Moving Charges and Magnetism - Result Question 80

82. A galvanometer of resistance 50Ω is connected to battery of 3V along with a resistance of 2950Ω 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

(a) 5050Ω

(b) 5550Ω

(c) 6050Ω

(d) 4450Ω

[2008]

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Answer:

Correct Answer: 82. (d)

Solution:

  1. (d) Total internal resistance =(50+2950)Ω =3000Ω

Emf of the cell, ε=3V

Current =εR=33000=1×103A=1.0mA

Current for full scale deflection of 30 divisions is 1.0mA.

Current for a deflection of 20 divisions,

I=(2030×1)mA or I=23mA

Let the resistance be xΩ. Then

x=εI=3V(23×103A)=3×3×1032Ω

=4500Ω

But the resistance of the galvanometer is 50Ω.

Resistance to be added

=(450050)Ω=4450Ω



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