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Q.

Two meters of voltage range 20.0 V and 30.0 V have to be constructed with a galvanometer. The resistance connected in series with the galvanometer is 1680 Ω for the 20.0 V range and 2930 Ω for the 30.0 V range. The resistance of the galvanometer and the full -scale current are respectively.

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a

320 Ω and 8 mA

b

70 Ω and 10 mA

c

820 Ω and 10 mA

d

820 Ω and 8 mA

answer is D.

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Detailed Solution

Given thatv=20.0V for R1=1680Ωv=30.0V for R2=2930ΩLet coil resistance = GIf full scale current is IgIgG+R1=20⇒Ig(G+1680)=20…..(1) And Ig(G+2930)=30…..(2) Dividing (1)&(2)G+1680G+2930=2030⇒3(G+1680)=2(G+2930)⇒G=5860−5040∴G=820Ω From (1) Ig=20820+1680=8×10−3A∴Ig=8mA

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