A galvanometer of resistance 50- 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 Given : #160; #160; R_G = 50Ω #160; #160; #160; #160; #160; #160; I_g = 5× 10^ 4 A #160 ...

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Byju's Answer

Standard XII

Physics

Galvanometer

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

595 Ω

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5050 Ω

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5059 Ω

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5950 Ω

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Solution

The correct option is C $5950 Ω$
Given : $R_{G} = 50 Ω$ $I_{g} = 5 \times 10^{- 4} A$ $V = 3$ volts
Let a resistance $R$ is connected in series to $R_{G}$ .
Thus $V = I_{g} (R_{G} + R)$
Or $3 = 5 \times 10^{- 4} (50 + R)$ $⟹ R = 5950 Ω$

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A galvanometer of resistance 50Ω 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 5950ΩGiven : RG=50Ω Ig=5×10−4A V=3 voltsLet a resistance R is connected in series to RG.Thus V=Ig(RG+R)Or 3=5×10−4(50+R) ⟹R=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 $5950 Ω$
Given : $R_{G} = 50 Ω$ $I_{g} = 5 \times 10^{- 4} A$ $V = 3$ volts
Let a resistance $R$ is connected in series to $R_{G}$ .
Thus $V = I_{g} (R_{G} + R)$
Or $3 = 5 \times 10^{- 4} (50 + R)$ $⟹ R = 5950 Ω$