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Ashok0210
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Asked: 2021-02-23T04:02:49+00:00 In:

Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

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Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Class 12 Physics
CBSE and UP Board
Current Electricity
Additional Exercise
Chapter-3 Exercise 3.24

NCERT Solutions Class 12 Physics Chapter 3 Question 24

2020-2021cbse and up boardclass 12 chapter-3 physicsclass 12 physicsncert class 12 current electricity
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    1 Answer

    1. Internal resistance of the cell = r

      Balance point of the cell in open circuit, l₁ = 76.3 cm

      An external resistance (R) is connected to the circuit with R = 9.5 Ω

      New balance point of the circuit, l₂ = 64.8 cm

      Current flowing through the circuit = I

      The relation connecting resistance and emf is,

      r = (l₁ -l₂ )/l₂ x R

      = (76.3 -64.8)64.8 x 9.5 =1.68 Ω

      Therefore, the internal resistance of the cell is 1.68Ω.

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