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

Infinity Learn · Accepted Answer

Current through the galvanometer,$$I=350$$  $$+2950$$  $$=$$  $$10$$−$$3$$  ACurrent for $$30$$ divisions  $$=$$  $$10$$−$$3$$  ACurrent for $$20$$ divisions $$=$$ $$10$$−$$330$$  ×  $$20=23$$  ×  $$10$$−$$3$$  AFor the same deflection to obtain for $$20$$ division, let resistance added be R.∴       $$23$$   ×   $$10$$−$$3$$  $$=$$  $$350+Ror$$     $$R=4450$$   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?

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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