i2(t))/dt - 6*d(i3(t))/dt = 0
10*d(i2(t))/dt + 1.25*d(i1(t))/dt + 7500*i1(t) = 0

The currents should go to zero, however, the equations show that i2(t)
and i3(t) approach 1.25 Amps. Am I doing something wrong?

By the way, OrCAD Pspice showed the same result.

Regards,
Stephen
Click to expand...
L2 and L3 can indeed circulate a current forever.

John
 
On Sat, 18 Oct 2003 15:57:09 -0700, "Baphomet"
<fandaDEATH2SPAMMERS@catskill.net> wrote:


L2 and L3 can indeed circulate a current forever.

Only if you believe in perpetual motion ;-) By the way, I have this
bridge........
Click to expand...

Ever heard of NMR? MRI? They both use big superconductive magnets.
They are shipped warm, and cooled down on-site with liquid nitrogen,
then liquid helium. Once it's cold, a big power supply is connected to
run the current up to a few kiloamps, then a shorting link is closed,
and the power supply is loaded on a truck and taken away. The current
circulates roughly forever, and the magnetic field is stable [1] as
long as the helium level is kept up. Helium is topped off every few
months to correct for boiloff.

John

[1] high-field superconductive magnets are subject to a bit of
pinned-domain slipping, which causes the field to droop very slightly
with time, generally in the 1 PPM per day sort of range. I think this
eventually levels out as things settle.

These things are hell on credit cards and hand tools.
 
"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:3un2pv42u2bta0i3okhna6016n5asr6t85@4ax.com...
On 17 Oct 2003 21:49:05 -0700, steven4443@yahoo.com (stephen hallacy)
wrote:

Hello,

I decided to review some electronics in my spare time. I came across
an inductor circuit in my electric circuit's text book. The circuit
is given below.

-----------R1 (7.5k)-------------
| |
| |
----L1 (1.25H)--------L3 (6H)----
| |
----L2 (10H)---
|
\\\

The initial currents for L1, L2 and L3 are 2A, 2A and 0A respectively,
and the currents are going from right to left in circuit shown above.
I went through the math and came up with

i1(t) = 2*exp(-1500*t)
i2(t) = 5/4+3/4*exp(-1500*t)
i3(t) = -5/4+5/4*exp(-1500*t)}

where i1(t) is the current in L1, i2(t) is the current in L2 and i3(t)
is the current in L3.

I used the following differential equations to come up with the
solution:

d(i1(t))/dt - d(i2(t))/dt - d(i3(t))/dt = 0
10*d(i2(t))/dt - 6*d(i3(t))/dt = 0
10*d(i2(t))/dt + 1.25*d(i1(t))/dt + 7500*i1(t) = 0

The currents should go to zero, however, the equations show that i2(t)
and i3(t) approach 1.25 Amps. Am I doing something wrong?

By the way, OrCAD Pspice showed the same result.

Regards,
Stephen

L2 and L3 can indeed circulate a current forever.
Click to expand...
Only if you believe in perpetual motion ;-) By the way, I have this
bridge........
 
"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:gga3pv0tde6uqfr2dd745qsa3uoom5q87k@4ax.com...
On Sat, 18 Oct 2003 15:57:09 -0700, "Baphomet"
fandaDEATH2SPAMMERS@catskill.net> wrote:


L2 and L3 can indeed circulate a current forever.

Only if you believe in perpetual motion ;-) By the way, I have this
bridge........



Ever heard of NMR? MRI? They both use big superconductive magnets.
They are shipped warm, and cooled down on-site with liquid nitrogen,
then liquid helium. Once it's cold, a big power supply is connected to
run the current up to a few kiloamps, then a shorting link is closed,
and the power supply is loaded on a truck and taken away. The current
circulates roughly forever, and the magnetic field is stable [1] as
long as the helium level is kept up. Helium is topped off every few
months to correct for boiloff.
Click to expand...

Point well taken John...but as long as energy from an outside source,
however infrequently, must be supplied, it ain't perpetual...pretty darn
close though ;-)


high-field superconductive magnets are subject to a bit of
pinned-domain slipping, which causes the field to droop very slightly
with time, generally in the 1 PPM per day sort of range. I think this
eventually levels out as things settle.

These things are hell on credit cards and hand tools.
Click to expand...
 
John Larkin <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in message news:<3un2pv42u2bta0i3okhna6016n5asr6t85@4ax.com>...
On 17 Oct 2003 21:49:05 -0700, steven4443@yahoo.com (stephen hallacy)
wrote:

Hello,

I decided to review some electronics in my spare time. I came across
an inductor circuit in my electric circuit's text book. The circuit
is given below.

-----------R1 (7.5k)-------------
| |
| |
----L1 (1.25H)--------L3 (6H)----
| |
----L2 (10H)---
|
\\\

The initial currents for L1, L2 and L3 are 2A, 2A and 0A respectively,
and the currents are going from right to left in circuit shown above.
I went through the math and came up with

i1(t) = 2*exp(-1500*t)
i2(t) = 5/4+3/4*exp(-1500*t)
i3(t) = -5/4+5/4*exp(-1500*t)}

where i1(t) is the current in L1, i2(t) is the current in L2 and i3(t)
is the current in L3.

I used the following differential equations to come up with the
solution:

d(i1(t))/dt - d(i2(t))/dt - d(i3(t))/dt = 0
10*d(i2(t))/dt - 6*d(i3(t))/dt = 0
10*d(i2(t))/dt + 1.25*d(i1(t))/dt + 7500*i1(t) = 0

The currents should go to zero, however, the equations show that i2(t)
and i3(t) approach 1.25 Amps. Am I doing something wrong?

By the way, OrCAD Pspice showed the same result.

Regards,
Stephen

L2 and L3 can indeed circulate a current forever.

John
Click to expand...

Thanks for the response guys.

I agree John. For this ideal component circuit, the current circulates forever.

Stephen
 
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