Question

A portable x-ray unit has a step-up transformer. The 120 V input is transformed to the 100 kV output needed by the x-ray tube. The primary has 50 loops and draws a current of 13.9 A when in use. What is the number of loops in the secondary

Answer 1

`N_s\approx41667 \hspace{3}lo ops`

Step-by-step explanation:

In an ideal transformer, the ratio of the voltages is proportional to the ratio of the number of turns of the windings. In this way:

`(V_p)/(V_s) =(N_p)/(N_s) \\\\Where:\\\\V_p=Primary\hspace{3} Voltage\\V_s=V_p=Secondary\hspace{3} Voltage\\N_p=Number\hspace{3} of\hspace{3} Primary\hspace{3} Windings\\N_s=Number\hspace{3} of\hspace{3} Secondary\hspace{3} Windings`

In this case:

`V_p=120V\\V_s=100kV=100000V\\N_p=50`

Therefore, using the previous equation and the data provided, let's solve for
`N_s` :

`N_s=(N_p V_s)/(V_p) =((50)(100000))/(120) =(125000)/(3) \approx41667\hspace{3}loo ps`

Hence, the number of loops in the secondary is approximately 41667.