# when a 12 ohm resistor is connected in series with a moving coil galvanometer than it subtraction reduces form 50 division to 10 division find the resistance of the galvanometer

## $HereletthevalueofcurrentfromonedivisionbeiA.\phantom{\rule{0ex}{0ex}}ThusthecurrentthroughammeterI=50iandthroughGalvanometerIg=10i\phantom{\rule{0ex}{0ex}}S=\frac{IgG}{I-Ig}\phantom{\rule{0ex}{0ex}}\Rightarrow G=\frac{(I-Ig)S}{Ig}\phantom{\rule{0ex}{0ex}}=\frac{50i-10i}{10i}\times 12\Omega =4\times 12\Omega =48\Omega $

Regards

**
**