A galvanometer of resistance $$50$$ Ω is connected to a battery of $$3$$ V along with a resistance of $$2950$$ Ω in series. A full scale deflection of $$30$$ divisions is obtained in the galvanometer. In order to reduce this deflection to $$20$$ divisions, the resistance in series should be

Question

A galvanometer of resistance $$50$$  Ω  is connected to a battery of $$3$$ V along with a resistance of $$2950$$ Ω  in series. A full scale deflection of $$30$$ divisions is obtained in the galvanometer. In order to reduce this deflection to $$20$$ divisions, the resistance in series should be

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initial current  $$i=33000=10$$−$$3A10$$−$$3A$$→$$30divisions20div$$→$$10$$−$$3$$×$$23A$$ For current  $$23$$×$$10$$−$$3$$ resistance supposed to be, $$10$$−$$3$$×$$23=3R+50R+50=92$$×$$103=4500R=4450$$Ω

A galvanometer of resistance 50 $Ω$ is connected to a battery of 3 V along with a resistance of 2950 $Ω$ in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

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A galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

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Ready to Test Your Skills?

free study material

A galvanometer of resistance 50 $Ω$ is connected to a battery of 3 V along with a resistance of 2950 $Ω$ in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be