Q 1 :    

To measure the internal resistance of a battery, a potentiometer is used. For R=10Ω, the balance point is observed at l= 500 cm and for R=1Ω the balance point is observed at l = 400 cm. The internal resistance of the battery is approximately:       [2024]

  • 0.2Ω

     

  • 0.4Ω

     

  • 0.1Ω

     

  • 0.3Ω

     

(4)

At balance point, the current in circuit is i=εr+R

VAB=ε-ir=ε-(εr+R)·r

VAB=εr+εR-εrr+R=εRr+R

also, VAB=λ,λ is potential gradient.

λ=εRr+R

Now, l1l2=R1r+R1×r+R2R2500400=10r+10×r+11

5(r+10)=40(r+1)  (r+10)=8(r+1)

7r=2  r=270.3