Phyics
posted by ALan on .
A resistor has a resistance R, and a battery has an internal resistance r. When the resistor is connected across the battery, ten percent less power is dissipated in R than would be dissipated if the battery had no internal resistance. Show that the ratio r/R is equal to 0.054
I let R = 1 ohms
r 0.054 ohms
*Without Internal resistance*
Vloop=0
emfIR=0
emf=IR
*With internal resisntace*
Vloop=0
emfIRIr=0
emf=IR+Ir
I get stuck after this, I am unsure where to head now. Can someone please write out the procedure? Please and thank you.

P1 = V^2/R is the power dissipated in R with no internal resistance.
With internal resistance r, the current is V/(R+r) and the power dissipated in R is
P2 = I^2*R = V^2*R/(R+r)^2
If P2/P1 = 0.9,
R^2/(R+r)^2 = 0.9
R/(R+r) = 0.9487
(R+r)/R = 1 + r/R = 1.054
r/R = 0.054 
Thanks a lot!