A galvanometer with a full-scale deflection of 20 mA is Converted to read 8 K by connecting a 395 \(\Omega\) resistor in series with it. Determine the internal resistance of the galvanometer
- 2.5\(\Omega\)
-
5.0\(\Omega\)
- 8.0\(\Omega\)
- 10.0\(\Omega\)
The correct answer is: B
Explanation
r = resistance of the galvanometer
l = current through galvanometer = 20mA or 0.02A
V1 = p.d across the galvanometer = I X r = 0.02 X r
: V1 = 0.02r
V2 = p.d across the multiplier = 8 - 0.02r
R = resistance of the multiplier = 395Ξ©
where R = \(\frac{V}{I}\)
--> 395 = \(\frac{8β0.02r}{0.02}\)
CROSS MULTIPLY
8β0.02r = 395 * 0.02
8β0.02r = 7.9
--> 0.02r = 8 - 7.9
β΄ r = \(\frac{0.1}{0.02}\)
= 5Ξ©