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.

Question

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.

Infinity Learn · Accepted Answer

Given $$thatv=20.0V$$ for $$R1=1680$$Ω$$v=30.0V$$ for $$R2=2930$$ΩLet 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)⇒$$G=5860$$−$$5040$$∴$$G=820$$Ω From (1) $$Ig=20820+1680=8$$×$$10$$−$$3A$$∴$$Ig=8mA$$

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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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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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.