# Is i1 equivalent to i2 when the switch opens?

Discussion in 'Homework Help' started by raddian, Jun 20, 2015.

Jun 20, 2015
6
0

I understand that i1 is not equal to i2 when the switch is closed. So I'm a little confused at what the problem is asking.

I was able to get

-12 + Ri + L di/dt = 0

di/dt + (R/L) i = 12/L

where i represents i1 when the circuit is open.

Last edited: Jun 20, 2015
2. ### WBahn Moderator

Mar 31, 2012
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If the switch is open, then how does the 12V enter into it? What is R? You have two resistors and we are not mind readers.

To answer the question in the thread title, i1 can't be equal to i2 after the switch is open because that would require the top center junction to be becoming increasingly negatively charged. But by KCL you know that i1 = -i2. Minus signs matter!

3. ### MrAl Well-Known Member

Jun 17, 2014
2,433
490
Hi,

This isnt that confusing is it?

When the switch is closed, i1 is equal to 1 amp, and i2 is equal to 2 amps.
When the switch opens, i2 is equal to -i1 and for the first instant that would be -1 amps.
After that i1 decreases exponentially and i2=-i1, and you should be able to find the equation right?

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,387
497
When the switch is closed, you have two meshes. Therefore two currents.

When the switch is open, you have one mesh. Therefore one current. So yes, when switch is open, i1=i2.

5. ### WBahn Moderator

Mar 31, 2012
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If i1=i2 when the switch is open, then there are a lot of electrons piling up somewhere and a lot of electrons being robbed from somewhere else.

6. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,387
497
When the switch is open, the inductor will start to discharge.

Jun 20, 2015
6
0
I assumed the 12V was stored as energy from the inductor, similar to how a capacitor stores energy. This, i think, is not right, but I'm not sure why. R is 18 Ohms.

I think I'm starting to understand.

So when the switch opens, I should apply KVL around the loop.

I have


\begin{align}i_1 * 12 + L \frac{di_1}{dt} + i_2 * 6 = 0 \\

i_2 = -i_1 \\

12i_1 + -i_1 *6 +L \frac{di_1}{dt} = 0 \\

L \frac{di_1}{dt} + 6i_1 = 0 \\

\frac{di_1}{dt} + \frac{6i_1}{L} = 0 \\
\end{align}

And the solution is the form

$
i(t) = i_o e^{\frac{-Rt}{L}}

i_1(t) = 1A * e^{\frac{-18t}{2}}
$

Thank you all!

8. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
Yes. So? That doesn't make i1 = i2.

For Pete's sake, look at the diagram!

9. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
A capacitor stored energy as a voltage, but an inductor stores energy as a current. This, in this problem, the current in the inductor just before the switch opens is the same as the current in the inductor just after the switch opens.

I'll look at the rest in a little bit when I get back to my computer.

10. ### Ramussons Active Member

May 3, 2013
557
92
i1 = i2 @ t=0

unless we are going to discuss sparking at the switch @ t=0

11. ### WBahn Moderator

Mar 31, 2012
17,743
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So why does KCL not hold on that node at t=0+?

Write KCL for that node after the switch is open.

You have i1 + i2 = 0.

If, as so many people are claiming, i1 = i2, then the only possible value for either is zero. But if i1 = zero at t=0+, then you are violating the requirement that the current in a conductor be continuous per conservation of energy.

12. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
Attention to detail. Attention to detail. Attention to detail. That is what so many people in this thread simply don't seem to be able to get a handle on.

You appear to want to sum up the voltage drops as you go around the loop clockwise (based on your first term). Fine, then do that.

As you go clockwise through the 12 Ω resistor the voltage drop is

$
V_{12 \Omega } \; = \; $$i_1$$ $$12 \Omega$$
$

As you go clockwise through the 2 H inductor the voltage drop is

$
V_{2 H} \; = \; $$2 H$$ $$\frac{d}{dt}i_1$$
$

As you go clockwise through the 6 Ω resistor the voltage drop is

$
V_{6 \Omega} \; = \; -$$i_2$$ (6 \Omega)
$

I said it all the way back in Post #2 -- Minus signs matter!!!!

The sum of these has to be zero, by KVL

$
V_{12\Omega} \; + \; V_{2 H} \; + \; V_{6 \Omega} \; = \; 0
\;
$$i_1$$$$12 \Omega$$ \; + \; $$2 H$$ $$\frac{d}{dt}i_1$$ \; + \; -$$i_2$$ (6 \Omega) \; + \; 0
$

Yes, despite what several other people that seem to be struggling with KCL are saying.

