# Does the frequency of AC current affect the induced emf in a transformer (mutual induction)?

Faraday’s law states that the induced emf is directly proportional to the rate of change of flux.

So if I have a constant input voltage and increase the frequency of an alternating current in a transformer, would the rate of the secondary coil cutting field lines and thus the emf induced increase and if not, why?

I’m a bit confused by the word “rate”, and the idea of frequency affecting voltage seems really absurd to me.

• When two coils are closely coupled, thinking in terms of just one coil will lead you astray. Instead of one equation for one coil, you two coupled equations for the pair. – Alfred Centauri Nov 9 '17 at 16:28
• Welcome to Physics.SE! Note that you need to input two line breaks to get a paragraph break in your post. The help also describes other formatting capabilities. – Alex Shpilkin Nov 9 '17 at 16:50

would the emf induced increase and if not, why?

When you have two coupled (ideal) inductors, you have to consider a coupled set of equations rather than one.

For example, and assuming sinusoidal excitation at angular frequency $\omega$, we can write

$$V_p = j\omega(L_p I_p + M I_s)$$ $$V_s = j\omega(M I_p + L_s I_s)$$

where $L_p, L_s, M$ are the primary, secondary, and mutual inductances respectively.

Consider the case that the secondary is open (not connected to a load) so that $I_s = 0$. Then

$$V_p = j\omega L_p I_p$$ $$V_s = j\omega M I_p$$

and it follows that

$$V_s = j\omega M\frac{V_p}{j\omega L_p} = \frac{M}{L_p}V_p$$

The angular frequency dependence has canceled out! In the case of unity coupling, (no flux leakage), we have that

$$M = \sqrt{L_p L_s}$$

and so

$$V_s = \sqrt{\frac{L_s}{L_p}}V_p$$

Since the inductances are proportional to the square of the number of turns $N$, we finally have the ideal transformer relation

$$V_s = \frac{N_s}{N_p}V_p$$

When the secondary is connected to a resistive load, Ohm's law yields

$$V_s = -R I_s$$

(The negative sign is there because we take $I_s$ to be positive if it is in to the more positive terminal of the secondary) and the second of the coupled equations becomes

$$-R I_s = j\omega(M I_p + L_s I_s) \Rightarrow I_s = -\frac{j\omega M}{R + j\omega L_s}I_p$$

Note that in the limit that $R \rightarrow 0$ and unity coupling, we get the ideal transformer relation

$$I_s = -\frac{N_p}{N_s}I_p$$

But for non-zero $R$, the secondary current is a function of the angular frequency $\omega$. Once again, assuming unity coupling, we get

$$I_s = -\frac{j\omega \frac{\sqrt{L_p L_s}}{R}}{1 + j\omega \frac{L_s}{R}}I_p$$

Note that as $\omega \rightarrow 0$, the secondary current goes to zero (for finite $L_p, L_s$) and so, there is no transformer action at DC.

On the other hand, for $\omega \gg \frac{R}{L_s}$, we recover the ideal transformer relation.

Having solved for $I_s$ in terms of $I_p$, we solve the first coupled equation for $V_p$ and get

$$V_p = \frac{j\omega L_p}{1+j\omega\frac{L_s}{R}}I_p$$

Again, note that as $\omega \rightarrow 0$, we have

$$V_p \rightarrow j\omega L_p I_p$$

and see that there is no transformer action at DC, i.e., the primary doesn't 'see' the resistance attached to the secondary.

Also, for $\omega \gg \frac{R}{L_s}$, we have

$$V_p \approx \left(\frac{N_p}{N_s}\right)^2R\,I_p$$

which is the ideal transformer relation for the transformation of impedances.

Finally, note that in the dual limit as $L_p, L_s \rightarrow \infty$, keeping the ratio $\frac{L_p}{L_s} = \frac{N_p}{N_s}$ constant, the ideal transformer equations become exact over all $\omega$, i.e., the ideal transformer 'works at DC'.