A galvanometer of resistance 50 - gives a full scale deflection for a current 5 - 10-4 A- The resistance that should be connected in series with the galvanometer to read 3 V isA- 5050 -B- 5059 - 595 -D- 5950 -

R = V/I_g G R=3/5 × 10^ 4 50 = 3 × 10^ 4/5 50 6000 50=5950 Ω ...

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Byju's Answer

Standard XII

Physics

Conductivity and Resistivity

A galvanomete...

Question

A galvanometer of resistance 50 $Ω$ gives a full scale deflection for a current $5 \times 10^{- 4} A$ . The resistance that should be connected in series with the galvanometer to read 3 V is

$5059 Ω$

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$595 Ω$

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$5950 Ω$

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$5050 Ω$

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Solution

The correct option is C
$5950 Ω$

$R = \frac{V}{I_{g}} - G R = \frac{3}{5 \times 10^{- 4}} - 50 = \frac{3 \times 10^{- 4}}{5} - 50 6000 - 50 = 5950 Ω$

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A galvanometer of resistance 50 Ω gives a full scale deflection for a current 5×10−4A. The resistance that should be connected in series with the galvanometer to read 3 V is

The correct option is C 5950Ω R=VIg−GR=35×10−4−50=3×10−45−506000−50=5950Ω

A galvanometer of resistance 50 $Ω$ gives a full scale deflection for a current $5 \times 10^{- 4} A$ . The resistance that should be connected in series with the galvanometer to read 3 V is

The correct option is C
$5950 Ω$

$R = \frac{V}{I_{g}} - G R = \frac{3}{5 \times 10^{- 4}} - 50 = \frac{3 \times 10^{- 4}}{5} - 50 6000 - 50 = 5950 Ω$