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There is a sudden inrush of the primary current, when the transformer is initially energized and the maximum value of the magnetic flux is more than twice the normal value of the flux. Hence, due to the very high peak of the magnetising current, the core is driven far into saturation.

Let a sinusoidal voltage is applied to the primary winding of a transformer is

$$\mathrm{𝑣_{1} = \nu_{1𝑚}\:sin(\omega 𝑡 + 𝜃) … (1)}$$

And the secondary winding of the transformer is kept open circuited. Here, θ is the angle of the voltage at *t* = 0.

Now, if the core losses and the primary resistance are neglected, then,

$$\mathrm{\nu_{1} = 𝑁_{1}\frac{𝑑φ}{𝑑𝑡}… (2)}$$

Where,

N

_{1}is the number of turns in the primary winding,Φ is the magnetic flux in the core of the transformer.

In the steady-state condition, the maximum value of the applied voltage to the primary is given by,

$$\mathrm{\nu_{1𝑚} = \omega \:φ_{𝑚}\:𝑁_{1} … (3)}$$

Now, from the equations (1) and (3), we get,

$$\mathrm{𝑁_{1}\:\frac{𝑑φ}{𝑑𝑡}= \nu_{1𝑚 }\:sin(\omega 𝑡 + 𝜃)}$$

$$\mathrm{\Rightarrow\:\frac{𝑑φ}{𝑑𝑡}=\frac{\nu_{1𝑚}}{𝑁_{1}}sin(\omega 𝑡 + 𝜃) … (4)}$$

From equations (3) and (4), we have,

$$\mathrm{\frac{𝑑φ}{𝑑𝑡}= \omega φ_{𝑚}\:sin(\omega 𝑡 + 𝜃) … (5)}$$

Integrating on both sides of the eqn. (4), we get,

$$\mathrm{φ = −φ_{𝑚}\:cos(\omega 𝑡 + 𝜃) + φ_{𝐶} … (6)}$$

Where, φC is the constant of integration and its value can be determined by the initial conditions i.e. at t = 0.

Assume that when the transformer is disconnected from the supply, a small residual flux (φ_{r}), remained in the core of it. Therefore, at initial conditions,

$$\mathrm{𝑡 = 0; φ = φ_{𝑟}}$$

$$\mathrm{∴ \: φ_{𝑟} = −φ_{m}\:cos 𝜃 + φ_{𝐶}}$$

$$\mathrm{\Rightarrow\:φ_{𝐶} = φ_{𝑟} + φ_{m}\:cos 𝜃 … (7)}$$

Substituting the value of φC in the eq. (6), we have,

$$\mathrm{φ = [−φ_{𝑚}\:cos(\omega 𝑡 + 𝜃)] + [φ_{𝑟} + φ_{𝑚}\:cos 𝜃] … (8)}$$

The eq. (8) shows that the flux consists of two components viz. the steady state component (φ_{ss}) and the transient component (φC).

$$\mathrm{Steady\:state\:component\:of\:flux,\:φ_{𝑠𝑠} = −φ_{𝑚}\:cos(\omega 𝑡 + 𝜃) … (9)}$$

$$\mathrm{Transient\:component\:of\:flux,\: φ_𝐶 = φ_{𝑟} + φ_{𝑚}\:cos 𝜃 … (10)}$$

From the eqn. (10), it can be seen that the transient component is the function of θ, where θ is the instant at which the transformer is switched on to the supply.

If the transformer is switched on at θ = 0, then,

$$\mathrm{φ_{𝐶} = φ_{𝑟} + φ_{𝑚}}$$

Under this condition, the total flux in the core becomes,

$$\mathrm{φ = −φ_{𝑚}\:cos(\omega 𝑡 + 𝜃) + φ_{𝑟} + φ_{𝑚}}$$

At ωt = π, we get,

$$\mathrm{φ = −φ_{𝑚}\:cos\:π + φ_{𝑟} + φ_{𝑚}}$$

$$\mathrm{\Rightarrow\:φ = φ_{𝑟} + 2φ_{𝑚} … (11)}$$

Hence, the magnetic flux in the core attains the maximum value of the flux which by given in the eq. (11). It can be seen that it is more than twice the normal value of the flux. This is known as * doubling effect*. As a result of this, the core goes into deep saturation.

The magnetising current required for producing such a large flux in the core may be as large as 10 times of the normal magnetising current. Sometimes, the RMS value of this magnetising current may be greater than the primary rated current of the transformer. This inrush in magnetising current may produce electromagnetic forces about 25 times of the normal value. Thus, the windings of the large transformers are strongly braced. These inrush currents may also cause abnormal operation of the protective devices and humming due to magnetostriction of the core.

In order to obtain no transient inrush current, the transient component of the flux should be zero, i.e.,

$$\mathrm{φ_{𝐶} = φ_{𝑟} + φ_{𝑚}\:cos 𝜃 = 0}$$

$$\mathrm{\Rightarrow\:cos 𝜃 = −\frac{φ_{𝑟}}{φ_{𝑚}}}$$

Since, the value of φ_{r} is usually very small, therefore,

$$\mathrm{cos 𝜃 ≅ 0;\:\:𝜃 =\frac{𝑛π}{2}}$$

Hence, if the transformer is connected to the supply source near a positive or negative voltage maximum, then the inrush current will be minimised.

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