Quick questions on Transformers and AC transmission: HSC Physics Module 6

A transformer is two coils (the primary and secondary) wound on the same ferromagnetic core. An alternating current in the primary produces a changing flux in the core. The same changing flux links the secondary, inducing an EMF in it (Faraday's law).

An ideal transformer has no losses. Power in equals power out:

Faraday's law requires $d\Phi / dt \neq 0$ to induce a secondary EMF. A DC primary current produces a constant flux, so the secondary EMF is zero (except during the brief switch-on transient). AC, by changing direction many times per second, produces the continuous flux change required.

Typical efficiencies: 95 percent for small transformers, above 99 percent for large grid transformers.

Transmitting electrical power $P = VI$ over a long line of resistance $R_{\text{line}}$ wastes power as $P_{\text{loss}} = I^2 R_{\text{line}}$. The loss depends on the current squared, not the voltage. So for the same transmitted power $P$, a higher transmission voltage means a smaller current and dramatically smaller line losses.

A small power station delivers $1.0$ MW to a town through a transmission line of total resistance $5.0$ ohms. Compare the line losses at $1000$ V transmission versus $100$ kV transmission.

$V_s / V_p = N_s / N_p$, but $I_p / I_s = N_s / N_p$. The voltage ratio and the current ratio are inverses of each other.

A common error in design questions; the device gives no output (no induced EMF) and may overheat. Always state the AC requirement when explaining transformer operation.

Eddy-current losses and hysteresis losses are both in the core; resistive losses are in the windings; flux leakage is a geometry issue. Markers expect you to distinguish them.

It is to reduce $I^2 R$ loss, which depends on $I^2$, not on $V$. Higher $V$ at the same $P$ means lower $I$.