A) 35 mA
B) 40 mA
C) 30 mA
D) 25 mA
Correct Answer: C
Solution :
Here, galvanometer current \[{{i}_{g}}=\frac{1mA}{10}\times 50=5mA\] \[S=4\Omega \]and\[G=20\Omega \] As, sheet\[S=\frac{{{i}_{g}}G}{i-{{i}_{g}}}\Rightarrow 4=\frac{5\times {{10}^{-3}}\times 20}{i-2\times {{10}^{-3}}}\] On solving, we get\[i=30\times {{10}^{-3}}=30mA\]You need to login to perform this action.
You will be redirected in
3 sec