AT Any Time Padhai Academy JEE Mains Special Potentiometer

POTENTIOMETER A potentiometer is a linear conductor of uniform cross-section with a steady current set up in it. This maintains a uniform potential gradient along the length of the wire. Any potential difference which is less than the potential difference maintained across the potentiometer wire can be measured using this. The wire should have high resistivity and low expansion coefficient. For example: Manganin or Constantine wire etc e, r r+R VA-VBR R+r Potentiometer wire B Resistance-R Potential gradient (x) Potential difference per unit length of A- wire VA - VB R+r L

Primary circuit of potentiometer is shown in figure determine: (A) current in primary circuit (B) potential drop across potentiometer wire AB (C) potential gradient (means potential drop per unit length of potentiometer wire) (D) maximum potential which we can measure above potentiometer Example R=10 , L = 10m

Example Primary circuit of potentiometer is shown in figure determine: (A) current in primary circuit (B) potential drop across potentiometer wire AB (C) potential gradient (means potential drop per unit length of potentiometer wire) (D) maximum potential which we can measure above potentiometer R = 10 , L = 10m 2 2 31 Solution: (a) i= Ans. r+R1 +R 1+2010 20 31 (b) VAB= iR=- 10 VABvo Ans. 31 xVAB 2 yolt/m 31 Ans. 20 31 (d) Maximum potential which we can measure by it-potential drop across wire ABvolt

Application of potentiometer (a) To find emf of unknown cell and compare emf of two cells. In case I, in figure, (2) is joint to (1) then balance length = 11 in case II, In figure, (3) is joint to (2) then balance length (2 e,r e,r 1 2 If any one of E or &2 is known the other can be found. If x is known then both 1 and 2 can be found 8 12

In an experiment to determine the emf of an unknown cell, its emf is compared with a standard cell of known emf 1.12 V. The balance point is obtained at 56cm with standard cell and 80 cm with the unknown cell. Determine the emf of the unknown cell Example 2 Solution: Here, 1.12V; Using equation -XL 11 56 cm; l2=80 cm we get 11 80) = 1.6 V Ans or 2 1.12 2

(b) To find current if resistance is known xe I-1 R, R2 Similarly, we can find the value of R2 also Potentiometer is ideal voltmeter because it does not draw any current from circuit, at the balance point.

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विजय त्रिपाठी

2 years ago

Thank u Sir
Aap daily current bhi share kre.
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Nisha Saha

10 months ago

sir please take communication system as well. thank you so much.

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Srivathsan V

10 months ago

sir wave optics please

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