12 Physics E1/E2 Using Potentiometer Sum & Difference Method XII Sci

  12 Physics 

 E1/E2 Using Potentiometer 

Sum & Difference Method     

Potentiometer

Compare the  e.m.f. of two given cells using potentiometer by connecting them separately.

Hence determine E1/E2.

Potentiometer

Aim : To compare E.M.F. of two given cell’s connecting them separately by using

potentiometer, Hence determine E1/E2.

Apparatus: Cell battery , potentiometer , resistance box plug key, galvanometer , connecting wires ,

etc.

Diagram:

Calculation

MCQs – Potentiometer (Sum & Difference Method: E₁/E₂)

1. The potentiometer compares two emf sources by the sum and difference method. To calculate E₁/E₂, the formula used is:

a) (l₁ + l₂) / (l₁ – l₂)
b) (l₁ – l₂) / (l₁ + l₂)
c) l₁ / l₂
d) (l₂ – l₁) / (l₂ + l₁)

Answer: a)

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2. In the sum and difference method, the null point balancing lengths for the two cells are measured:

a) Separately
b) When cells are connected in series and opposition
c) Only when cells are connected in parallel
d) Only with a single cell at a time

Answer: b)

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3. The sum balancing length corresponds to:

a) When both cells assist each other
b) When both oppose each other
c) When only E₁ is used
d) When only E₂ is used

Answer: a)

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4. The difference balancing length corresponds to:

a) Two cells assist
b) Two cells oppose
c) Only stronger cell
d) Only weaker cell

Answer: b)

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5. The potentiometer works on which principle?

a) Ohm’s law
b) Kirchhoff's law
c) Null deflection method
d) Wheatstone bridge concept

Answer: c)

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6. The ratio E₁/E₂ = (l_sum + l_diff) / (l_sum – l_diff) is valid when:

a) E₁ > E₂
b) E₁ < E₂
c) Both cells are identical
d) For any two cells

Answer: d)

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7. If the balancing length for sum is 180 cm and for difference is 20 cm, then E₁/E₂ is:

a) 10
b) 5
c) 2
d) 1.29

Answer: b)
(180 + 20) / (180 – 20) = 200 / 160 = 1.25 (Apologies, correct answer is 1.25, not in options → closest 1.29 (d))

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8. A potentiometer is preferred over a voltmeter for comparing EMFs because:

a) It draws no current from the cell
b) It is cheaper
c) It uses fewer components
d) It is more compact

Answer: a)

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9. For accurate readings, the potentiometer wire must have:

a) High resistance & short length
b) Low resistance & uniform cross-section
c) High resistance & non-uniform cross-section
d) Variable resistance

Answer: b)

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10. If the null point shifts towards the right end of the wire, it indicates:

a) Driving cell emf decreased
b) Test cell emf increased
c) Resistance increased
d) Potentiometer wire heated up

Answer: a)

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Quiz