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A potential difference of 2V exists across a potentiometer wire of 2m length. When the potential difference across a 2Ω resistance of a second circuit is measured by this potentiometer wire, it amounts to 5mm balancing length. The current in the second circuit is
  • a)
    250A
  • b)
    25 x 10-4 A
  • c)
    1.5 A
  • d)
    2.5 A
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A potential difference of 2V exists across a potentiometer wire of 2m ...
2.5 mA
Potential gradient in the wire
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A potential difference of 2V exists across a potentiometer wire of 2m ...
Ohm resistor in the circuit is 1V, the length of the wire from one end of the potentiometer needed to balance the circuit is:

We can use the formula for the potential difference across a potentiometer wire:

V = (l/L) * V_pot

where V is the potential difference across a length l of the wire, L is the total length of the wire, and V_pot is the total potential difference across the wire.

In this case, we know that V_pot = 2V and L = 2m. We also know that when the potential difference across the 2 Ohm resistor is 1V, the circuit is balanced. This means that the potential difference across the rest of the circuit (including the potentiometer wire) is also 1V.

Let x be the length of the wire from one end of the potentiometer. Then, the potential difference across this length of wire is:

V = (x/2) * 2V

We want this potential difference to be 1V, so we can set up the equation:

(x/2) * 2V = 1V

Simplifying:

x = 1m

Therefore, the length of the wire from one end of the potentiometer needed to balance the circuit is 1m.
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A potential difference of 2V exists across a potentiometer wire of 2m length. When the potential difference across a 2Ω resistance of a second circuit is measured by this potentiometer wire, it amounts to 5mm balancing length. The current in the second circuit isa)250Ab)25 x 10-4 Ac)1.5 Ad)2.5 ACorrect answer is option 'B'. Can you explain this answer?
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