Imagine two rubber balloons, A and B, where B's rubber wall is twice as thick as balloon A.
They are both inflated by injecting air into them, but balloon B's thicker wall will resist inflation more. To inflate them both with 1 litre of air would take much more physical effort for B than A would require, and yet it's the same amount of air in both cases.
In other words, you expend more Joules of energy inflating balloon B than for A, to the same volume, to move the same amount of air from outside the balloons to inside.
To extend this analogy to power, remember that power is the rate of doing work, the number of Joules of "exertion" made each second. The unit is Watts, which means Joules per second. If you intend to inflate both balloons to the same volume of 1L in the same amount of time, obviously you will have to work harder blowing into balloon B than A. In this scenario you are doing more Joules of work per second (power, in Watts) for ballon B than for A.
Conversely, if you don't care how long it takes, then you can make the same effort to inflate both balloons, but for balloon B you will have to make that effort for longer. In this scenario you are delivering the same power to both balloons, but for far longer in the case of balloon B.
The rate of air flow here is analagous to electric current, which is charge flow. Balloon volume would be analogous to total electric charge moved. I hope it's clear by this point that you can't measure power, or work done by knowing only how much air/charge moved, since in all cases it has been 1 litre.
Current tells you how much charged moved, in the same way volume tells you how much air you moved, but it doesn't tell you anything about how much effort the battery or lungs made in order to do the moving. To know that, you also need to know how hard you blew into the balloon, or how hard the battery/mains pushed charge through some device.
For the air analogy, clearly the measure of "how hard" is pressure, and for electricity, that measure is voltage.
If you have to blow twice as hard (twice the pressure) to inflate ballon B to 1L in 10s, as you do to inflate balloon A to 1L in the same time, then you've done twice the amount of work over the same duration, and therefore delivered twice the power. That's why fluid power is proportional to pressure, and electrical power is proportional to voltage.
Now understand that to inflate any balloon to 1L, you can control the flow of air into it, and that rate of flow will determine how long it takes to deliver each Joule of energy. In other words, doubling the air-flow/current will also double the the number of Joule's of work done each second. That's why electrical power is also a function of electric current.
If you double electric current, you double the power, the number of Joules of energy delivered each second to whatever that current is flowing through. If you double the voltage across something, then you also double the power delivered to that thing.
Crucially, what happens if you double both? You quadruple the power. For a fixed electrical resistance, something interesting happens. Let's say you double the voltage across a resistor. Ohm's law tells us what happens to the current:
$$
\begin{aligned}
\frac{V}{R} &= I \\ \\
\frac{2V}{R} &= 2I
\end{aligned}
$$
Doubling the voltage across the resistor will also double the current through it, and the combined effect on power is a quadrupling. Similarly, if you double the current through a resistor, you also consequently double the voltage across it. Either case therefore results in a quadrupling of power, a relationship which can be clearly seen in the "square" terms of the two other famous power equations:
$$
\begin{aligned}
P = \frac{V^2}{R} \\ \\
P = I^2 \times R
\end{aligned}
$$
TLDR; To get to your question, current on its own cannot tell you about energy delivered per second. Voltage alone cannot tell you this either. Only together can you know the power:
$$ P = I \times V $$
Few household appliances have constant electrcal resistance, though an electric kettle comes close. For a kettle, doubling the voltage supply from its usual 110V to 220V would (if the kettle survived) cause the kettle to boil the same amount of water in a quarter of the time, because four times as many Joules of heat energy are delivered to the water each second.
Most electrical appliances are not so simple, and may not be treated as constant resistances. As an example, if you had a multi-voltage TV, rated at 100W, doubling the supply voltage would halve the current, to maintain 100W. That requires some complex switch-mode power supply magic, but it is possible.
Most appliances would just die if you doubled the voltage. Unless they are designed to cope with the sudden doubling or quadrupling of power delivery (which entails some kind of dynamic control of their "apparent resistance"), they usually deal with it by not doing anything any more. Like when you put your back out trying to deliver too many Joules of gravitational potential energy to something in too short a time.