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NEET 2002 (AIPMT) · Cells: EMF, Internal Resistance & Combinations

Cells: EMF, Internal Resistance & Combinations question from NEET 2002 (AIPMT)

Q51NEET 2002 · AIPMTEasy

Cells: EMF, Internal Resistance & Combinations

For a cell terminal potential difference is 2.2V when circuit is open and reduce to 1.8 V when cell is connected to a resistance of R = 5 Ω\Omega. Determine internal resistance of cell (r)

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Answer: (a)

  1. Open-circuit voltage = emf: E=2.2 VE = 2.2\text{ V}
  2. I=VR=1.85=0.36 AI = \frac{V}{R} = \frac{1.8}{5} = 0.36\text{ A}
  3. r=E−VI=0.40.36=109 Ωr = \frac{E - V}{I} = \frac{0.4}{0.36} = \frac{10}{9}\ \Omega. So the answer is (a).
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