Understanding AC Power: Maximum and Minimum Currents on Resistor and Inductance

[0fibonacci1]

Homework Statement

I have some questions about powers in AC circuit. What is the maximum and minimum current power on resistor and inductance.

Homework Equations

Um = U / V2
Im = I / V2

The Attempt at a Solution

prmax = U*I=(Um/V2)*(Im/V2)=(Um*Im)/2. But answer seems to actually be that prmax = Um*Im, how can that be? And for plmax = 1/2 Um * I am plmin=- 1/2 Um *Im

 

[tiny-tim]

hi 0fibonacci1!

your notation is difficult to understand

here's an extract from the pf library on impedance …

Power:

Power = work per time = voltage times charge per time = voltage times current:

$$P = VI =\ V_{max}I_{max}\cos(\omega t + \phi/2)\cos(\omega t - \phi/2)$$
$$=\ V_{max}I_{max}(\cos\phi + \cos2\omega t)/2$$​

(because $2cosAcosB = cos(A-B) + cos(A+B))$)

$$=\ V_{rms}I_{rms}(\cos\phi + \cos2\omega t)$$
$$=\ \left(V_{rms}^2\left/|Z|\right.\right)(\cos\phi + \cos2\omega t)$$​

So (instantaneous) power is the constant part, $P_{av} = V_{rms}I_{rms}\cos\phi$ (the average power), plus a component varying with double the circuit frequency, $V_{rms}I_{rms}\cos2\omega t$ (so a graph of the whole power is a sine wave shifted by a ratio $\cos\phi$ above the x-axis).​

 

[NascentOxygen]

The Attempt at a Solution

prmax = U*I=(Um/V2)*(Im/V2)=(Um*Im)/2.

It almost seems that you might be using V2 as denoting √[/size]2 ?