The correct option is (a) True
The best I can explain: Yes, voltage and current sensitivity are related to each other. Current sensitivity
IS = \(\frac {NAB}{k}\); Voltage sensitivity = \(\frac {\theta }{V} = \frac {\theta }{IR} = \frac {NAB}{Kr}\)
Therefore, voltage sensitivity ➔ VS = \((\frac {1}{R})\) x IS.