A galvanometer with full-scale deflection of 10 mA is to be converted to a voltmeter with full-scale deflection of 5v. if a series resistance of 498Ξ© is used for the conversion, the resistance of the galvanometer is ?
- 1β¦
-
2β¦
- 5β¦
- 10β¦
The correct answer is: B
Explanation
r = resistance of the galvanometer
l = current through galvanometer = 1mA or 0.01A
V1 = p.d across the galvanometer = I X r = 0.01 X r
: V1 = 0.01r
V2 = p.d across the multiplier = 5 - 0.01r
R = resistance of the multiplier = 498Ξ©
where R = \(\frac{V}{I}\) --> 498 = \(\frac{5 - 0.01r}{0.01}\)
CROSS MULTIPLY --> 498 X 0.01 = 5 - 0.01r
0.01r = 5 - 4.98
β΄ r = \(\frac{0.02}{0.01}\) = 2Ξ©
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