Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

How to calculate the current in the inductor~~~~~~

Status
Not open for further replies.

ZengLei

Full Member level 1
Joined
Jan 24, 2006
Messages
99
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Location
WuHan China PR
Activity points
2,143
fisrtly the switch is short,i know the current in the inductors are 5A

and then the switch is open,so i got confused about calculating the current in the circuit~~~~~~,how to calculate it???any idears???
Thanks a lot~!
 

Sal

Full Member level 4
Joined
Nov 29, 2005
Messages
237
Helped
53
Reputation
106
Reaction score
13
Trophy points
1,298
Activity points
3,034
Hi

After the switch is open, it can be considered L1 and L2 as current sources, then just apply superposition and transient analysis to the remaining circuits. This.
1. Assuming that the internal resistance of the inductances is negligible (0 ohms), the initial current in both branch is the same and equal to 5A.
2. Consider L1 as a short circuit. then by transient analysis the current is I2(t) = 5*exp(-t/tau2), where tau2 = L2/(R1+R2).
3. Do the same to L2 and get I1(t)=-5*exp(-t/tau1), where tau1 = L1/(R1+R2). The (-) in I1 is arbitrary, just to show the current direction is opposite to I2(t).
4. Add I2(t)+I1(t) and get I(t).

If you think this info is useful click on the "helped me" button for statistics

Sal
 

mateuko

Junior Member level 1
Joined
Mar 22, 2007
Messages
16
Helped
2
Reputation
4
Reaction score
0
Trophy points
1,281
Activity points
1,403
Sal said:
Hi

After the switch is open, it can be considered L1 and L2 as current sources, then just apply superposition and transient analysis to the remaining circuits. This.
1. Assuming that the internal resistance of the inductances is negligible (0 ohms), the initial current in both branch is the same and equal to 5A.
2. Consider L1 as a short circuit. then by transient analysis the current is I2(t) = 5*exp(-t/tau2), where tau2 = L2/(R1+R2).
3. Do the same to L2 and get I1(t)=-5*exp(-t/tau1), where tau1 = L1/(R1+R2). The (-) in I1 is arbitrary, just to show the current direction is opposite to I2(t).
4. Add I2(t)+I1(t) and get I(t).

If you think this info is useful click on the "helped me" button for statistics

Sal
Without calculating we can see that i(t) = 5*exp(-t/tau) where tau is 1H/2ohm (in this case, we can substitute two inductors by one of inductance 2H-1H=1H). We have interesting case when we assume that both inductances are identical (pure theoretical case). In this case the current i(t) = 0 and the voltages on inductors are Dirac's pulses.
 

ZengLei

Full Member level 1
Joined
Jan 24, 2006
Messages
99
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Location
WuHan China PR
Activity points
2,143
Hi Sal
i'm afriad i will not click on the "helped me" button,because u are incorrect:

EN~~~~

YES mateuko,you're right,voltages on inductors are Dirac's pulses.!!!But the current is not 0Ampere

Yes ,The current in the inductor will not change instantaneously,But it's not always correct,we could see in the formula:

v=Ldi/dt
so we have :(1/L)*∫udt + i(0)=i(t)

we can know if the u is a impulse the current in the inductor will change instantaneously!!!!!

Then we turn to the curcuit above ,On that conditon the current will change!(i read it in our book,and our teach tell me that on that condition the current will change!!!!)

we should calculate like this(Assuming that there is no mutual coupling between the individual inductors):

L1*i1-L2*i2=(L1+L2)*i
so we get :i=(1*5-2*5)/(1+2)=3/5!!!!!!

i don't know how this comes from:
L1*i1-L2*i2=(L1+L2)*i
 

the_edge

Full Member level 3
Joined
Feb 18, 2007
Messages
181
Helped
18
Reputation
36
Reaction score
5
Trophy points
1,298
Location
Belgrade, Serbia
Activity points
2,264
L1*i1-L2*i2=(L1+L2)*i
so we get :i=(1*5-2*5)/(1+2)=3/5!!!!!!

i don't know how this comes from:
L1*i1-L2*i2=(L1+L2)*i

This is not good... You are trying to multiply Amperes and Henryes, you can't do that....



Here is complete solution....
 

