Answer:
The power dissipated is quadrupled
Step-by-step explanation:
The power dissipated by a circuit is given by
![P=I^2 R](https://img.qammunity.org/2020/formulas/physics/high-school/4htslhg8uce1y0sravvkgvdb5hjhj3ylyv.png)
where
I is the current
R is the resistance
In this problem, the resistance is kept constant, while the current is doubled:
I' = 2I
So, the new power dissipated is
![P'=I'^2 R = (2I)^2 R= 4I^2 R=4P](https://img.qammunity.org/2020/formulas/physics/high-school/qtpl7s053sg2ceiltdl4v07mojo9yaea34.png)
So, the power dissipated is quadrupled.