MCQ
Current sensitivity of a moving coil galvanometer is $5\,div/mA$ and its voltage sensitivity (angular deflection per unit voltage applied) is $20\,div/V$. The resistance of the galvanometer is ................. $\Omega$
  • A
    $40$
  • B
    $25$
  • C
    $500$
  • D
    $ 250$

Answer

Current sensitivity

$\mathrm{I}_{\mathrm{s}}=\frac{\mathrm{NBA}}{\mathrm{c}}$

Voltage sensitivity

$\mathrm{v}_{\mathrm{s}}=\frac{\mathrm{NBA}}{\mathrm{CR}_{\mathrm{G}}}$

So, resistance of galvanometer

$R_{G}=\frac{l_{s}}{V_{s}}=\frac{5 \times 1}{20 \times 10^{-3}}=\frac{5000}{20}=250\, \Omega$

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