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Moving Charges And Magnetism

Question
CBSEENPH12040065
Wired Faculty App

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

  • 5050 Ω

  • 5550 Ω

  • 6050 Ω

  • 4450 Ω

Solution

D.

4450 Ω

Current through the galvanometer 

space straight I thin space equals space fraction numerator 3 over denominator left parenthesis 50 plus 2950 right parenthesis end fraction space equals space 10 to the power of negative 3 end exponent space straight A Current space for space 30 space divisions space equals space 10 to the power of negative 3 end exponent space straight A Current space for space 20 space divisions space equals space 10 to the power of negative 3 end exponent over 30 space straight x space 20 equals space 2 over 3 space straight x space 10 to the power of negative 3 end exponent space straight A For space the space same space deflection space to space obtain space for space 20 space divisions comma let space resistance space added space be space straight R therefore comma 2 over 3 space straight x space 10 to the power of negative 3 end exponent space equals space fraction numerator 3 over denominator left parenthesis 50 plus 1 straight R right parenthesis end fraction Or space straight R space equals space 4450 space straight capital omega

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω   N1 = 30,
A1 = 3.6 x 10–3 m2,  B1 = 0.25 T
R2 = 14 Ω;  N2 = 42,
A2 = 1.8 x 10–3 m2, B2 = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

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