Power vs Resistance: 2.35(9+.05)^2-2(r+.5)/(r+0.5)^4

  • Thread starter Thread starter sammiyahc0
  • Start date Start date
  • Tags Tags
    Power Resistance
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
2 replies · 2K views
sammiyahc0
Messages
9
Reaction score
0
1.
The power delivered by a battery to an apparatus of resistance R (in ohms) is P = 2.35R /(R + 0.5)^2
watts.Find the rate of change of power with respect to resistance for R = 9.(Round your answer to four decimal places

2. quotient rule
3. 2.35(9+.05)^2-2(r+.5)/(r+0.5)^4
 
Physics news on Phys.org

Discussion

You've not applied the quotient rule correctly. You dropped a factor in the second term, and a 2.35 has been reduced to 2. it would help if you included all appropriate parentheses. Why have you substituted r = 9 in some places but not in others?
 
What you have written is (I think - you appear to have substituted R=9 in some places but not others):[tex]2.35(R+0.5)^2-\frac{2(R+0.5)}{(R+0.5)^4}[/tex]which I don't think is right. Even assuming that you meant [tex]\frac{2.35(R+0.5)^2-2(R+0.5)}{(R+0.5)^4}[/tex]I still don't think this is right. I was taught that the quotient rule as[tex]d\frac{u}{v}=\frac{vdu-udv}{v^2}[/tex]Perhaps you could show us what you think [itex]u, v, du, dv[/itex] are?