# Power transfer in a transformer

Can it be proved using the concept of induced emf that power supplied at the primary coil equals power consumed at the secondary. I tried following. Let primary coil be called 1 and secondary be called 2. Assuming there is no resistance in the primary and a purely resistive load on the secondary. Kirchoff's eqns for the two coils are:

$V_1-L_1dI_1/dt+MdI_2/dt=0$

$MdI_1/dt-L_2dI_2/dt-I_2R=0$

Multiplying 1st eqn by $I_1$ and 2nd by $I_2$ and adding the two eqns one gets$V_1I_1-L_1I_1dI_1/dt+MI_1dI_2/dt+MI_2dI_1/dt-L_2I_2dI_2/dt-I_2^2R=0$

$L_1I_1dI_1/dt$ represents rate of change of energy of magnetic field due to coil 1 $L_2I_2dI_2/dt$ represents rate of change of energy of magnetic field due to coil 2 $I_2^2R$ represents power dissipated in the resistor. The sum of these three terms should equal $V_1I_1$. but I am getting the extra term of $MI_1dI_2/dt+MI_2dI_1/dt$. Can anybody help me out.

• Sorry if this is obvious, but what is $M$? Also does it help that $M I_1 dI_2/dt + M I_2 dI_1/dt = M d/dt (I_1 I_2)$? Commented May 17, 2013 at 6:29
• M is the mutual inductance of the two coils. No that does not help. Commented May 17, 2013 at 9:21
• @SUper: this just proves that the rate of change of magnetic energy is also dependent on the other coil, I can't find anything wrong with it as if the potential of both coil are dependent on each other, it shouldn't be a big deal that even their energies or powers are interrelated! Commented Apr 29, 2014 at 15:09

Assuming you've done the math right, this is true.

Instantaneous power supplied = rate of energy stored in magnetic field + rate of energy dissipated in resistor.

For low frequencies, you can neglect radiation. So most of the magnetic field energy is conservative, meaning that it is typically returned to the circuit. Just like how you may take a loan from a bank and return it at a later time(without interest!).

So, on an average:

Average power supplied = average rate of energy dissipation in resistor.

Everything is almost ok.

The first problem is that if you analyse transient states, you should use lower case letters like $i$ and $u$ for variables.

Henceforth I keep lowercase.

If you agree that $$Mi_1\frac{di_2}{dt}+Mi_2\frac{di_1}{dt} = M\frac{d}{dt}\left(i_1 i_2\right)$$ and know that $M$ unit is the same as $L$, this factor is representing energy that was stored in magnetic field of both coils.

You must not omit this factor. For example, if you turn $v_1$ off, there still flows current because the energy was stored. It will be dissipated on the resistor.

Considering the AC (sinusoidal currents) this element is responsible for taking so called reactive power by transformer. If you solve your equations in complex space (so we are now using capital letters), then you will see, that this has imaginary unit $j$ before, so cannot be compared to $I^2 R$. So $U_1 I_1 \neq I_2^2 R$ but $U_1 I_1 cos \phi = I_2^2 R$.

You cannot say $i_2^2 R$ is power of the circuit. The power in AC is $I_2^2 R$ where $I_2$ is an rms value of current, defined as ... Er, please look here I'm not good in this formatting stuff, sorry.

Usually, however, the reactance that is related to mutual inductance $\omega M$ is very large if compared to $\omega L$.

In general, instantaneous power into the primary does not equal the power delivered by the secondary: some goes to change the internal energy stored in the transformer.

That internal energy is not just that from the self-inductances $L_1$ and $L_2$. The mutual inductance $M$ terms arise because the primary and secondary magnetic circuits are coupled, that is, their magnetic fields interfere.

From the circuit point of view, imagine that the primary and secondary coils are driven by current sources $i_1$ and $i_2$, with positive current flowing into the coils, and associated voltage polarities $v_1$ and $v_2$, so that:

$$v_1 = L_1 \frac{di_1}{dt} + M \frac{di_2}{dt}$$ $$v_2 = M \frac{di_1}{dt} + L_2 \frac{di_2}{dt}$$

Then $v_1 i_1 + v_2 i_2$ is the instantaneous power flowing into the transformer, and the same calculation you performed gives:

$$v_1 i_1 + v_2 i_2 = L_1 i_1 \frac{di_1}{dt} + M \left( i_1 \frac{di_2}{dt} + i_2 \frac{di_1}{dt} \right) + L_2 i_2 \frac{di_2}{dt}$$ $$= L_1 i_1 \frac{di_1}{dt} + M \frac{d (i_1 i_2)}{dt} + L_2 i_2 \frac{di_2}{dt}$$

Integrating from a zero initial state, the stored energy $E$ in the transformer is

$$E = \frac{1}{2} L_1 i_1^2 + M i_1 i_2 + \frac{1}{2} L_2 i_2^2$$

Alternatively, from the field perspective, the stored energy in this linear transformer is the volume integral of the magnetic energy density $\mu H^2 /2$, where the vector magnetic field $\boldsymbol{H}$ can be decomposed (by superposition) into components $\boldsymbol{H_1}$ generated by (and proportional to) $i_1$ and $\boldsymbol{H_2}$ by $i_2$:

$$E = \frac{\mu}{2} \int dv \, {H^2}= \frac{\mu}{2} \int dv \, {\boldsymbol{H} \cdot \boldsymbol{H}} = \frac{\mu}{2} \int dv \, {(\boldsymbol{H_1} + \boldsymbol{H_2}) \cdot (\boldsymbol{H_1} + \boldsymbol{H_2})}$$

$$= \frac{\mu}{2} \int dv \, \left( H_1^2 + 2 \boldsymbol{H_1} \cdot \boldsymbol{H_2} + H_2^2 \right)$$

Comparing the two results, one sees that the $M$ term arises when the dot product of the two constituent magnetic fields is non-zero, that is, when the fields interfere.

A very useful idealization results from taking two limits:

1. perfect coupling: $M= \sqrt{L_1 L_2}$
2. The self-inductances go to infinity such that the ratio $L_2/L_1 = n^2$ is finite.

In this limit, the so-called "ideal transformer", very simple input/output relations hold:

$$\frac{v_2}{v_1} = n \,\,\, , \,\,\, \frac{i_2}{i_1} = -\frac{1}{n}$$

For such an ideal transformer the power supplied to the primary does indeed equal the power delivered by the secondary.