Answer:
No.
Explanation:
[tex]P_r[/tex] = Power rating = 1 W
R = Resistance = [tex]350\ \Omega[/tex]
V = Voltage = [tex]24\ \text{V}[/tex]
Power is given by
[tex]P=\dfrac{V^2}{R}\\\Rightarrow P=\dfrac{24^2}{350}\\\Rightarrow P=1.65\ \text{W}[/tex]
[tex]1.65\ \text{W}>1\ \text{W}[/tex]
So
[tex]P>P_r[/tex]
Hence, the resistor is not operating within its power rating.
