A galvanometer with its coil resistance 25,Om | vedclass

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(D) Given:Galvanometer resistance,$R_g = 25\,\Omega$Full-scale deflection current,$I_g = 1\,mA = 10^{-3}\,A$Maximum current to be measured,$I = 2\,A$T

Vedclass · Accepted Answer

$1.25 	imes 10^{-2}\,\Omega$

Vedclass · Answer

(D) Given:Galvanometer resistance,$R_g = 25\,\Omega$Full-scale deflection current,$I_g = 1\,mA = 10^{-3}\,A$Maximum current to be measured,$I = 2\,A$To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.The potential difference across the galvanometer and the shunt must be equal:$I_g R_g = (I - I_g) S$Substituting the values:$10^{-3} 	imes 25 = (2 - 10^{-3}) S$$0.025 = (2 - 0.001) S$$0.025 = 1.999 S$$S = \frac{0.025}{1.999} \approx \frac{0.025}{2} = 0.0125\,\Omega$Thus,$S = 1.25 	imes 10^{-2}\,\Omega$.

$A$ galvanometer with its coil resistance $25\,\Omega$ requires a current of $1\,mA$ for its full-scale deflection. In order to construct an ammeter to read up to a current of $2\,A$,the approximate value of the shunt resistance should be:

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