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A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of 10Ω. Its internal resistance is
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- 1.0 ohm
- 0.5 ohm
- 2.0 ohm
- zero
Correct Option: A
Here
E > | ||
R + r |
hence the lengths 110 cm and 100 cm are interchanged.
Without being short-circuited through R, only the battery E is balanced.
E = | × l1 = | × 110 ................(i) | ||
L | L |
When R is connected across E, Ri = | × l2 | |
L |
Or, R | = | × 100..............(ii) | ||||
R + r | L |
Dividing (i) by (ii), we get
= | |||
R | 100 |
or, 100 R + 100 r = 110 R