Question 23.5: A portable x-ray unit has a step-up transformer, the 120 V i...

Strategy and Solution for (a)
We solve $\frac{V_{ s }}{V_{ p }}=\frac{N_{ s }}{N_{ p }}$ for $N_{ s }$, the number of loops in the secondary, and enter the known values. This gives

$N_{ s }=N_{ p } \frac{V_{ s }}{V_{ p }}$                 (23.30)

$=(50) \frac{100,000 V }{120 V }=4.17 \times 10^{4}$.

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 $\frac{I_{ s }}{I_{ p }}=\frac{N_{ p }}{N_{ s }}$ for $I_{ s }$ and entering known values. This gives

$I_{ s }=I_{ p } \frac{N_{ p }}{N_{ s }}$                  (23.31)

$=(10.00 A ) \frac{50}{4.17 \times 10^{4}}=12.0 mA$.

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 $P_{ p }=I_{ p } V_{ p }=(10.00 A )(120 V )=1.20 kW$. This equals the power output $P_{ p }=I_{ s } V_{ s }=(12.0 mA )(100 kV )=1.20 kW$, as we assumed in the derivation of the equations used.