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The fact that transformers are based on Faraday’s law of induction makes it clear why we cannot use transformers to change DC voltages. If there is no change in primary voltage, there is no voltage induced in the secondary. One possibility is to connect DC to the primary coil through a switch. As the switch is opened and closed, the secondary produces a voltage like that in [link] . This is not really a practical alternative, and AC is in common use wherever it is necessary to increase or decrease voltages.

The first part of the figure shows a graph of DC voltage input. The graph shows a variation of voltage V p along the Y axis and time t along the X axis. The wave is a pulsed wave nearly square in nature with the vibrations only in positive half cycle. The negative half cycles are not present in the wave. The second part of the figure shows a spike wave graph. The graph shows a variation of voltage V s along the Y axis and time t along the X axis. The wave has both positive and negative half cycles shown as sharp spikes of uniform amplitude.
Transformers do not work for pure DC voltage input, but if it is switched on and off as on the top graph, the output will look something like that on the bottom graph. This is not the sinusoidal AC most AC appliances need.

Calculating characteristics of a step-down transformer

A battery charger meant for a series connection of ten nickel-cadmium batteries (total emf of 12.5 V DC) needs to have a 15.0 V output to charge the batteries. It uses a step-down transformer with a 200-loop primary and a 120 V input. (a) How many loops should there be in the secondary coil? (b) If the charging current is 16.0 A, what is the input current?

Strategy and Solution for (a)

You would expect the secondary to have a small number of loops. Solving V s V p = N s N p size 12{ { {V rSub { size 8{s} } } over {V rSub { size 8{p} } } } = { {N rSub { size 8{s} } } over {N rSub { size 8{p} } } } } {} for N s size 12{N rSub { size 8{s} } } {} and entering known values gives

N s = N p V s V p = ( 200 ) 15 . 0 V 120 V = 25 . alignl { stack { size 12{N rSub { size 8{s} } =N rSub { size 8{p} } { {V rSub { size 8{s} } } over {V rSub { size 8{p} } } } } {} #" "= \( "200" \) { {"15" "." "0 V"} over {"120 V"} } ="25" {} } } {}

Strategy and Solution for (b)

The current input can be obtained by solving I s I p = N p N s size 12{ { {I rSub { size 8{s} } } over {I rSub { size 8{p} } } } = { {N rSub { size 8{p} } } over {N rSub { size 8{s} } } } } {} for I p size 12{I rSub { size 8{p} } } {} and entering known values. This gives

I p = I s N s N p = ( 16 . 0 A ) 25 200 = 2 . 00 A . alignl { stack { size 12{I rSub { size 8{p} } =I rSub { size 8{s} } { {N rSub { size 8{s} } } over {N rSub { size 8{p} } } } } {} #" "= \( "16" "." "0 A" \) { {"25"} over {"200"} } =2 "." "00"" A" {} } } {}

Discussion

The number of loops in the secondary is small, as expected for a step-down transformer. We also see that a small input current produces a larger output current in a step-down transformer. When transformers are used to operate large magnets, they sometimes have a small number of very heavy loops in the secondary. This allows the secondary to have low internal resistance and produce large currents. Note again that this solution is based on the assumption of 100% efficiency—or power out equals power in ( P p = P s size 12{P rSub { size 8{p} } =P rSub { size 8{s} } } {} )—reasonable for good transformers. In this case the primary and secondary power is 240 W. (Verify this for yourself as a consistency check.) Note that the Ni-Cd batteries need to be charged from a DC power source (as would a 12 V battery). So the AC output of the secondary coil needs to be converted into DC. This is done using something called a rectifier, which uses devices called diodes that allow only a one-way flow of current.

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Transformers have many applications in electrical safety systems, which are discussed in Electrical Safety: Systems and Devices .

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Test prep for ap courses

Which of the following statements is true for a step-down transformer? Select two answers.

  1. Primary voltage is higher than secondary voltage.
  2. Primary voltage is lower than secondary voltage.
  3. Primary current is higher than secondary current.
  4. Primary current is lower than secondary current.

(a), (d)

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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