Magnetic Effects Of Current Question 455
Question: A galvanometer of resistance 50 Q is connected to battery of 3V along with a resistance of 2950 Q 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
Options:
A) $ 5050\Omega $
B) $ 5550\Omega $
C) $ 6050\Omega $
D) $ 4450\Omega $
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Answer:
Correct Answer: D
Solution:
[d] Total internal resistance $ =(50+2950)\Omega $ $ =3000\Omega $ Emf of the cell, $ \varepsilon =3V $
$ \therefore $ Current $ =\frac{\varepsilon }{R}=\frac{3}{3000}=1\times {{10}^{-3}}A=1.0,mA $
$ \therefore $ Current for full scale deflection of 30 divisions is 1.0mA.
$ \therefore $ Current for a deflection of 20 divisions, $ I=( \frac{20}{30}\times 1 )mA $ or $ I=\frac{2}{3}mA $ Let the resistance be x $ \Omega $ . Then $ x=\frac{\varepsilon }{I}=\frac{3V}{( \frac{2}{3}\times {{10}^{-3}}A )}=\frac{3\times 3\times 10^{3}}{2}\Omega =4500\Omega $ But the resistance of the galvanometer is $ 50\Omega $ .
$ \therefore $ Resistance to be added $ =(4500-50)\Omega $ $ =4450\Omega $
BETA
