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  • Compare the current sensitivity and voltage sensitivity of the following moving coil galvanometers <img alt="Meter A: n=25, A=1.3 \times 10^{-3} \mathrm{~m}^2 B=0.45 \mathrm{~T}, R=25 \Om

Compare the current sensitivity and voltage sensitivity of the following moving coil galvanometers

Meter A: n=25, A=1.3 \times 10^{-3} \mathrm{~m}^2 B=0.45 \mathrm{~T}, R=25 \Omega,\\ Meter B: n=65, A=4.0 \times 10^{-3} \mathrm{~m}^n B=0.35 T, \quad R=30 \Omega

You are given that the springs in the two meters have the same tonsional constants

Option: 1

0.1285, 0.1542


Option: 2

0.4325, 0.47


Option: 3

0.1542, 0.2842


Option: 4

0.4325, 0.476


Answers (1)

best_answer

For\: meter\: A,\: n_1=25 \begin{array}{ll} \\A_1=1.3 \times 10^{-3} \mathrm{~m}^2 & B_1=0.45 \mathrm{~T} \\ R_1=25 \Omega & \end{array}

For\: meter\: B,\: n_2=65 \begin{array}{ll} \\A_2=4.0 \times 10^{-5} \mathrm{~m}^2 & B_2=0.35 \mathrm{~T} \\ R_2=30 \Omega & \end{array}

Current sensitivity of a meter=\frac{n B A}{K}

\begin{aligned} \frac{\text { Current sensitivity of meter } A}{\text { Current sensitivith of meter } B} & =\frac{n_1 B_1 A_1}{K_1} \times \frac{K_2}{n_2 B_2 A_2} \\ & =\frac{n_1 B_1 A_1}{n_2 B_2 A_2}\left(\because K_1=K_2\right) \\ = & \frac{25 \times 0.45 \times 1.3 \times 10^{-3}}{65 \times 0.35 \times 4.0 \times 10^{-3}}=12.85 \times 10^{-2} \end{aligned}

Now, voltage sensitivity of a meter=\frac{n B A}{K}

\frac{Voltage\: sensitivity\: of\: meter\: A}{Voltage\: sensitivity\: of\: meter\: B}

=(Current \:sensitivity)=\frac{R_2}{R_1}

\begin{aligned} & =0.1285 \times \frac{30}{25} \\ & =0.1542 \end{aligned}

Posted by

Divya Prakash Singh

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