This is a common misconception among high-school students. We, students normally think that a formula will be applicable anywhere. Infact there are some terms and conditions for each and every formula in physics.
Let's take this case.
Here, the formula you're trying to use is:.
$$V=IR$$
Do you remember from which law this formula was derived from?
It's the Ohm's law and the formula $V=IR$ is used for DC circuits. The circuit we're dealing here is an AC one. Though the Ohm's law is applicable for both DC and AC, the formula has some changes. In AC circuit the formula takes this form:
$$V=IZ$$
Where, $Z$ is the impedence.
$$Z=R+jX$$
Where, $R$ is resistance and $X$ is reactance.
You'll learn more of this in Higher Secondary.
Now comes the main point.
According to Ohm's law , "if there's no change in the physical state of the conductor (such as temperature), then the ratio of the potential difference applied at its ends and the current flowing through it is constant."
Which means,
$$V/I=R=constant$$
But in the case of transformer, the power is constant, not the resistance. Which means,
$$VI=P=constant$$
In the formula $V=IR$ , there are no constant terms. Therefore, the current you obtain will be wrong.
To check if your answer is wrong, just reverse the procedure. See if you can obtain the power given in the question from the current you've found and the potential difference in the question.
$P_{input} = 10kW,$
$R = 1Ω,$
$V = 1000V$
$$Since, V = IR,$$
$$\implies I = V/R$$
$$\implies I = 1000V/1Ω$$
$$\implies I = 1000A$$
Now, let's find the input power from the current obtained.
$$P_{input} = VI$$
$$\implies P_{input} = 1000V×1000A$$
$$\implies P_{input} = 1000kW$$
$$But, P_{input} = 10kW$$
$$Therefore, I ≠ 1000A$$
This is also applicable to the $V=IR$ formula. Let's try this question:
Consider a DC circuit with a resistor $R$ of resistance $20Ω$ . The current in the circuit is $2A$ and the power is $100W$ . Find the potential difference applied.
$V=IR$ is the formula to be used here. But let's try to find potential difference using the formula $V=P/I$ . Note that the formula $V=P/I$ contains no constant ($R$ is the constant here).
$R = 20Ω, P = 100W, I = 2A$
$$Since, V = P/I,$$
$$\implies V = 100W/2A$$
$$\implies V = 50V$$
Let's reverse check this.
$$R = V/I$$
$$\implies R = 50V/2A$$
$$\implies R = 25Ω$$
$$But, R = 20Ω$$
$$Therefore, V ≠ 50V$$
See? The power given in my question and the resistance given in your question are unwanted information. The only intention behind putting these unwanted information in the question is to confuse us and make us lose marks. So, only take what's necessary from the question.
First I will try to answer your second question. A transformer consists of one solenoid producing a magnetic field that affects a second solenoid. The magnetic field produced in a large solenoid is calculated from:
$$B=\mu_0IN/L$$
where $\mu_0$ is the permeability constant, $I$ is current, $N$ is the number of turns and $L$ is the length of the solenoid. As you can see, it depends only on the current, because the magnetic field according to Maxwell's laws is caused by the movement of changes, not by the potential in which those charges are.
The process is much more complex than it seems. The current of the primary coil generates a magnetic field, which generates a voltage in the second coil (due to Maxwell's laws, and yes it seems confusing), as I have just said. This voltage divided by the resistance of the second coil will give you the current of the second coil. The difficulty comes when you find that the current of the primary coil depends on the resistance of the first coil, the resistance of the second coil, the inductance of both coils, the voltage applied to the first coil, and you will get a time-depending current. Why this dependence? Because the magnetic fields act like very weird resistors. Apart from that, the problem would be even more complex as there is a magnetic field caused by the secondary coil, but I won't go into that.
About the inductor (your first question, I think), I am not sure I have understood it correctly but you have to consider the inductor when calculating the current of a circuit as Ohm's law has to be modified. You don't have that $V=RI$ anymore, but that $V=RI-L\frac{dI}{dt}$
being L the inductance, which depends on the number of turns of the inductor, its area, etc. You will find that this means that there the voltage drop is caused both by the resistor and by the inductor.
Best Answer
The step-up transformer boosts voltage on the secondary. Since the power must be preserved on primary and secondary, the current is reduced by the same factor:
$$V_P \cdot I_P = V_S \cdot I_S$$
The loads are rarely characterized as resistances. In reality they have some complex impedance. For example, motors are modelled as inductances, air conditioning devices even have negative impedance etc.
For transmission you should think of it like this: there is some power $P$ required by loads; if the voltage on the transmission line is $V$ then the current will be $I = \frac{P}{V}$. The total power output equals loads plus ohmic losses in the transmission line.
If the load is purely resistive as in your example, then power increases with voltage since $P_R = \frac{U^2}{R}$.
See related discussion about power transmission: https://physics.stackexchange.com/a/682024/149541