Testwiki:Reference desk/Archives/Mathematics/2013 September 3

From testwiki
Revision as of 15:50, 25 February 2022 by imported>MalnadachBot (Fixed Lint errors. (Task 12))
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to navigation Jump to search

Template:Error:not substituted

{| width = "100%"

|- ! colspan="3" align="center" | Mathematics desk |- ! width="20%" align="left" | < September 2 ! width="25%" align="center"|<< Aug | September | Oct >> ! width="20%" align="right" |Current desk > |}

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


September 3

generalized eigenvalue maximization using convex feasibility

Find a family of convex functions ϕt:𝕊k×k×𝕊++k×k such that supukuTAuuTButϕt(A,B)0.

AnalysisAlgebra (talk) 04:27, 3 September 2013 (UTC)

capacitive vars

if a 1 kw motor operates at 80% power factor ,how many cap vars are required to increase the power factor to 90% ??cant seem to find the formoula?? — Preceding unsigned comment added by 204.152.90.10 (talk) 19:40, 3 September 2013 (UTC)

If this is a homework question, then it's probably intended that you work it out rather than look up a formula. I suggest you assume that either the current or the voltage is known: so (for example) assume that the voltage is V. Then work out:
  1. What is the in-phase (with V) component of the current (we have the power remember)?
  2. What is the magnitude of the current for 80% power factor?
  3. What is the quadrature (to V) component of the current (Pythagoras may help you here)?
  4. What would the quadrature current be if the power factor was actually 90%?
  5. How much must you reduce the quadrature current to change the power factor from 80% to 90%
  6. Remembering that the voltage was assumed to be V, how many volt-amps is that?

--catslash (talk) 21:20, 3 September 2013 (UTC)