Posted by **ALan** on Wednesday, February 10, 2010 at 12:07am.

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

emf-IR=0

emf=IR

*With internal resisntace*

Vloop=0

emf-IR-Ir=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.

- Phyics -
**drwls**, Wednesday, February 10, 2010 at 1:09am
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

- Phyics -
**ALan**, Wednesday, February 10, 2010 at 11:16am
Thanks a lot!

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