Q13. A Potentiometer wire of length L and a resistance r are connected in series with a battery of e

Q: What is the correct answer for CUET UG Physics Q13?

The correct answer is C. In a potentiometer circuit, the main current (I) flowing through the potentiometer wire (resistance r) and the series resistance (r_a) connected to the battery (E₀) is given by Ohm's law: I = E₀ / (r_a + r). The potential gradient (k) along the potentiometer wire is the potential drop across the wire divided by its total length L. The potential drop across the wire is V_wire = I * r. So, k = V_wir

Q: Q13. A Potentiometer wire of length L and a resistance r are connected in series with a battery of emf Eo and a resistance r(a) An unknown emf E is balanced at a length I of the potentiometer wire.…

The correct answer is C. In a potentiometer circuit, the main current (I) flowing through the potentiometer wire (resistance r) and the series resistance (r_a) connected to the battery (E₀) is given by Ohm's law: I = E₀ / (r_a + r). The potential gradient (k) along the potentiometer wire is the potential drop across the wire divided by its total length L. The potential drop across the wire is V_wire = I * r. So, k = V_wir

Q: What are the options for this CUET UG Physics question?

A. E₀r / (r+r_a)l · B. E₀r / I r_a · C. E₀r l / ((r_a + r) L) · D. E₀l / L

Question

Q13. A Potentiometer wire of length L and a resistance r are connected in series with a battery of emf Eo and a resistance r(a) An unknown emf E is balanced at a length I of the potentiometer wire. The emf E will be:
(a) LEor / (r+r₁)l
(b) Bor / Ir1
(c) Eor l / (r+r₁) L
(d) Fol / L

FirstInTest · Accepted Answer

In a potentiometer circuit, the main current (I) flowing through the potentiometer wire (resistance r) and the series resistance (r_a) connected to the battery (E₀) is given by Ohm's law: I = E₀ / (r_a + r). The potential gradient (k) along the potentiometer wire is the potential drop across the wire divided by its total length L. The potential drop across the wire is V_wire = I * r. So, k = V_wire / L = (I * r) / L. Substituting the expression for I: k = (E₀ * r) / ((r_a + r) * L). An unknown emf (E) is balanced at a length 'l' of the potentiometer wire, meaning E = k * l. Substituting the expression for k, we get E = (E₀ * r * l) / ((r_a + r) * L). This matches option (c), assuming r(a) is represented by r_a in the options.

Answer

E₀r / (r+r_a)l

Answer

E₀r / I r_a

Answer

E₀r l / ((r_a + r) L)

Answer

E₀l / L

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