Current Electricity

JEE Main level, single correct, medium

365 of 380

Question 365 of 380

A galvanometer has a 50 division scale. Battery has no internal resistance. It is found that there is deflection of 40 divisions when R.B.=2400Ω.R.B.=2400\Omega .Deflection becomes 20 divisions when resistance taken from resistance box is 4900 Ω. Then we can conclude :



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Answer: (A) Current sensitivity of galvanometer is 20μA/division20\mu A/division

Current in the circuit,

I=V2400+RGandI2=V4900+RGI=\dfrac{V}{2400+{R}_{G}}and\dfrac{I}{2}=\dfrac{V}{4900+{R}_{G}}

⇒RG=100Ω\Rightarrow {R}_{G}=100\Omega

I=22500=8×10−4AI=\dfrac{2}{2500}=8\times {10}^{-4}A

Current Sensitivity
=8×10−440=\dfrac{{8\times 10}^{-4}}{40}

20μA20\mu Aper division

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