question_answer

A galvanometer of resistance \[240\text{ }\Omega \] allows only 4% of the main current after connecting a shunt resistance. The value of the shunt resistance is

A)  \[10\text{ }\Omega \]                  

B)  \[20\text{ }\Omega \]

C)  \[8\text{ }\Omega \]                                    

D)  \[5\text{ }\Omega \]

Correct Answer: A

Solution :

Given, galvanometer resistance \[G=240\text{ }\Omega \] Shunt resistance S = ?                 \[{{I}_{G}}=\frac{4}{100}I\] From figure voltage through the circuit.                 \[(I-{{I}_{G}})S={{I}_{G}}G\] or            \[\left( 1-\frac{4I}{100} \right)S=\frac{4I}{100}\times 240\] or            \[S=\frac{4\times 240}{96}=10\,\Omega \]