Question

A portable x-ray unit has a transformer, the 120V input of which is transformed to the 100kV output needed by the x-ray tube. The primary has 50 loops and draws a current of 10A when in use. What is ratio of the number of loops in the secondary to the number of loops in the primary

Answer 1

2500 : 3

Step-by-step explanation:

To solve this, let's build some relationships among the power, voltage, current and number of loops in an ideal transformer.

For an ideal transformer, the powers generated by the primary and secondary coils are the same. In other words, the input power and output power remain the same.

Input power = Output power

Where;

Input power =
`V_(p)` x
`I_p`

Output power =
`V_(s)` x
`I_s`

=>
`V_(p)` x
`I_p` =
`V_(s)` x
`I_s`

This can be re-written as:

`(V_p)/(V_s)` =
`(I_s)/(I_p)` ------------------(i)

Also, the transformer turns ratio (r) is the quotient of the number of turns/loops in the primary coil and the number of turns/loops in the secondary coil. i.e

r =
`(N_p)/(N_s)`

This ratio is equal to the ratio given in equation (i) i.e

r =
`(N_p)/(N_s)` =
`(V_p)/(V_s)` =
`(I_s)/(I_p)`

=> r =
`(N_p)/(N_s)` =
`(V_p)/(V_s)`

=>
`(N_p)/(N_s)` =
`(V_p)/(V_s)`

This can be re-written as

=>
`(N_s)/(N_p)` =
`(V_s)/(V_p)` -----------(ii)

From the question,

`V_(p)` = 120V

`V_(s)` = 100kV = 100000V

Substitute these values into equation (ii) as follows;

`(N_s)/(N_p)` =
`(100000)/(120)`

`(N_s)/(N_p)` =
`(10000)/(12)`

`(N_s)/(N_p)` =
`(2500)/(3)`

Therefore, the ratio of the number of loops in the secondary to the number of loops in the primary is 2500 : 3