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