Answer:
c) quadruple in magnitude
Step-by-step explanation:
The power dissipated in the circuit is given by:
![P=I^2 R](https://img.qammunity.org/2020/formulas/physics/high-school/4htslhg8uce1y0sravvkgvdb5hjhj3ylyv.png)
where
I is the current in the circuit
R is the total resistance of the circuit
In this problem:
- The current is doubled: I' = 2 I
- The resistance is kept constant: R' = R
So, the power dissipated is
![P' = (I')^2 R' = (2I)^2 R=4 I^2 R=4 P](https://img.qammunity.org/2020/formulas/physics/high-school/deyylys28thp51gjpfr62fkqju03wfysht.png)
so, the power dissipated increase by a factor 4 (quadruples).