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Question

The p.d. between the terminal of a cell in an open circuit was 2.2V. When it was measured across a resistor of 5 it was found to be 1.8V. The internal resistance of the cell is


A

109Ω

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B

910Ω

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C

127

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D

712

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Solution

The correct option is A

109Ω


Step 1: Give data:

Open circuit emf of the cell is E=2.2V

When measured across the 5Ω resistor the Voltage becomes V=1.8V

Step 2: Calculating the internal resistance,

We have a relation

V=E-IR......(i)

Also, we know

l=ER+r

Where r = internal resistance putting in (i)

V=E-ER+rr

Now we substitute the given values

1.8=2.2-2.25+rr1.8=2.2(5+r)-2.2r5+r1.8=2.2×55+r5+r=111.8r=111.8-5r=109

Thus, the correct answer is option A,109Ω.


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