Nope, because you didn't pay attention to signs when you set up the problem.

Then, how do you go from this

to this

You have no link from one step to the next. How does that 6 in the first equation magically become an 18 in the second? This is magical methods -- and it gets my students a negative grade on the entire assignment.

Jun 20, 2015
6
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I realize I did make a few mistakes. I'll correct them starting from here.
$
$$i_1$$$$12 \Omega$$ \; + \; $$2 H$$ $$\frac{d}{dt}i_1$$ \; + \; -$$(-i_1)$$ (6 \Omega) \; = \; 0
\;
$$i_1$$$$12 \Omega + 6 \Omega$$ \; + \; $$2 H$$ $$\frac{d}{dt}i_1$$ \; = \; 0
\;
$$i_1$$$$18 \Omega$$ \; + \; $$2 H$$ $$\frac{d}{dt}i_1$$ \; = \; 0
$

Now, in my book, if the differential equation is in the form :

$
L \frac{di}{dt} + R i = V_s
$

then the solution to the differential equation is:

$
i(t) = \frac{V_s}{R} + e^{\frac{-Rt}{L}}
$

(from the textbook)

In this case,
V_s = 0,
R = 18 Ohms,
L = 2 H,
and i_0 is the initial value of i1 when t=0-, which is 1 A.

Plugging all this in yields

$
i(t) = 0 + 1e^{\frac{-18 \Omega t}{2H}} = 1e^{-9t (s^{-1})}
$

Last edited: Jun 21, 2015
14. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
Definitely making progress.

Hopefully you see that the differential equation you had before led to a different time constant.

But now look at that result for i(t) and ask if it makes sense. The first term is Vs/R which is a current. That's good. But the second term is just a number and it is a number that starts at 1 and then decays to zero. Does that make sense? To be more blunt, at t=0 you are saying that the current is

$
i(t=0) \; = \; \frac{V_s}{R} + 1
$

and after a long time you are saying that

$
i(t=\infty) \; = \; \frac{V_s}{R}
$

Do either of those make sense?

15. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
I strongly recommend that you take the time to solve this differential equation for yourself instead of relying on what your book says the answer is. It's one thing to rely on a provided form of the solution WHEN you fully understand WHERE that form comes from. It's quite another to just accept it because it's there.

Jun 20, 2015
6
0
It makes sense only assuming that V_s = 0. I think this is true BECAUSE the voltage source is DISCONNECTED when the switch is open. In other words, The solution to the diff eq stems from the fact that it was an RL circuit, connected to a DC voltage source. Because the source is disconnected when the switch opens, V_s is zero. Is this the right?

When V_s = 0, the current going through i(t=0) will be 1 A. I assume there is a current of +1 A because the inductor is a "voltage source" because it stores magn. energy. Ohm's Law relates the voltage of this inductor with the current passing through the circuit. As t reaches infinity, the energy stored in the inductor will be dissipated by the resistors, causing the inductor to lose all its energy, and thus providing no voltage. Ohm's Law affirms that no voltage = no current. Thus i(t=inf) = V_s/R = 0/R = 0.

17. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
It doesn't make sense at all! The 1 is NOT a current! You CAN NOT add 1 to a Vs/R. You can only add quantities that have the same units. If your work results in an expression that would add two quantities that do not have the same units, then you KNOW that your results are wrong!

18. ### Feng Yu New Member

Jun 20, 2015
5
0
When t<0,i1=12/12=1 A
Since the i1 is the current of the L, i1 (t<0) = i1 (t=0+)
When t>0 ,i1=0. find the time constant =2/(12+6)=1/9
Then you can write,(i1 (t=0+)- 0)*exp(-t/(1/9)) + 0 = 1*exp(-9t)
----------------------
when the moment switch open(t=0+),the inductor could be regarded as a current generator with current =i1,and i1 = -i2.
when t tends to infinity(t>0),i1 becoms smaller and smaller,but you always have i1=-i2.

19. ### Ramussons Active Member

May 3, 2013
557
92
So you have a spark at the switch points @ t=0.

20. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
Why would there be a spark? The inductor has not been open-circuited.

LOOK AT THE DIAGRAM!!!!

The currents i1 and i2 are defined to be in OPPOSITE DIRECTIONS!!!!!!!!!

BOTH i1 and i2 are flowing AWAY from the top center node!

APPLY KCL to that node when the switch is open.

i1 + i2 = 0

That means that

i1 = -i2

This is VERY different than i1 = i2. Like I said in Post #2, minus signs matter!