# Is alternating current or alternating voltage supplied in the primary coil in a transformer?

In an ideal transformer, I want to know whether a alternating current or alternating voltage is supplied in the primary coil.

If AC current is supplied, then there is change in magnetic flux and an alternating voltage is induced in the secondary coil. However, the current in the secondary coil depends on the resistance of the load, which is $$I_2=\frac{V_2}{R_L}$$ and not necessarily being $$I_2=\frac{V_1I_1}{V_2}$$.

• If you connect an ideal current source, then $V_1$ is free to be whatever value is necessary to make both of your equations true at the same time. Sep 4, 2018 at 0:31
• ie. Both V and I Sep 4, 2018 at 0:42
• In my opinion, it is an alternating electric field that is applied. That electric field is what produces current flow and voltage drop across circuit elements. Sep 4, 2018 at 1:16

Normally, an alternating voltage is applied to the primary. This, in turn, causes current to flow in the primary. The alternating primary current creates a varying magnetic flux in the core of the transformer. Ideally all of the magnetic flux is coupled to the secondary winding inducing a time varying voltage in the secondary. Moreover in an ideal transformer the windings have no impedance (inductance, capacitance, or resistance) and thus there are no losses.

Consequently, for an ideal (lossless) transformer, power in must equal power out (conservation of energy), that is

$$V_1 I_1 = V_2 I_2$$ where 1 denotes the primary and 2 denotes the secondary. And $$I_2 = I_1\frac{V_1}{V_2}$$ and in addition

$$V_2 = \frac {V_1}{a}$$ where a is the ratio of primary turns to secondary turns. $$a= \frac{N_1}{N_2}$$ Any impedance connected to the secondary of an ideal transformer needs to result in the conservation of power, whether it be apparent power, reactive power, or real power (the last being the case if the load is a resistor). In this respect the secondary of an ideal transformer can be considered as either an ideal current or voltage source such that $$I_2 =\frac{V_2}{R_L}$$ for all values of $R_L$.

For a real transformer, however, the secondary current/voltage would only be independent of the load for a limited range of load depending on the capacity of the transformer.

Hope this helps.

• "... the secondary of an ideal transformer behaves like an ideal current source" This statement should not be applicable to a transformer fed by a voltage source? Did you mean to say "ideal voltage source"?
– V.F.
Sep 4, 2018 at 11:26
• @V.F. Both the primary and secondary are inductors, therefore they behave as current sources in the context of lumped element circuits (Kirchhoff's equations). This is also why within the same approximation the proper drive is a voltage source in the primary. (e.g., you cannot drive one current source with another in series) Sep 4, 2018 at 11:48
• As I understand it, the equivalent circuit for an ideal transformer can be represented using a pair of controlled voltage and current sources, where the current or voltage source can be in either the primary or secondary depending on whether the control factor is either the turns ratio, or its reciprocal. See the section on equivalent circuit from this fileIdealTrans.pdf Sep 4, 2018 at 12:04
• $V_2=\frac {V_1} a$, right? So, if $V_1$ is fixed (voltage source), $V_2$ must be fixed as well, i.e., $V_2$ must act as a voltage source.
– V.F.
Sep 4, 2018 at 12:09
• Search web for “ideal transformer equivalent circuit” and the treatment by the University of Toledo. I will edit my answer to reflect that the secondary may be considered an ideal current or voltage source. Sep 4, 2018 at 12:28

and not necessarily being

$$I_2=\frac{V_1I_1}{V_2}$$

In fact, assuming an ideal transformer, this is necessary. There is no contradiction for a resistive load.

Assuming an ideal transformer of turns ratio $n = \frac{V_p}{V_s} = \frac{I_s}{I_p}$ and a resistive load with resistance $R_L$, there is only one independent variable.

If the independent variable is the primary voltage then:

$$I_p = \frac{V_p}{n^2 R_L}$$

$$I_s = nI_p = \frac{V_p}{nR_L}$$

$$V_s = I_sR_L = \frac{V_p}{n}$$

If the independent variable is the primary current then:

$$V_p = I_pn^2 R_L$$

$$V_s = \frac{V_p}{n} = I_pnR_L$$

$$I_s = \frac{V_s}{R_L} = nI_p$$

Normally, do we apply an alternating voltage on the primary coil or an alternating current?

In the case of AC power transformers, the AC mains are very good approximations to a voltage source with very low output impedance.

In the case of audio output transformers used in vacuum tube amplifiers, the primary typically sees the relatively high output impedance looking into the anode of the output tube(s).

So, the answer is it depends on the application.

But, as stated above, for a resistive load the voltage across and current through the primary are proportional and, as The Photon has noted, "you can't have one without the other".

• Normally, do we apply an alternating voltage on the primary coil or an alternating current? Sep 4, 2018 at 4:41
• @superfrankie, you can't have one without the other. Sep 4, 2018 at 5:54

In most AC power applications, the primary coil is driven by a fixed voltage.

If the input voltage of a transformer is fixed, the output voltage depends on the transformer ratio, but does not depend on the load current (within the operating range of the transformer), which is important for most devices using AC power.

If the input current of a transformer is fixed, the output voltage depends on the load current, so such arrangement would be more appropriate for a fixed load or other cases, where the output voltage does not have to be fixed.

Your second equation follows from the conservation of power, which (ignoring various losses) is applicable to both cases.

In electricity we must have a potential difference, i.e. voltage before anything (like current) can happen in a circuit. For example you plug in a 60w light and the 120VAC voltage drives about 0.5 amp because of the resistance is high or you can plug in a toaster and the current is 10 amps.

Transformers need to be large for larger currents because the wire is so long and the resistance gets high. So for the V times I(primary) to be equal to the V times I secondary we need large gauge wire to get the current up in the primary for higher power.