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A potentiometre wore has a resistance 40 ohm and it's length 10m. it is connected to a resistance of 760ohm in series. If emf of battery is 2V then what will be potential gradient?
Verified Answer
A potentiometre wore has a resistance 40 ohm and it's length 10m. it ...
Given in the question :-
Refer to the attachment
Wire resistance = 40 ohm
length = 10 m.
R = 760 ohm
v = 2
We know the formula ,
i = v/R
i = 2/(760 +40)
i = 2/800 A
i = 1/400 A .
Now, Potential drop at wire = i x 40
= 1/10 v.
Potential gradient = Potential drop / length
Potential gradient  = (1/10) / 10 v/m
Potential gradient  = 0.01 v/m.
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Most Upvoted Answer
A potentiometre wore has a resistance 40 ohm and it's length 10m. it ...
Given:
Resistance of potentiometer wire (R) = 40 ohm
Length of potentiometer wire (L) = 10m
Resistance in series (r) = 760 ohm
EMF of the battery (E) = 2V

To Find:
Potential gradient (∆V/∆L)

Solution:

Step 1: Calculate the total resistance in the circuit.
Total resistance (Rt) = R + r

Step 2: Calculate the current flowing in the circuit using Ohm's law.
I = E/Rt

Step 3: Calculate the potential difference across the potentiometer wire using Ohm's law.
∆V = I * R

Step 4: Calculate the potential gradient (∆V/∆L).
Potential gradient (∆V/∆L) = ∆V / L

Step 1: Calculate the total resistance in the circuit.
Rt = R + r
= 40 + 760
= 800 ohm

Step 2: Calculate the current flowing in the circuit using Ohm's law.
I = E/Rt
= 2/800
= 0.0025 A

Step 3: Calculate the potential difference across the potentiometer wire using Ohm's law.
∆V = I * R
= 0.0025 * 40
= 0.1 V

Step 4: Calculate the potential gradient (∆V/∆L).
Potential gradient (∆V/∆L) = ∆V / L
= 0.1 / 10
= 0.01 V/m

Answer:
The potential gradient in the given circuit is 0.01 V/m.
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A potentiometre wore has a resistance 40 ohm and it's length 10m. it is connected to a resistance of 760ohm in series. If emf of battery is 2V then what will be potential gradient?
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