Question

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?

Answer 1

If there is an increase in current sensitivity then there is also change in voltage sensitivity. So, using data in the equation for voltage sensitivity we get,
$$\begin{gathered} \frac{\alpha \text{'}}{V}=\frac{N\text{'}B\text{'}A\text{'}}{kR\text{'}} \\ \frac{\alpha \text{'}}{V}=\left(\frac{20}{100}+1\right)\frac{NBA}{k\left(1.5R\right)} \\ \frac{\alpha \text{'}}{V}=\left(1.2\right)\frac{NBA}{k\left(1.5R\right)} \\ \frac{\alpha \text{'}}{V}=0.8\frac{\alpha }{V} \\ \frac{\alpha \text{'}}{V}=\left(80\%\right)\times \frac{\alpha }{V} \end{gathered}$$
So, there is an increase in 80% of original value in voltage sensitivity.