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

  • A 2.5\(\Omega\)
  • B 5.0\(\Omega\)
  • C 8.0\(\Omega\)
  • D 10.0\(\Omega\)
waec 2022physics

The correct answer is B. 5.0\(\Omega\)

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

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