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Current Electricity question

2002 · Q173
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Current Electricity question

2002 · Q173

NEETPhysicsCurrent ElectricityMCQ+4 / −1
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\OmegaΩ. Determine internal resistance of cell (r)
  1. A
    109Ω{{10} \over 9}\Omega910​Ω
  2. B
    910Ω{{9} \over 10}\Omega109​Ω
  3. C
    119Ω{{11} \over 9}\Omega911​Ω
  4. D
    59Ω{{5} \over 9}\Omega95​Ω
View written solutionFree

Correct answer: A

AIPMT 2002 Physics - Current Electricity Question 50 English Explanation

Terminal potential difference is 2.2 V when circuit is open.

∴\therefore∴ e.m.f. of the cell = E = 2.2 volt

Now, when the cell is connected to the external resistance, circuit current I is given by

I=ER+r=2.25+rI = {E \over {R + r}} = {{2.2} \over {5 + r}}I=R+rE​=5+r2.2​ ampere, where r is the internal resistance of the cell.

Potential difference across the cell = IR

=2.25+r×5=1.8 = {{2.2} \over {5 + r}} \times 5 = 1.8=5+r2.2​×5=1.8

∴\therefore∴ 5 + r = 11/1.8.

∴r=111.8−5=110−9018=109Ω\therefore r = {{11} \over {1.8}} - 5 = {{110 - 90} \over {18}} = {{10} \over 9}\Omega∴r=1.811​−5=18110−90​=910​Ω

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