A galvanometer of resistance 5.0Ω has full-scale deflection for a current of 100mA. How would its range be extended to 1.0A? By placing a resistance of
\(\frac{5}{9}\)\(\Omega\) in parallel
\(\frac{9}{5}\)\(\Omega\) in series
45\(\Omega\) in parallel
45\(\Omega\) in series
\(\frac{9}{5}\)\(\Omega\) in parallel
Correct answer is A
Given Data:
I\(_g\) = 100ma = 0.1A
R\(_g\)=5
where I\(_s\) = I - I\(_g\) → 1 - 0.1
I\(_s\) = 0.9
I\(_g\)R\(_g\) = I\(_s\)R\(_s\)
0.1 * 5 = 0.9 * R\(_s\)
R\(_s\) = \(\frac{0.1 * 5}{0.9}\)
R\(_s\) = \(\frac{5}{9}\)