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.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.
When the switch is open, the inductor will start to discharge.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.
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.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.
Yes. So? That doesn't make i1 = i2.When the switch is open, the inductor will start to discharge.
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 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.
So why does KCL not hold on that node at t=0+?i1 = i2 @ t=0
unless we are going to discuss sparking at the switch @ t=0
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.So when the switch opens, I should apply KVL around the loop.
I have
\(
i_1 * 12 + L \frac{di_1}{dt} + i_2 * 6 = 0
\)
Yes, despite what several other people that seem to be struggling with KCL are saying.\(
i_2 = -i_1
\)
Nope, because you didn't pay attention to signs when you set up the problem.\(
12i_1 + -i_1 *6 +L \frac{di_1}{dt} = 0
\)
to this\(
\frac{di_1}{dt} + \frac{6i_1}{L} = 0
\)
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.\(
i_1(t) = 1A * e^{\frac{-18t}{2}}
\)
I realize I did make a few mistakes. I'll correct them starting from here.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
\)
Definitely making progress.I realize I did make a few mistakes. I'll correct them starting from here.
\(
L \frac{di}{dt} + R i = V_s
\)
then the solution to the differential equation is:
\(
i(t) = \frac{V_s}{R} + e^{\frac{-Rt}{L}}
\)
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.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:
\(
i(t) = \frac{V_s}{R} + e^{\frac{-Rt}{L}}
\)
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?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?
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!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.
So you have a spark at the switch points @ t=0.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.
Why would there be a spark? The inductor has not been open-circuited.So you have a spark at the switch points @ t=0.
by Jake Hertz
by Jake Hertz
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