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I am reading about the characteristics of an ideal transformer and it is mentioned that self inductance of each coil is infinite. Could someone help me to understand the physics behind it? When self inductance goes to infinity that means that the voltage across the coil goes to infinity and I have a large magnetic flux and current. I was wondering if this is a right conclusion.

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    \$\begingroup\$ Infinite resistance doesn't imply infinite voltage and neither should infinite inductance. \$\endgroup\$ – Andy aka Jan 29 '16 at 14:10
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If self inductance is infinite, then any rate of change of flux would generate infinite voltage, but look at it the other way. Any amount of back emf can be generated with infinitesimally small rate of change of flux. When we apply a voltage to the primary of a transformer (with no secondary load), an equal and opposite back emf is initially generated and no current flows. In order to maintain the back emf, there must be a rate of change of flux, which requires an increasing primary current. The larger the self inductance, the smaller the rate of current rise required to maintain the emf. As self inductance approaches infinity the rate of rise of current approaches zero. So in the ideal transformer one could apply a voltage to the primary for ever with no current flow. The voltage across the socondary will be the primary voltage times the turns ratio. The current in the primary will be the secondary current / turns ratio.

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Could someone help me to understand the physics behind it?

If we mathematically allow the primary and secondary self-inductances of an otherwise ideal transformer go to infinity, the low frequency response of the transformer goes to zero frequency. That is to say, an ideal transformer transforms DC voltage and current just as it does AC voltage and current.

Put another way, if the primary and secondary inductances are finite, the otherwise ideal transformer has a low frequency corner frequency - it cannot be used as a transformer to arbitrarily low frequencies.

In the case of a uncoupled inductor, letting the inductance go to infinity essentially results in an ideal current source.

You may know that an ideal voltage or current source has 'infinite' (unlimited) energy. Since the energy stored in an inductor is

$$W_L = \frac{1}{2}Li_L^2$$

it follows that, for non-zero current through, an inductance of infinity implies infinite energy stored.

Further,

$$v_L = L \frac{di_L}{dt}$$

implies that an inductance of infinity permits no rate of change of current through for a finite voltage across.

In other words, such an inductor will maintain a constant current through for any finite voltage across; it is an ideal current source.

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