Why is the coils of secondary turn thinner than that of coils of primary turns of step up transformer?

Expert Answers
valentin68 eNotes educator| Certified Educator

In a step up transformer the rapport n between the secondary voltage `V_s` and the primary voltage `V_p` is greater than 1.

`n = V_s/V_n > 1`

If we consider the transformer ideal (no power losses), the electric power in the secondary coil need to be equal to the electric power in the primary coil `P_s =P_p` . Since the electric power is by definition

`P = V*I` this means `V_p*I_p = V_s*I_s` . By rearranging this relation we get

`I_p/I_s =V_s/V_n = n >1 `

which means the current in the primary coil is n times bigger than the current in the secondary coil.

To keep the wire cold, the density of current (`J = I/S` ) in both primary and secondary coil wires must not exceed `J= 2.5 -: 3 A/(mm^2)` .

Therefore because `I_p > I_s` the diameter `d_p` of the primary coil wire will be bigger than the diameter `d_s`  of the secondary coil wire.

`S_p =n*S_s`  which implies `d_p^2 =n*d_s^2 `

Access hundreds of thousands of answers with a free trial.

Start Free Trial
Ask a Question