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*
*With internal resisntace*
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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