Transformer step-up to the cell

What voltage reaches the cell for a given turns ratio, and what current does that cost?

U = V \times \frac{s}{p} \qquad J = I \times \frac{p}{s}

Method

1. Divide the secondary turns by the primary turns. That ratio is the whole of the transformer’s behaviour; the absolute number of turns matters for saturation and losses, not for the ratio.
2. Multiply the primary voltage by the ratio. That is the secondary voltage.
3. Divide the primary current by the same ratio for the secondary current. Voltage up means current down, by exactly the same factor.
4. Multiply volts by amps on each side. For an ideal transformer the two products are equal — the device moves power between windings, it does not create it.
5. In a VIC this secondary voltage is the starting point, not the finish: the resonant rise across the cell multiplies it again by the circuit’s Q.

Assumptions

• The transformer is ideal — perfect coupling, no leakage inductance, no winding resistance and no core loss. A real VIC transformer with a gapped core has significant leakage, and the secondary voltage under load falls below this.
• The core is not saturating. Past saturation the inductance collapses, the primary current rises steeply and the secondary voltage stops following the ratio at all.
• The secondary is lightly loaded. These relations describe an unloaded or lightly loaded winding; a heavy load pulls the secondary voltage down through the leakage impedance.
• This is the transformer alone. Any resonant rise across the cell happens after this and is a separate multiplication — see the Q factor calculation.
• Power is conserved. A step-up transformer does not produce energy; any claim that the output power exceeds the input has to be met somewhere other than in this equation.