Does a higher voltage always mean a higher electric field strength? In a step-up transformer, the output voltage is higher than the input voltage, while the output current is lower than the input current. Basically, since P = VI is conserved, the current has to decrease proportionately in the secondary coil if the voltage increases. But as we know that the voltage in the secondary coil is higher, does this mean that the electric field strength is also higher there? (Normally, when we increase the voltage by adding more cells in series, we are basically increasing the electric field strength that causes the current, right?)
If E is higher in the secondary coil, then shouldn't the electrons experience more drift velocity, and hence shouldn't the circuit produce more current?
 A: I do not understand your statement, So now, the potential difference between the secondary coil of the step-up transformer is 9V, but this time, the current will decrease.
$9\rm V$ ac RMS means that the average power dissipated over a cycle is the same as that dissipated when a $9\rm V$ dc supply is connected and so the RMS current should equal the dc current.
If it is a sinusoidal voltage variation then the instantaneous power varies from zero to being proportional to $(9\sqrt 2)^2 = 162$ where $9\sqrt 2$ is the peak voltage.
In terms of what is happening to the electrons and the electric fields inside the resistor they will suffer a variation equal to the frequency of the supply, sometime being zero and sometimes reaching a maximum value.
A: This an answer to the amended question.
You are mixing up two ideas here.
If the transformer is $100\%$ efficient then $V_{\rm primary}I_{\rm primary} =V_{\rm secondary}I_{\rm secondary}$.
If you keep the same value resistor connected to the secondary and increase the voltage across it, $V_{\rm secondary}'$, by increasing the primary voltage, $V_{\rm primary}'$, then the secondary current, $I_{\rm secondary}'$, will increase as $V_{\rm secondary}'=R\,I_{\rm secondary}'$.
This will result in the primary current,$I_{\rm primary}'$, increasing.
So now you will have more input power and more output power, ie $V_{\rm primary}' \,I_{\rm primary}'=V_{\rm secondary}' \,I_{\rm secondary}'$
