When a 100-? resistor is connected to a battery of emf E and internal resistance

Question

When a 100-? resistor is connected to a battery of emf E and internal resistance r, the battery delivers a power P = 0.794 W. When a 200 ? resistance is connected to same battery it delivers a power P = 0.401 W. What are the emf and internal resistance of the battery?

Answer

E   = 9 V, r = 2 ?

E   = 4.5 V, r = 4 ?

E   = 18 V, r = 4 ?

E   = 10 V, r = 5 ?

E   = 12 V, r = 6 ?

  

     

E   = 9 V, r = 2 ?

     

     

E   = 4.5 V, r = 4 ?

     

     

E   = 18 V, r = 4 ?

     

     

E   = 10 V, r = 5 ?

     

     

E   = 12 V, r = 6 ?

  

Explanation / Answer

P = E^2/ R

0.794 = E^2 / (100+r)

0.401 = E^2/( 200 +r)

Dividing we get

0.794/0.401 = (200+r)/(100+r)

79.4 + 0.794 r = 80.2 + 0.401 r

0.393 r = 0.8

r = 8/3.93

= 2 ohms

E^2 = 0.794 * ( 102)

E = 9 V

E = 9 V, r = 2 ?

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