How Can Energy Change Instantaneously in a Coupled Circuit?

[cnh1995]

Homework Statement

Homework Equations

Kirchhoff's laws

The Attempt at a Solution

Applying KVL on primary,
di₁/dt+i₁=5+2di₂/dt...1)

Similarly for secondary,
2di₁/dt= 4di₂/dt + i₂...2)

After solving these equations using Laplace Transform, I got
I₁= 5-e^-0.2t A and
I₂ = 2e^-0.2t-2e^-0.25t.

I have verified these answers using initial and final conditions and I believe they are correct.

However, at t=0, the energy in the system comes out to be 8J (E= 0.5L₁i₁²+0.5L₂I₂²-Mi₁i₂, since there is magnetic opposition).

How is this possible? How can the energy change instantaneously here?
For E to be zero at t=0, i₂ at t=0 should be 2A. But here, i₂|_t=0= 0A.

What am I missing here?

 

[Grinkle]

I1 (0) looks to be 4A per your solution to I1, which does not seem right. I1(0) and I2(0) both need to be zero, yes?

 

[gneill]

After solving these equations using Laplace Transform, I got
I1= 5-e-0.2t A and
I2 = 2e-0.2t-2e-0.25t.

Your I₁ looks good, but I₂ is suspicious. How does a second time constant arise?

However, at t=0, the energy in the system comes out to be 8J (E= 0.5L1i12+0.5L2I22-Mi1i2, since there is magnetic opposition).

How is this possible? How can the energy change instantaneously here?
For E to be zero at t=0, i2 at t=0 should be 2A. But here, i2|t=0= 0A.

Blame it on the mutual inductance. Since $\frac{dI}{dt}$ will be nonzero at $t = 0$ for both loops, there will be induced currents regardless of the "rule of thumb" behavior that we assume for isolated inductors.

 

[cnh1995]

How does a second time constant arise?

I will check the math again. But I₂ should be zero at t=0 and at t=∞. I believe a single time constant can't make that happen.

Also, in the KVL equation, while taking LT of di₁/dt, I wrote sI₁(s)-i₁(0) and put i₁(0)=0. How come it is not that same in the final expression for i₁?

 

[cnh1995]

I will check the math again. But I₂ should be zero at t=0 and at t=∞. I believe a single time constant can't make that happen.

Also, in the KVL equation, while taking LT of di₁/dt, I wrote sI₁(s)-i₁(0) and put i₁(0)=0. How come it is not same in the final expression of i₁?

Plus, di/dt at t=0 in a normal series R-L circuit is E/L (non-zero). I didn't understand your point about di/dt being non-zero here.

 

[gneill]

The fact that the currents will be changing means that there will be changing flux though the loops each caused by the changing current in the other loop. Then, regardless of the usual "I = 0 at t = 0 for an inductor", Faraday's law says that a changing flux in a loop will induce an EMF.

When I do these sorts of problems I usually re-draw the circuit and write in the Laplace Domain values for the circuit components. Then write the equations by inspection. So for your circuit,

hence:

By the way, I've confirmed the results of the above analysis using LTSpice to simulate the circuit:

As you can see, $i_1$ does indeed start off at 4 amps, and $i_2$ at 2 amps. Goes against intuition, but the math bears it out.

 

[cnh1995]

I also ran a simulation on my phone and it is giving me the same result.

Goes against intuition, but the math bears it out.

Well, I believe I₂ must start off at 2A as per the energy equation in the OP.

I am getting the exact same equations in the s-domain as you are, but my answer for i₂ has two time constants.

Also,

Also, in the KVL equation, while taking LT of di1/dt, I wrote sI1(s)-i1(0) and put i1(0)=0. How come it is not that same in the final expression for i1?

Why is this happening?

 

[cnh1995]

As you can see, i1i1i_1 does indeed start off at 4 amps, and i2i2i_2 at 2 amps. Goes against intuition, but the math bears it out.

On the secondary side, as per the KVL equations, mutually induced emf is the "source" and L₂ and R₂ behave as the load.
At t=0, this mutually induced emf Mdi₁/dt equals 0.4V.
How come the secondary current is 2A? That would require at least 2V in the secondary.

This circuit is crazy!

 

[gneill]

I am getting the exact same equations in the s-domain as you are, but my answer for i2 has two time constants.

Let me cheat and use Mathcad to do all the work:

Also, in the KVL equation, while taking LT of di1/dt, I wrote sI1(s)-i1(0) and put i1(0)=0. How come it is not that same in the final expression for i1?

Again, Faraday's law says that the induced EMF will be spread out all along the loop. While we model mutual inductance as a simple source in series with the inductor, it's really not. The current induced by the EMF is not truly the $I_1$ of the diagram.

Edit:
What I meant to say is, the 5 V source is not the only source of EMF driving current $I_1$. And the induction based contribution is driven by Faraday.

 

[cnh1995]

Yes, when you solve both the s-domain equations simultaneously, you get i₂=2A and that doesn't violate the physics of circuits!

Perhaps my thinking below is incorrect because the Laplace equations have something to do with discontinuity of the current function at t=0.

On the secondary side, as per the KVL equations, mutually induced emf is the "source" and L₂ and R₂ behave as the load.
At t=0, this mutually induced emf Mdi₁/dt equals 0.4V.
How come the secondary current is 2A? That would require at least 2V in the secondary.

This circuit is crazy!

I found the expression for i₁(t), then found the expression for Mdi₁/dt and then took the LT of the equation
Mdi₁/dt= i₂+4di₂/dt.
If you solve this, you get

I2 = 2e-0.2t-2e-0.25t.

Why is this method incorrect?

 

[gneill]

I found the expression for i1(t), then found the expression for Mdi1/dt and then took the LT of the equation
Mdi1/dt= i2+4di2/dt.
If you solve this, you get

I2 = 2e-0.2t-2e-0.25t.

Why is this method incorrect?

I'm not sure, but perhaps because you lose the fact that there is an initial "DC" current of 5 amps when you differentiate in the time domain? If I do the differentiation in the Laplace domain (simply by multiplying $I_1(s)$ by $s$), then I have

$2 \frac{5 (4 s + 1)}{(5 s + 1)} = I_2(s) + 4 s I_2(s)$

and

$I_2(s) = \frac{10}{5 s + 1}$

 

[cnh1995]

I'm not sure, but perhaps because you lose the fact that there is an initial "DC" current of 5 amps when you differentiate in the time domain? If I do the differentiation in the Laplace domain (simply by multiplying $I_1(s)$ by $s$), then I have

$2 \frac{5 (4 s + 1)}{(5 s + 1)} = I_2(s) + 4 s I_2(s)$

and

$I_2(s) = \frac{10}{5 s + 1}$

I agree, that must be it!
I had removed the 5A dc simply by taking its derivative in the time domain.

Thanks for all your help!

 

[rude man]

It may be noted that it's fortunate that the coefficient of coupling is unity.
If for example M=1 instead of 2 we would get I1(s) = (4s+1)/[s(3s²+5s+1)] with pretty tortuous inversion math. Ditto for I2.