Question

An rms voltage of 29 V is required for a device whose impedance is 12 12. (a) What should the turn ratio of a transformer be, so the device can be operated from a 125 V line? N2 / N1 = 0.232 (b) Suppose the transformer is accidentally connected in reverse with the secondary winding across the 125 V-rms line and the 12-12 load across the primary. How much rms current will then flow in the primary winding? ХА

Answer 1

(a) Rms voltage of primary coil = V₁ = 125 V.

Rms voltage of secondary coil = V₂ = 29 V.

So, N₂ / N₁ = V₂ / V₁ = 29 V / 125 V

or, N₂ / N₁ = 0.232.

(b) Rms voltage across the secondary coil of turns N₂ is now V₂^/ = 125 V.

So, rms voltage V₁^/ across the primary coil with turns N₁ is given by : V₂^/ / V₁^/ = N₂ / N₁ = 0.232

or, V₁^/ = V₂^/ / 0.232 = 125 V / 0.232 = 538.8 V.

Impedance of the load is R = 12 $\Omega$ .

So, rms current in the primary winding is V₁^/ / R = 538.8 V / 12 $\Omega$ = 44.9 A.