What Is Step-Down Transformer and How It Works?

[ epsilon = -frac{d}{dt} phi_m] 

where, ( E_{epsilon mf} ) is the induced emf in the closed loop,

( frac{d}{dt} phi_m ), is the rate of change of magnetic flux passing through the closed loop. 

The negative sign in the Faraday’s Law equation indicates the if the magnetic flux,( frac{d}{dt} phi_m ), is increasing then induced emf, ( epsilon ) , is decreasing and vice versa. This is mathematical interpretation of Lenz’s Law which states that “The direction of any magnetic induction effect is such as to oppose the cause of the effect.”

Mutual Induction elucidates how a current is induced in a coil by
another current-carrying coil in close proximity, experiencing varying
magnetic flux. The induced current is directly linked to changes in the
initial current’s rate.

Consider the following picture of demonstration of mutual inductance:

Here, the ac current or changing current i1 flowing in the coil 1 with N1 turns produces a magnetic field B1 which flows through the coil 2. This in effect generates an induced and induced current emf in the coil 2. The induced emf in coil 2 is given by the Faraday’s Law,

Consider two coils, coil 1 and coil 2, one primary and one secondary, with $N_1$ and $N_2$ turns, respectively, positioned such that the changing magnetic flux in the primary coil induces an EMF in the secondary coil. To calculate primary and secondary inductance of a transformer see toroidal transformer calculation.

The magnetic flux through the secondary coil due to the current in the primary coil is given by:

${\mathrm{\Phi }}_2=M_{21}{I}_1$

Where:

• ${\mathrm{\Phi }}_2$
• $M_{21}$
• $I_1$

According to Faraday’s law, the induced EMF in the secondary coil is:

${\mathcal{E}}_2=-\frac{d{\mathrm{\Phi }}_2}{dt}$

Now, if the current in the primary coil is changing with time ($\frac{d{I}_1}{dt}$

${\mathcal{E}}_2=-\frac{d{\mathrm{\Phi }}_2}{dt}=-\frac{d(M_{21}{I}_1)}{dt}$

By applying the product rule of differentiation:

${\mathcal{E}}_2=-M_{21}\cdot \frac{d{I}_1}{dt}-I_1\cdot \frac{d{M}_{21}}{dt}$

Comparing this with the equation for induced EMF in the secondary coil (${\mathcal{E}}_2=-\frac{d{\mathrm{\Phi }}_2}{dt}$

Therefore, we get:

${\mathcal{E}}_2=-M_{21}\cdot \frac{d{I}_1}{dt}$

Finally, if we consider the total induced voltage ($V_2=N_2\cdot {\mathcal{E}}_2$

$V_2=-M_{21}\cdot N_2\cdot \frac{d{I}_1}{dt}$

This equation relates the induced voltage in the secondary coil ($V_2$) to the rate of change of current in the primary coil ($\frac{d{I}_1}{dt}$

 The calculate primary current of transformer and transformer output current calculator can be used to calculate the transformer input and output current.

Transformer Construction