Conservation of energy when placing two inductors next to each other

[sudera]

Homework Statement

How does conservation of energy apply when placing an inductor next to another inductor with a current $i(t) = at$ going through it?

Relevant Equations

$\varepsilon = -\dfrac{d\Phi_B}{dt}$

This is more like a theoretical question of my own than actual homework.

Say there is a circuit with a current source and an inductor. There is a current $i(t)=at$ going through the inductor. We now place a new circuit with an inductor and a resistor next to it. The current $i(t)$ causes a changing magnetic flux - and thus an emf - through this new circuit. Since $i(t)$ is linear, the emf is constant, and the current in the new circuit will also be constant and will not create a changing magnetic flux through the original circuit, so the current in the original circuit doesn't change. In other words, the second circuit received current and energy for free.

Obviously, this can't be true because of the law of conservation of energy. But where does this reasoning go wrong? I know there is a back emf in the transient state when the second inductor is introduced, but what happens in the steady state?

 

[mitochan]

Hi.
Do you consider reaction from the new nearby circuit to the original circuit ?
Current in new circuit generate magnetic flux that reduces magnetic flux of the original circuit.
In order to maintain magnetic flux of the original circuit you may need more current.

 

[sudera]

Hi.
Do you consider reaction from the new nearby circuit to the original circuit ?
Current in new circuit generate magnetic flux that reduces magnetic flux of the original circuit.
In order to maintain magnetic flux of the original circuit you may need more current.

Since the induced emf of the second circuit is a constant, the current will also be constant, and thus will not create a changing magnetic flux through the first circuit and will not change the current in the first circuit, so the power in the first circuit is unaffected while the power in the second circuit is increased. Is this reasoning wrong?

 

[cnh1995]

Homework Statement: How does conservation of energy apply when placing an inductor next to another inductor with a current $i(t) = at$ going through it?
Homework Equations: $\varepsilon = -\dfrac{d\Phi_B}{dt}$

In other words, the second circuit received current and energy for free.

Obviously, this can't be true because of the law of conservation of energy.

Yes, this can't be true. The energy absorbed by the second circuit has to come from the current source placed in the first circuit.

Since i(t)i(t)i(t) is linear, the emf is constant,

True.

Homework Statement: How does conservation of energy apply when placing an inductor next to another inductor with a current $i(t) = at$ going through it?
Homework Equations: $\varepsilon = -\dfrac{d\Phi_B}{dt}$

and the current in the new circuit will also be constant and will not create a changing magnetic flux through the original circuit,

Not true. This is a magnetically coupled system. You are not considering the effect of mutual inductance between the two coils. The current in the second circuit does affect the first circuit. To know exactly how, how much and for how long, you need to solve the differential equation of this system.

 

[sudera]

Yes, this can't be true. The energy absorbed by the second circuit has to come from the current source placed in the first circuit.True.

Not true. This is a magnetically coupled system. You are not considering the effect of mutual inductance between the two coils. The current in the second circuit does affect the first circuit. To know exactly how, how much and for how long, you need to solve the differential equation of this system.

Would this system look something like this?

$v_1(t) = L_1 \dfrac{di_1}{dt} - M\dfrac{di_2}{dt}$
$v_2(t) = M \dfrac{di_1}{dt} - L_2\dfrac{di_2}{dt}$

Can I use $i_1(t)=at$ and $i_2(t)=0$ or is this not allowed because the current becomes different from the original current supply because of the emf? When I do this, and I compare this solution (with certain values for $a,L_1,L_2,M$ filled in) to the solution when the second circuit isn't there, the voltage over the first inductor remains the same.

 

[mitochan]

Hi.

We now place a new circuit with an inductor and a resistor next to it.

With a register does another term $i_2 R$ appear in RHS of $v_2(t)$ ?

 

[sudera]

Hi.

With a register does another term $i_2 R$ appear in RHS of $v_2(t)$ ?

Hi, I use $v_2(t)$ only to denote the voltage over the inductor, not the entire circuit.

 

[rude man]

Would this system look something like this?

