Transformer Current Ratio⚠ unverified

Understanding the Parameters

• $I_p$, $I_s$ — the winding currents. Where voltage is lower, current is proportionally higher.
• $a$ — equals the turns ratio $n = N_p/N_s$ (see Transformer Turns Ratio); the secondary carries $n\times$ the primary current when stepping voltage down by $n$.

Derivation (Approaching a Proof)

An ideal transformer dissipates no power, so the power in equals the power out:

$$V_p I_p = V_s I_s.$$

Rearranging and using the voltage/turns ratio $V_p/V_s = N_p/N_s = n$ (see Transformer Turns Ratio):

$$\frac{I_s}{I_p} = \frac{V_p}{V_s} = \frac{N_p}{N_s} = n.$$

So currents scale inversely to voltages. A useful corollary is impedance transformation: a load $Z_s$ on the secondary appears from the primary as $Z_p = n^2 Z_s$ — transformers scale impedance by the square of the turns ratio, the basis of impedance matching.

History

The inverse current relation follows from Faraday's law plus energy conservation and was understood as soon as the transformer itself was (1880s). Stepping current up while stepping voltage down is what lets a small transmission current deliver large load current locally — and, in reverse, what makes current transformers practical for metering.

Related Concepts: Transformer Turns Ratio, Transformer Efficiency, Watt's Law

Notes: Registry calculator transformer-current-ratio (unverified). Ideal device; magnetizing current and losses make the real ratio slightly off.