Get Answers to all your Questions

header-bg qa

The current sensitivity of moving coil galvanometer is increased by 25%.This increase is achieved only by changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be :

Option: 1

+25%


Option: 2

-25%


Option: 3

-50%


Option: 4

\mathrm{Zero}


Answers (1)

best_answer

\begin{aligned} & \tau=\mathrm{mB} \quad A=\text { area of coil } \\ \end{aligned}

\begin{aligned} & \mathrm{K} \theta=\mathrm{IANB} \quad \mathrm{B}=\text { magnetic field } \\ \end{aligned}

\begin{aligned} & \frac{\theta}{\mathrm{I}}=\frac{\mathrm{ANB}}{\mathrm{K}} \quad \text { Currect senstivity } \\ \end{aligned}

 \begin{aligned} & \text { Voltage sensitivity }=\frac{\theta}{\mathrm{V}}=\frac{\theta}{\mathrm{IR}}=\frac{\text { Current sensitivity }}{\mathrm{R}} \\ \end{aligned}

\begin{aligned} & \mathrm{R}=\text { constant } \\ \end{aligned}

\begin{aligned} & \text { Voltage sensitivity } \propto \text { current sensitivity } \\ & \end{aligned}

 so the percentage change in voltage sensitivity  is also 25 \%

Posted by

chirag

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE