SELECT * FROM question_mgmt as q WHERE id=2255 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=8 AND chapterId=117 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th physics Current Electricity figure shows a 2 0 v potentiometer used for the de

Question:

Figure 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.

Answer:

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,

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


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SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='8' AND ex_no!=0 AND status=1 and id!=2255 ORDER BY last_viewed_on desc limit 0,10

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  • Answered by Ekta Mehta
  • 4 months ago

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