In a step-up transformer, if the voltage in the secondary is increased, then the current in the primary:

For an ideal transformer, the input power is equal to the output power.

$V_p I_p = V_s I_s$

Let a constant load resistance $R$ be connected across the secondary coil. The current in the secondary coil is given by $I_s = \dfrac{V_s}{R}$.

Substituting this into the power equation, we get:

$V_p I_p = V_s \left(\dfrac{V_s}{R}\right)$

$V_p I_p = \dfrac{V_s^2}{R}$

$I_p = \dfrac{V_s^2}{V_p R}$

For a constant primary supply voltage $V_p$ and a fixed load resistance $R$, the primary current $I_p$ is directly proportional to the square of the secondary voltage ($I_p \propto V_s^2$).

Therefore, if the voltage in the secondary is increased, the current in the primary also increases.

Answer: Increases