# 23.7 Transformers  (Page 2/8)

 Page 2 / 8
${V}_{p}=-{N}_{\text{p}}\frac{\Delta \Phi }{\Delta t}\text{.}$

The reason for this is a little more subtle. Lenz’s law tells us that the primary coil opposes the change in flux caused by the input voltage ${V}_{\text{p}}$ , hence the minus sign (This is an example of self-inductance , a topic to be explored in some detail in later sections). Assuming negligible coil resistance, Kirchhoff’s loop rule tells us that the induced emf exactly equals the input voltage. Taking the ratio of these last two equations yields a useful relationship:

$\frac{{V}_{\text{s}}}{{V}_{\text{p}}}=\frac{{N}_{\text{s}}}{{N}_{\text{p}}}\text{.}$

This is known as the transformer equation    , and it simply states that the ratio of the secondary to primary voltages in a transformer equals the ratio of the number of loops in their coils.

The output voltage of a transformer can be less than, greater than, or equal to the input voltage, depending on the ratio of the number of loops in their coils. Some transformers even provide a variable output by allowing connection to be made at different points on the secondary coil. A step-up transformer    is one that increases voltage, whereas a step-down transformer    decreases voltage. Assuming, as we have, that resistance is negligible, the electrical power output of a transformer equals its input. This is nearly true in practice—transformer efficiency often exceeds 99%. Equating the power input and output,

${P}_{\text{p}}={I}_{\text{p}}{V}_{\text{p}}={I}_{\text{s}}{V}_{\text{s}}={P}_{\text{s}}\text{.}$

Rearranging terms gives

$\frac{{V}_{\text{s}}}{{V}_{\text{p}}}=\frac{{I}_{\text{p}}}{{I}_{\text{s}}}\text{.}$

Combining this with $\frac{{V}_{\text{s}}}{{V}_{\text{p}}}=\frac{{N}_{\text{s}}}{{N}_{\text{p}}}$ , we find that

$\frac{{I}_{\text{s}}}{{I}_{\text{p}}}=\frac{{N}_{\text{p}}}{{N}_{\text{s}}}$

is the relationship between the output and input currents of a transformer. So if voltage increases, current decreases. Conversely, if voltage decreases, current increases.

## Calculating characteristics of a step-up transformer

A portable x-ray unit has a step-up transformer, the 120 V input of which is transformed to the 100 kV output needed by the x-ray tube. The primary has 50 loops and draws a current of 10.00 A when in use. (a) What is the number of loops in the secondary? (b) Find the current output of the secondary.

Strategy and Solution for (a)

We solve $\frac{{V}_{\text{s}}}{{V}_{\text{p}}}=\frac{{N}_{\text{s}}}{{N}_{\text{p}}}$ for ${N}_{\text{s}}$ , the number of loops in the secondary, and enter the known values. This gives

$\begin{array}{lll}{N}_{\text{s}}& =& {N}_{\text{p}}\frac{{V}_{\text{s}}}{{V}_{\text{p}}}\\ & =& \left(\text{50}\right)\frac{\text{100,000 V}}{\text{120 V}}=4\text{.}\text{17}×{\text{10}}^{4}\text{.}\end{array}$

Discussion for (a)

A large number of loops in the secondary (compared with the primary) is required to produce such a large voltage. This would be true for neon sign transformers and those supplying high voltage inside TVs and CRTs.

Strategy and Solution for (b)

We can similarly find the output current of the secondary by solving $\frac{{I}_{\text{s}}}{{I}_{\text{p}}}=\frac{{N}_{\text{p}}}{{N}_{\text{s}}}$ for ${I}_{\text{s}}$ and entering known values. This gives

$\begin{array}{lll}{I}_{\text{s}}& =& {I}_{\text{p}}\frac{{N}_{\text{p}}}{{N}_{\text{s}}}\\ & =& \left(\text{10}\text{.}\text{00 A}\right)\phantom{\rule{0.10em}{0ex}}\frac{\text{50}}{4\text{.}\text{17}×{\text{10}}^{4}}\phantom{\rule{0.10em}{0ex}}=\phantom{\rule{0.10em}{0ex}}\text{12.0 mA}\text{.}\end{array}$

Discussion for (b)

As expected, the current output is significantly less than the input. In certain spectacular demonstrations, very large voltages are used to produce long arcs, but they are relatively safe because the transformer output does not supply a large current. Note that the power input here is ${P}_{\text{p}}={I}_{\text{p}}{V}_{\text{p}}=\left(\text{10}\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{A}\right)\left(\text{120}\phantom{\rule{0.25em}{0ex}}\text{V}\right)=1\text{.}\text{20}\phantom{\rule{0.25em}{0ex}}\text{kW}$ . This equals the power output ${P}_{\text{p}}={I}_{\text{s}}{V}_{\text{s}}=\left(\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{mA}\right)\left(\text{100}\phantom{\rule{0.25em}{0ex}}\text{kV}\right)=1\text{.}\text{20}\phantom{\rule{0.25em}{0ex}}\text{kW}$ , as we assumed in the derivation of the equations used.

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.

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