A galvanometer of $50\, ohm$ resistance has $25$ divisions. A current of $4 \times 10^{-4}$ ampere gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $25\, volts$, it should be connected with a resistance of
A$2500 \,\Omega$ as a shunt
B$2450 \,\Omega$ as a shunt
C$2550 \,\Omega$ in series
D$2450 \,\Omega$ in series
AIPMT 2004, Medium
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D$2450 \,\Omega$ in series
d Full deflection current ${i_g} = 25 \times 4 \times {10^{ - 4}}$$ = 100 \times {10^{ - 4}}A$
Using $R = \frac{V}{{{I_g}}} - G$$ = \frac{{25}}{{100 \times {{10}^{ - 4}}}} - 50$$ = 2450\,\Omega $ in series.
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