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MHT CET 2025 (19 Apr, Shift 1) · Kirchhoff's Laws & Measuring Instruments

Kirchhoff's Laws & Measuring Instruments question from MHT CET 2025 (19 Apr, Shift 1)

Q35MHT CET 2025 · 19 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments

The scale of a galvanometer is divided into 160 equal divisions. The galvanometer shows full scale deflection of 16 mA and maximum voltage is 80 mV. Now the range is changed so that galvanometer reads 160 V. The required resistance to be connected is

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Answer: (a)

$G = \dfrac{80\text{ mV}}{16\text{ mA}} = 5\ \Omega$ , $I_g = 0.016\text{ A}$ .
To read $160\text{ V}$ , add series $R = \dfrac{V}{I_g} - G = \dfrac{160}{0.016} - 5$
$R = 10000 - 5 = 9995\ \Omega$ in series
So the answer is (a).

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