If the power rating of a 400-Ω resistor is 0.800 W, what is the maximum voltage that can safely be connected across the resistor? Option 1: 4.00 V. Option 2: 8.00 V. Option 3: 12.00 V. Option 4: 16.00 V.

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.

Learn more about resistors

brainly.com/question/37375337

#SPJ11