Finding the Peak Value of Current: A Puzzling Solution

[temaire]

Homework Statement

Determine the peak value of current of:

\underline{I} = \frac{120\angle0^0}{10+j10}

The Attempt at a Solution

I change this into polar form and got:

\underline{I} = 12.0 \angle -45^0

Therefore, the peak value of the current should be 12.0 A. However, the solution key says 8.4853 A. How can this be so?

 

[temaire]

I just confirmed with my prof that the solution was wrong. The answer is 12.0 A.

Mods, you can lock this thread.

 

[vk6kro]

I like the old answer better.

The bottom part of the fraction has a magnitude of SQRT( 200) or 14.12.

Divide 120 by this to get 8.485.

 

[temaire]

I like the old answer better.

The bottom part of the fraction has a magnitude of SQRT( 200) or 14.12.

Divide 120 by this to get 8.485.

But you would have to change that to polar form to get the peak current.

 

[vk6kro]

No,

The 120 is already the peak voltage and the 14.12 ohms is the impedance so the peak current is 120 / 14.142 or 8.485 amps.