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Why is the coils of secondary turn thinner than that of coils of primary turns of step...
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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 `
Posted by valentin68 on September 19, 2013 at 11:02 AM (Answer #1)
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