Answer:
The power dissipated is reduced by a factor of 2
Explanation:
The power dissipated by a resistor is given by:
[tex]P=I^2 R[/tex]
where
I is the current
R is the resistance
by using Ohm's law, [tex]I=\frac{V}{R}[/tex], we can rewrite the previous equation in terms of the voltage applied across the resistor (V):
[tex]P=(\frac{V}{R})^2R=\frac{V^2}{R}[/tex]
In this problem, the resistance of the element is doubled, while the voltage is kept constant. So we have [tex]R'=2R[/tex] while V remains the same; substituting into the formula, we have:
[tex]P'=\frac{V^2}{2R}=\frac{1}{2}\frac{V^2}{R}=\frac{P}{2}[/tex]
so, the power dissipated is reduced by a factor 2.
