Q. A galvanometer of resistance 240$\Omega$allows only 4% of the main current after connecting a shunt resistance. The value of the shunt resistance is
A) 10$\ \Omega$
B) 20$\ \Omega$
C) 8$\ \Omega$
D) 5$\ \Omega$
Explanation
(a) Given, galvanometer resistance G=24$0\Omega$
Shunt resistance $S = ?$
$I_{G} = \frac{4}{100}I$
From figure voltage through the circuit.
$\left( I - I_{G} \right)S = I_{G}G\ \ $
$or\ \ \ \ \ \left( I - \frac{4I}{400} \right)S = \frac{4I}{100} \times 240$
$or\ \ \ \ \ \ S = \frac{4 \times 240}{96} = 10\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.