Transformers (Circuits)

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Transformer Concept Map

A transformer makes use of Faraday's law and the ferromagnetic properties of an iron core to efficiently raise or lower AC voltages. It cannot increase power so that if the voltage is raised, the current is proportionally lowered and vice versa.

The Main Idea

Transformer and Faraday's Law

From Faraday's law as well as conservation of energy we see that an ideal transformer the voltage ratio is equal to the turns ratio, and power in equals power out. Transformers uses both of these to convert from either high to low or low to high voltages.

A Mathematical Model

For a "step-down" transformer (one that converts from high to low voltage and increases current):

If a solenoid is built wrapping [math]\displaystyle{ {N}_{1} }[/math] turns around a hollow cylinder for the primary coil, and wrapping [math]\displaystyle{ {N}_{2} }[/math] turns around the outside of the secondary coil, and then connecting the primary coil to a an AC power supply, the emf that will develop in the secondary coil will be as follows:

The magnetic field made by the primary coil: [math]\displaystyle{ B = \frac{\mu_0IN_1}{d} }[/math] The cross-sectional area of the solenoid is A, so the emf in one turn of the secondary coil is: [math]\displaystyle{ \frac{AdB}{dt} }[/math] The total emf in the secondary coil is [math]\displaystyle{ {N}_{2} }[/math] times the emf in one turn, so the potential difference across the secondary coil is: [math]\displaystyle{ {N}_{2}A(mu_0{N}_{1}/d)dI/dt }[/math] .

The potential difference across the primary coil is [math]\displaystyle{ \frac{LdI}{dt} }[/math], where [math]\displaystyle{ L = \frac{\mu_0AIN_1^2}{d} }[/math], so the potential difference across the primary coil is: [math]\displaystyle{ A({\mu_0}IN_1^2/d)dI/dt }[/math]

Comparing [math]\displaystyle{ emf_2={N}_{2}A(mu_0{N}_{1}/d)dI/dt }[/math] with [math]\displaystyle{ emf_1= A(mu_0{N}_{1}^2/d)dI/dt }[/math], we see that [math]\displaystyle{ emf_2= ({N}_{2}/{N}_{1})emf_1 }[/math]. The ratio of the number of turns determines the change in voltage.

Faraday's law applied to a transformer can be written as: [math]\displaystyle{ \frac{V_s}{V_p}= \frac{N_s}{N_p} }[/math], where the subscripts refer to primary and secondary coils.

Because energy is conserved and power is [math]\displaystyle{ I \Delta {E} }[/math], the smaller voltage in the secondary coil is accompanied by a larger current. This can be written as: [math]\displaystyle{ P_p= V_pI_p=V_sI_s = P_s }[/math].

In the case of a "step-up" transformer, the primary coil has few turn and the secondary many, therefore increasing the voltage and decreasing the current.

A Computational Model

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Examples

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Simple

A transformer has a primary coil with 102 turns and a secondary coil of 360 turns. The AC voltage across the primary coil has a maximum of 124 V and the AC current through the primary coil has a maximum of 3 A. What are the maximum values of the voltage and current for the secondary coil?

[math]\displaystyle{ {V_s}=? {V_p}= 124V {N_s}= 360 {N_p}= 102 }[/math].

[math]\displaystyle{ \frac{V_s}{V_p}= \frac{N_s}{N_p} }[/math].

[math]\displaystyle{ \frac{V_s}{124V}= \frac{360}{102} }[/math].

[math]\displaystyle{ {V_s}= 438V }[/math].

[math]\displaystyle{ P_p= V_pI_p=V_sI_s = P_s }[/math].

[math]\displaystyle{ V_p=124V I_p=3A V_s= 438V I_s = ? }[/math].

[math]\displaystyle{ (124V)(3A)= (438V)I_s }[/math].

[math]\displaystyle{ I_s = 0.85A }[/math].

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References

Chabay, R., & Sherwood, B. (2015). Electric Potential. In Matter & interactions (4th ed., Vol. Two, pp. 920). Danvers, Massachusetts: J. Wiley & sons. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/transf.html