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Question 65 of 102

Q.A galvanometer of resistance 100 Ω which can measure a maximum current of 1 mA is converted into an ohmmeter by connecting a battery of emf 1 V and a fixed resistance of 900 Ω in series with the galvanometer. When an external resistance is measured the current reading is 0.1 mA. Calculate the value of the resistance.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2017Subjective· 3mImportance★★★★★
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Treating the ohmmeter as a single series loop (battery, galvanometer, fixed resistor, unknown resistor), the measured current of 0.1 mA gives the total loop resistance, from which the unknown resistance is found to be 9000 Ω9000\,\Omega.

Setting up the circuit equation

In an ohmmeter of this type, the galvanometer (resistance GG), a fixed series resistance RsR_s, and the unknown external resistance XX to be measured, are all connected in a single series loop with a battery of emf ε\varepsilon. By Ohm's law applied to the whole loop, the current flowing is

I=εG+Rs+XI=\frac{\varepsilon}{G+R_s+X}

Given data

  • Galvanometer resistance, G=100 ΩG=100\,\Omega …

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