$v_1(t) = L_1 \dfrac{di_1}{dt} - M\dfrac{di_2}{dt}$
$v_2(t) = M \dfrac{di_1}{dt} - L_2\dfrac{di_2}{dt}$

Can I use $i_1(t)=at$ and $i_2(t)=0$ or is this not allowed because the current becomes different from the original current supply because of the emf? When I do this, and I compare this solution (with certain values for $a,L_1,L_2,M$ filled in) to the solution when the second circuit isn't there, the voltage over the first inductor remains the same.

With a step input voltage $v_1$ and a resistor at the output 2, substitute $v_2(t) = i_2R$ and solve your two equations.

If you do you wll find that $i_1$ has a constant (dc) term plus a ramp term, while $i_2$ will be a constant (dc) current.

You can also solve the equations assuming a ramp $i_1$ but this requires some setup imagination.

BTW the convention is for i2 to flow into the transformer, just as i1 does, changing your - signs to +.

 

[TSny]

The magnetic field stores energy. For the scenario of this problem, I think you can show that the rate that energy goes into the B field is less when the second circuit is in place compared to when the second circuit is not in place. The source supplies the same energy with or without the second circuit in place. When the second circuit is in place, the difference in magnetic field energy appears as the energy dissipated in the second circuit.

 

[alan123hk]

I think post #4 and post #8 have explained the problem.

My personal understanding and reasoning are as follows:

The constant current in coil 2 only means that it will not induce EMF in coil 1, which does not mean that the current in coil 1 will not be affected.

The coil 2 does not have to generate a change in the magnetic flux and the induced voltage in the coil 1 because it does not transfer power back to the circuit of coil 1.

Due to the mutual inductance, the additional current in the coil 1 will be created naturely to provide power to the load of the coil 2, namely I1(additional) = (N2/N1)*I2, where N1 and N2 are the number of turns of coil 1 and coil 2, respectively

In order to obtain a complete and correct answer, we must go through a formal and rigorous procedure to solve the differential equations of the system with appropriate initial conditions.

 

[TSny]

Let’s compare case A and case B shown below. In case A we have just circuit 1. In case B we have circuits 1 and 2 with mutual inductance.

In both cases, the source is assumed to provide a current $I_1=at$. In both cases, the same amount of Joule heating occurs in circuit 1. In case B, there is additional Joule heating in circuit 2 caused by the time-independent induced current $I_2$.

The question is whether the source must provide additional energy for case B compare to case A. You might be inclined to answer “yes” because of the additional Joule heating in case B. But you also might be inclined to answer “no” because the steady current $I_2$ does not cause any induced emf in circuit 1. Therefore, it is not any “harder” for the source to provide the current $I_1$ in case B.

I think the correct answer is that the source does not provide additional energy for case B. We can’t forget that the magnetic field itself stores energy. Standard textbooks derive the basic equation for the magnetic field energy for a system of two circuits which have self-inductance and mutual inductance. The formula is $$U_m = \frac {1}{2}L_1 I_1^2 + \frac {1}{2}L_2 I_2^2 - MI_1I_2$$ where the negative sign for the last term takes into account that the direction of currents $I_1$ and $I_2$ are opposite. (For case A, only the first term would be present.)

Taking the time derivative and noting that $I_2$ is time independent in our situation, $$\dot U_m = L_1 I_1 \dot I_1– M I_2 \dot I_1$$ In case A, the second term is not present. So, the rate at which energy goes into the magnetic field in case B is less than in case A by an amount $M I_2 \dot I_1$. But this is exactly the rate of Joule heating in circuit 2 in case B. To see this, note that $M \dot I_1$ is the induced emf in circuit 2. So, the Joule heating in circuit 2 is $(M \dot I_1) I_2 = M I_2 \dot I_1$.

So, in case B, the source provides the same amount of energy as in case A. In case B, the source supplies less energy to the magnetic field by just enough to account for the Joule heat in circuit 2.

 

[cnh1995]

I agree with @TSny's analysis. Excellent as always!

@sudera, if you are interested in the real world application of the above problem (current-excited magnetically coupled circuits):
"current transformers" work on exactly the same principle of energy transfer as explained by TSny.