A) 240
B) 120
C) 60
D) 480
Correct Answer: D
Solution :
Let the value of current for one division of galvanometer is i. Then, the current through the ammeter \[(I)=50i\] and current through the galvanometer \[({{I}_{g}})=10i\] As we know that, \[S=\frac{{{I}_{g}}}{(I-{{I}_{g}})}G\] \[\Rightarrow \] \[G=S\,\left( \frac{I-{{I}_{g}}}{{{I}_{g}}} \right)\] \[=12\times \,\left[ \frac{50i-10i}{10i} \right]\] \[=12\times \,\left[ \frac{40i}{10i} \right]=40\,\Omega \]You need to login to perform this action.
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