IBO

Full Member level 3
Joined
Mar 1, 2007
Messages
153
Helped
2
Reputation
4
Reaction score
1
Trophy points
1,298
Activity points
1,980
adding a point current doesnt change instantly in an inductor ( from di/dt term)
 

ZengLei

Full Member level 1
Joined
Jan 24, 2006
Messages
99
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Location
WuHan China PR
Activity points
2,143
Hi~Guys!!!!!

who said that from di/dt term the current in the inductor will not change???

according to the Calculus, if there is a impulse voltage,the current will change~~~~~

the_edge:

"This is not good... You are trying to multiply Amperes and Henryes, you can't do that...."

L*I=magnetic flux!!

maybe u're wrong from the beginning,u said I1=I2=5A,
and i think I1=I2!=5A~~~~~~~~~
 

the_edge

Full Member level 3
Joined
Feb 18, 2007
Messages
181
Helped
18
Reputation
36
Reaction score
5
Trophy points
1,298
Location
Belgrade, Serbia
Activity points
2,264
L*I=magnetic flux!!
Sorry, you are right, but thi is not important for this problem, you dont need this.


maybe u're wrong from the beginning,u said I1=I2=5A,
and i think I1=I2!=5A~~~~~~~~~

When the switch is closed, when you calculate bias current then I1=I2=5A... This is used as the initial condition for open switch calculation...
I1=I2=5A is due to resistances, as the resistance of L1 and L2 is 0ohms and R1=R2=1ohm... So, the equivalent resistance of this circuit is 0.5ohms as the R1 and R2 are in paralell... Then the current tru switch is Isw=5V/0.5ohm=10A. As R1=R2 => I1=I2=5A... And this is right...

Or other way, I1=5V/R1=5A, I2=5V/R2=5A... This is easyer way...


The rest of solution is good too, exept maybe the solution of that differential equation as I said that I am not good at solveing them. If you simulate this circuit in PSpice or other, you will see.... I can send you the results of simulation if you like....
 

ZengLei

Full Member level 1
Joined
Jan 24, 2006
Messages
99
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Location
WuHan China PR
Activity points
2,143
yes,i know the initial current is 5A in L1 and L2,

But when the switch is open,the current with 5A can't be the real initial current anymore,we should calculate the real value of the initial current in the circuit,When the switch is open ,the four elements are in series with each other,~~~~
the way to calculate the current is based on the theory:The flux in both L1 L2 will keep as a constent in this moment!~!!

So we must calulate as:L1*i1-L2*i2=(L1+L2)*i
after that we could calculate as the normal way ~~~

BTW:i'm not well up in english~~so ````~~~
MAYBE IT'S a bit difficult for us to talk~~~~~~

Added after 11 minutes:

ADD:
SWITCH CLOSEd:I1=I2=5A
SWITCH OPEN:the current will change at this moment ! the flux stored in the two inductors will re-distribute but the value of flux will keep a constent at this moment~~,so the current will change,and then there will be a Dirac's pulse voltage between the two inductors each~~~~

so the current will be (2*5-1*5)/3=5/3A~~~~

Added after 3 minutes:

the_dage:
i'm very hunger for the results of simulation you send me

Thanks~~~~~~~~~~~
 

the_edge

Full Member level 3
Joined
Feb 18, 2007
Messages
181
Helped
18
Reputation
36
Reaction score
5
Trophy points
1,298
Location
Belgrade, Serbia
Activity points
2,264
Circuit:
0_1175509478.jpg


Current tru L1:


Current tru L2:



It looks like you were right about initial condition, they are 5/3A... After a couple minutes of thinking I realised that it is because of flux... From the picture it looks like the rest of my calculation is good, but I am not so sure about it any more...






BTW:i'm not well up in english~~so ````~~~
MAYBE IT'S a bit difficult for us to talk~~~~~~
Dont mind this, your English is better than mine....
 

    ZengLei

    Points: 2
    Helpful Answer Positive Rating

ZengLei

Full Member level 1
Joined
Jan 24, 2006
Messages
99
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Location
WuHan China PR
Activity points
2,143
So,we have a common idear that the flux will keep as a constent,the current is 5/3A~~

My questionis why the flux will keep as a content at the momnet of opening the switch????
 

the_edge

Full Member level 3
Joined
Feb 18, 2007
Messages
181
Helped
18
Reputation
36
Reaction score
5
Trophy points
1,298
Location
Belgrade, Serbia
Activity points
2,264
As the inductance is reactive element, it memorize flux. And it cannot be changed momentarily, so flux(0-)=flux(0+). This is analog to capacitors who memorize voltage.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top