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

Answer:

 $N_s\approx41667 \hspace{3}lo ops$

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:

$\frac{V_p}{V_s} =\frac{N_p}{N_s} \\\\Where:\\\\V_p=Primary\hspace{3} Voltage\\V_s=V_p=Secondary\hspace{3} Voltage\\N_p=Number\hspace{3} of\hspace{3} Primary\hspace{3} Windings\\N_s=Number\hspace{3} of\hspace{3} Secondary\hspace{3} Windings$

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 $N_s$ :

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

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