Q18.A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5 Ω . When a resistance of 15 Ω is used for shunting null point moves to 300 cm . The internal resistance of the cell is ______ Ω .
What This Question Tests
This question requires applying the potentiometer principle to compare EMF and terminal voltage under different shunting resistances to determine the internal resistance of the cell.
Concepts Tested
Formulas Used
E = k L_null
V = k l_null
E/V = (R + r)/R
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📋 Question Details
- Chapter
- Current Electricity
- Topic
- Internal resistance of a cell using potentiometer
- Year
- 2023
- Shift
- 29 Jan Shift 2
- Q Number
- Q18
- Type
- Numerical
- NCERT Ref
- Class 12 Physics Ch 3: Current Electricity
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