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# potential gradient for potentiometer depends on internal resistance of driver cell . Explain the statement Grade:Upto college level

## 2 Answers

9 years ago

Potentiometer consists of a long resistive wire AB of length L (about 6 m to 10 m long) made up of mangnine or constantan and a battery of known voltage e and internal resistance r called supplier battery or driver cell. Connection of these two forms primary circuit.

One terminal of another cell (whose emf E is to be measured) is connected at one end of the main circuit and the other terminal at any point on the resistive wire through a galvanometer G. This forms the secondary circuit. Other details are as follows J = Jockey

K = Key

R = Resistance of potentiometer wire,

r = Specific resistance of potentiometer wire.

Rh = Variable resistance which controls the current through the wire AB

(i) The specific resistance (r) of potentiometer wire must be high but its temperature coefficient of resistance (a) must be low.

(ii) All higher potential points (terminals) of primary and secondary circuits must be connected together at point A and all lower potential points must be connected to point B or jockey.

(iii) The value of known potential difference must be greater than the value of unknown potential difference to be measured.

(iv) The potential gradient must remain constant. For this the current in the primary circuit must remain constant and the jockey must not be slided in contact with the wire.

(v) The diameter of potentiometer wire must be uniform everywhere.

Potential gradient (x): Potential difference (or fall in potential) per unit length of wire is called potential gradient i.e. x = V/L volt/m where V = iR = (e/R+Rn+r)R.

So x = V/L = iR/L = ip/A = e/(R+Rh+r) . R/L

(i) Potential gradient directly depends upon

(a) The resistance per unit length (R/L) of potentiometer wire.

(b) The radius of potentiometer wire (i.e. Area of cross-section)

(c) The specific resistance of the material of potentiometer wire (i.e. r)

(d) The current flowing through potentiometer wire (i)

(ii) Potential gradient indirectly depends upon

(a) The emf of battery in the primary circuit (i.e. e)

(b) The resistance of rheostat in the primary circuit (i.e. Rh)

Working: Suppose jockey is made to touch a point J on wire then potential difference between A and J will be V = xl

At this length (l) two potential difference are obtained

(i) V due to battery e and

(ii) E due to unknown cell If V > E then current will flow in galvanometer circuit in one direction If V < E then current will flow in galvanometer circuit in opposite direction If V = E then no current will flow in galvanometer circuit this condition to known as null deflection position, length l is known as balancing length.

In balanced condition E = xl

or E = xl = V/L l = iR/L l = (e/R+Rh+r) × R/L × l

If V is constant then L ∝ l ⇒ x1/x2 = L1L2 = l1/l2 Kushagra Madhukar
askIITians Faculty 629 Points
11 months ago
Dear student,
Please find the answer to your question below.

R = Resistance of potentiometer wire,
r = Specific resistance of potentiometer wire.
Rh = Variable resistance which controls the current through the wire AB

Now, the potential gradient is obtained as
x = V/L = iR/L = ip/A = e/(R+Rh+r) . R/L
Clearly, potential gradient for potentiometer depends on internal resistance of driver cell.

Thanks and regards,
Kushagra

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