When two resistors with equal resistances are wired in series the voltage across each resistor is equal to half the voltage of the power supply. Is this also the case the instant the power supply is turned on? At the instaneous moment when the electric current has just reached the first resistor in series is the voltage across that resistor equal to half the voltage of the battery or all the voltage of the battery? In the case of the former being true, how is the current "aware" of the second resistor in the circuit before encountering it? In the case of the later, how does the circuit eventually reach equilibrium?
Is this also the case the instant the power supply is turned on?
For ideal resistors in circuit theory, it's true as soon as the power supply is turned on.
In the real world, no power supply actually turns on instantly. And each resistor has a small parasitic capacitance. And the wires connecting the power supply to the resistors have parasitic indcutance. And each wire has a parasitic capacitance to each other wire in the circuit. If you introduce the effects of these parasitics into your ideal circuit model, you will find that the current does not equalize perfectly through the main wires because some goes through the parasitic capacitors. But this effect is miniscule, due to the non-zero turn-on time of the supply and the wire inductances slow the current turn-on and allow it to stay pretty darn near equal around the loop as it turns on.