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Basically, an electric charger consists of a transformer, which transfers electrical energy from one inductance to another.. Text

َAs i imagine the process physically the first circuit (inductance) should produce the same amount of magnetic flux, either the second inductance exist or not.. so, theoretically the electric consumption in (smartphone) charger should be the same (either we place a smartphone in the other side or not)..

But when i measured the current in the case of (no smartphone plugged in) i found it zero!

How we can interpret this physically?

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In an ideal transformer (which has no DC resistance), and which has no current flowing from the secondary coil, the primary current will be small and out of phase with the applied voltage. Average power in and out are both zero. When AC current flows from the secondary, it causes a shift in the phase and magnitude of the flux and in the primary current. I have a digital meter that shows what is being supplied by a wall plug. With my phone charger plugged in at 120 volts without the phone it's showing 0.09 amps, 1 watts and a power factor of 0.09. Plugged into the phone it shows 0.11 amps 3.5 watts and a PF of 0.27 . Note: Average power = (rms) amps x volts x PF (where the PF is the cosine of the phase difference).

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  • $\begingroup$ the magnetic field in the core of the first solenoid is directly proportional to the current flowing around the solenoid, and this magnetic field will produce in the second circuit an other current (but also proportional to magnetic field), so how you can proof theoretically the phase shift? $\endgroup$
    – Okba
    Commented Jul 25, 2020 at 17:35
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    $\begingroup$ A current flows in the secondary only if is connected to a load. The phase shift occurs when the load is connected. Notice what happened to the power factor when I connected my phone to the charger. $\endgroup$
    – R.W. Bird
    Commented Jul 26, 2020 at 21:16
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A transformer is not a charger. A transformer changes the voltage and current but it is of the same frequency, in the US 60Hz. An ac source cannot charge a battery because if in one half cycle of the 60Hz the source charges the battery then in the next half cycle when the polarity flips it will drain it. A battery charger needs a dc voltage: between the transformer and the battery there is a rectifier that is essentially a diode + a capacitor with a lot of other things. The transformer is there to change the input 110V (in the US) to a lower voltage that is "friendlier" to a diode and more importantly to a voltage near that of the battery to minimize losses in the rectification. When the primary is connected to an ac source but the secondary is open then the source sees an inductive load but it is still ac, so only an ac meter will work when you try to measure the current through it.

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  • $\begingroup$ It's right, but my question is about how the transformer don't consume the same current in the two cases (connected phone or not)? the primary circuit is independent of the second one, so theoretically it will produce the same magnetic field, so it will consume the same power! $\endgroup$
    – Okba
    Commented Jul 25, 2020 at 17:42
  • $\begingroup$ that is a different question but since ideally (assuming no dissipative losses) $V_1I_1=V_2I_2$, if the voltage is lower in the secondary (2) $V_2 <V_1$ then the current is lower in the primary (1) $I_1 < I_2$ $\endgroup$
    – hyportnex
    Commented Jul 25, 2020 at 18:28

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