Question

31.6 transformers
the resistance of the primary coil of a well - designed, 1:10 step - down transformer is 1 ω. with the secondary circuit open, the primary is connected to a 12 v ac generator. the primary current is
depends on the actual number of turns in the primary coil.
depends on the core material.
about 12 a.
essentially zero.
about 120 a.

Explanation:

Step1: Analyze the transformer operation when secondary is open

In a well - designed transformer, when the secondary circuit is open, there is no power transfer to the secondary. According to the principle of transformers, the primary current is mainly the magnetizing current. For an ideal (well - designed) transformer, the magnetizing current is very small (ideally zero in an ideal lossless transformer model). The resistance of the primary coil ($R = 1\ \Omega$) is not the main factor here because the induced back - emf in the primary coil (due to mutual induction) opposes the applied voltage. If we were to wrongly apply Ohm's law ($I=\frac{V}{R}$) without considering the transformer action, we would get $I=\frac{12\ V}{1\ \Omega}=12\ A$, but this is incorrect for a transformer in no - load (secondary open) condition.

Step2: Consider the nature of the primary current in no - load

The primary current in a well - designed transformer with secondary open is essentially the current required to establish the magnetic flux in the core. In a high - quality (well - designed) transformer, the losses (hysteresis and eddy current losses, and the small magnetizing current) are minimized. The induced emf in the primary ($e=-N_1\frac{d\Phi}{dt}$) is very close to the applied voltage ($V_1$). The small difference between $V_1$ and $e$ (due to small resistance and leakage reactance drops) results in a very small primary current.

Answer:

essentially zero.