Question

Derive the relationship V_P/V_S = I_S/I_P for a transformer.

Answer 1

An alternating emf VP from an ac source is applied across the primary coil of a transformer. This sets up an alternating current IP in the primary circuit and also produces an alternating magnetic flux through the primary coil such that

V_p = - Np \(\cfrac{dΦ_P}{dt}\),

where N_p is the number of turns of the primary coil and Φ_p is the magnetic flux through each turn.

Parts of a step down transformer

Assuming an ideal transformer (i.e., there is no leakage of magnetic flux), the same magnetic flux links both the primary and the secondary coils,

i. e., Φ_p = Φ_s

As a result, the alternating emf induced in the secondary coil,

V_S = N_S \(\cfrac{dΦ_S}{dt}\) = - N_S \(\cfrac{dΦ_P}{dt}\)

where N_S is the number of turns of the secondary coil. If the secondary circuit is completed by a resistance R, the secondary current is I_S = V_S/R, assuming the resistance of the coil to be far less than R. Ignoring power losses, the power delivered to the primary coil equals that taken out of the secondary coil, so

V_pI_p = V_sI_s .

∴ \(\cfrac{V_p}{V_s}\) = \(\cfrac{I_s}{I_p}\)

which is the required expression.