If the current sensitivity of a moving coil galvanometer is increased by 20%, its resistance also increases by 1.5 times.How will the voltage sensitivity of galvanometer be affected?

$\frac{\alpha \text{'}}{V}=\frac{N\text{'}B\text{'}A\text{'}}{kR\text{'}}\phantom{\rule{0ex}{0ex}}\frac{\alpha \text{'}}{V}=\left(\frac{20}{100}+1\right)\frac{NBA}{k\left(1.5R\right)}\phantom{\rule{0ex}{0ex}}\frac{\alpha \text{'}}{V}=\left(1.2\right)\frac{NBA}{k\left(1.5R\right)}\phantom{\rule{0ex}{0ex}}\frac{\alpha \text{'}}{V}=0.8\frac{\alpha}{V}\phantom{\rule{0ex}{0ex}}\frac{\alpha \text{'}}{V}=\left(80\%\right)\times \frac{\alpha}{V}$

So, there is an increase in 80% of original value in voltage sensitivity.

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