The rms current in a 41-Ω resistor is 0.69 A. What is the peak value of the voltage across this resistor?

Answer: V(rms) = A(ohms)(sqrt2)

To find the peak value of the voltage across the resistor, we can use the formula:

V(peak) = V(rms) * sqrt(2)

Given that the rms current is 0.69 A, and the resistor is 41 Ω, we can substitute the values into the formula:

V(peak) = 0.69 A * sqrt(2) * 41 Ω

Now, we can calculate the peak value of the voltage across the resistor:

V(peak) = 0.69 A * 1.414 * 41 Ω

V(peak) = 38.07 V (rounded to two decimal places)

Therefore, the peak value of the voltage across the 41-Ω resistor is approximately 38.07 V.

To find the peak value of the voltage across the resistor, we can use the formula:

V(peak) = V(rms) * sqrt(2)

Given that the rms current in the resistor is 0.69 A, we need to multiply it by the resistance value of 41 Ω and then multiply by sqrt(2) to get the peak voltage.

V(peak) = 0.69 A * 41 Ω * sqrt(2)

Let's calculate it:

V(peak) ≈ 0.69 A * 41 Ω * 1.414 (approximating sqrt(2) to 1.414)

V(peak) ≈ 30.139 V (rounded to three decimal places)

Therefore, the peak value of the voltage across the 41-Ω resistor is approximately 30.139 V.