23.6 Transformers  (Page 2/8)

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

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

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 $P_{\text{p}}=I_{\text{p}}{V}_{\text{p}}=(\text{10}\text{.}\text{00}\text{A})(\text{120}\text{V})=1\text{.}\text{20}\text{kW}$ . This equals the power output $P_{\text{p}}=I_{\text{s}}{V}_{\text{s}}=(\text{12}\text{.}0\text{mA})(\text{100}\text{kV})=1\text{.}\text{20}\text{kW}$ , as we assumed in the derivation of the equations used.