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The Figure below shows a potentiometer circuit in which the driver cell D has an emf of 6 V and internal resistance of 2 Ω. The potentiometer wire AB is 10 m long and has a resistance of 28 Ω.

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Question

The Figure below shows a potentiometer circuit in which the driver cell D has an emf of 6 V and internal resistance of 2 Ω. The potentiometer wire AB is 10 m long and has a resistance of 28 Ω. The series resistance RS is of 2 Ω.

  1. The current Ip flowing in the potentiometer wire AB when the jockey (J) does not touch the wire AB.
  2. emf of the cell X if the balancing length AC is 4.5 m.
Numerical
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Solution

  1. Resistance of 10 m wire AB = 28 Ω
    Rs = 20 Ω
    r = 2 Ω

    ∴ Total resistance (R) = (28 + 20 + 2) Ω = 50 Ω
    `I_p = V/R`
    = `6/50  A`
    = 0.12 A
  2. VAB = IP × RAB
    = 0.12 × 28 V
    = 3.36 V
    emf of cell `X = Kl   ...(("Here"  K = 3.36/10  Vm^-1 = 0.336  Vm^-1),(and l = 4.5  m))`
    X = 0.336 × 4.5 V
    X = 1.512 V
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