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\(\Omega\)
Correct answer is B
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 = 395Ω
where R = \(\frac{V}{I}\)
--> 395 = \(\frac{8−0.02r}{0.02}\)
CROSS MULTIPLY
8−0.02r = 395 * 0.02
8−0.02r = 7.9
--> 0.02r = 8 - 7.9
∴ r = \(\frac{0.1}{0.02}\)
= 5Ω