Transformers (Circuits)

Electricity sent through power lines is transmitted with high voltages through long thick power lines because wires have a resistance that causes power loss at a rate proportional to the current squared. By transmitting at a high voltage, energy loss is minimized. Home appliances however operate at much lower voltages. Something is needed to convert the power to a high current, low voltage power that home appliances can use. This conversion from high voltage to low voltage, and vice versa, is accomplished by a transformer.

Background

Inductance

Currents can be induced (produced) by changing the current through a coil. This is due to the changing magnetic field [math]\displaystyle{ \textstyle (dB/dt) }[/math] produced by varying the current through the coil. We know from the Maxwell-Faraday Law of Maxwell's Equations:

[math]\displaystyle{ |emf| = \oint \overrightarrow{E}_{NC} \cdot d\overrightarrow{l} = \left | \frac{d\phi_{mag}}{dt} \right \vert }[/math]

Or that a changing magnetic field through an area produces a non-Coloumb electric field.

Mathematical Formulae

Before moving on to a discussion of the mathematics of transformers, here are some formulas it will be helpful to recall:

• Magnetic Field Inside a Solenoid: [math]\displaystyle{ B=\frac{\mu_0 N I}{d} }[/math]

Where [math]\displaystyle{ \textstyle N }[/math] is the number of coils and [math]\displaystyle{ \textstyle d }[/math] is the length of the solenoid.

• Magnitude of self-induced emf: [math]\displaystyle{ \textstyle \left|emf_{ind}\right \vert=L\left|\frac{d I}{d t} \right \vert }[/math]

Where [math]\displaystyle{ L }[/math] is the proportionality constant called the "inductance" or "self-inductance" which equals [math]\displaystyle{ \textstyle \frac{\mu_0 N^2}{d}\pi R^2 }[/math]

• Expanding this, we get the self-induced emf in a solenoid is: [math]\displaystyle{ \textstyle emf= \frac{\mu_0 N^2}{d}\pi R^2 \frac{d I}{d t} }[/math]

• Finally, remember your units. [math]\displaystyle{ emf }[/math] is measured in volts, self-inductance is [math]\displaystyle{ \textstyle(V•s/A) }[/math] or the "henry" (H), and [math]\displaystyle{ B }[/math] is measured in Tesla (T) or [math]\displaystyle{ \textstyle(\frac{kg}{s^2 A}) }[/math]

How They Work

Conversion from high to low, or low to high voltage can be accomplished using the principles discussed above. Consider a solenoid with [math]\displaystyle{ N_1=100 }[/math] coils around a hollow cylinder of length [math]\displaystyle{ d=.3 m }[/math]. Now wrap [math]\displaystyle{ N_2 = 200 }[/math] coils around this solenoid to form the secondary coil. If an alternating current is run through the primary coil, we get a non-zero [math]\displaystyle{ \textstyle\frac{d I}{d t} }[/math]We can now calculate the potential difference across each coil.

Primary Coil

As stated above, the induced emf in the primary coil is [math]\displaystyle{ L\left|\frac{d I}{d t} \right \vert }[/math]. Expanding this and substituting [math]\displaystyle{ A=\pi R^2 }[/math] for the area, we get a potential difference across the primary coil of [math]\displaystyle{ \textstyle A(\mu_0 N_1^2 /d)dI/dt }[/math].

Secondary Coil

A current is induced in the secondary coil by the changing magnetic field produced by the primary coil. The magnetic field is [math]\displaystyle{ \textstyle B = \mu_0 N_1 I/d }[/math] and it is changing across area [math]\displaystyle{ A }[/math] (which is only the area of the inner coil, not the outer secondary coil. So the emf in one turn of the secondary coil is [math]\displaystyle{ A dB/dt }[/math]. We have [math]\displaystyle{ N_2 }[/math] secondary coils, so the emf is [math]\displaystyle{ N_2 AdB/dt }[/math]. If we expand out our [math]\displaystyle{ dB/dt }[/math] term, we can get the emf across the second coil in a formula similar to the emf across the primary coil: [math]\displaystyle{ emf_{sec} = N_2A(\mu_0N_1/d)dI/dt }[/math].

Voltage Ratio

We now can see that the ratio of the secondary to primary emf is [math]\displaystyle{ emf_{pri}/emf_{sec} }[/math]. This yields:

[math]\displaystyle{ \frac {N_2A(\mu_0N_1/d)dI/dt}{A(\mu_0 N_1^2 /d)dI/dt} }[/math] which leaves us with [math]\displaystyle{ \frac {N_2}{N_1} }[/math] or in this case [math]\displaystyle{ \frac{200}{100}\lt /math. So this transformer would create a emf 2 times the emf in the primary coil. Because we can't create energy from nothing, power (\lt math\gt I\Delta V }[/math]) must be conserved, so the double voltage in the secondary coil is accompanied by a current of half the strength of the primary coil.

Circuits

Connectedness

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See also

Faraday's Law

This will give you a general understanding of Faraday's Law, which is the basis behind transformer technology.

Inductance

Inductance is another property of an electrical conductor derived from Faraday's law.

Gauss's Flux Theorem

Changing the flux of a magnetic field around a coil will induce voltage.

Further reading

Books, Articles or other print media on this topic

External links

http://www.edisontechcenter.org/Transformers.html

References

Chabay, R., & Sherwood, B. (2015). Electric Potential. In Matter & interactions (4th ed., Vol. Two, pp. 920). Danvers, Massachusetts: J. Wiley & sons.