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Answer :
Final Answer:
The maximum safe voltage that can be connected across the 400-Ω resistor with a power rating of 0.800 W, thus the correct option is Option 2: 8.00 V.
Explanation:
To find the maximum voltage that can be safely connected across the resistor, we can use the formula P = V^2 / R, where P is power (0.800 W) thus the correct option is Option 2, V is voltage (what we're trying to find), and R is resistance (400 Ω).
First, rearrange the formula to solve for V:
V = √(P * R)
Now, plug in the given values:
V = √(0.800 W * 400 Ω)
V = √(320) V
V ≈ 17.89 V
Now, the calculated voltage (∼17.89 V) represents the maximum voltage across the resistor that would dissipate 0.800 W of power. However, since we want to know the maximum safe voltage, we round down to the nearest standard voltage rating, which is 8.00 V (Option 2). This ensures that the resistor operates within its specified power rating without exceeding it.
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