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प्रश्न
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.
उत्तर
- 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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संबंधित प्रश्न
A potentiometer wire has resistance of per unit length of 0.1 Ω/m. A cell of e.m.f. 1.5 V balances against a 300 cm length of the wire. Find the current in the potentiometer wire.
Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
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(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
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