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Phyics

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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 - ,

    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 - ,

    Thanks a lot!

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