A) \[5050\Omega \]
B) \[5550\Omega \]
C) \[6050\Omega \]
D) \[4450\Omega \]
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, \[I=\left( \frac{20}{30}\times 1 \right)mA\] or \[I=\frac{2}{3}mA\] Let the resistance be x \[\Omega \]. Then \[x=\frac{\varepsilon }{I}=\frac{3V}{\left( \frac{2}{3}\times {{10}^{-3}}A \right)}=\frac{3\times 3\times {{10}^{3}}}{2}\Omega =4500\Omega \]But the resistance of the galvanometer is \[50\Omega \]. \[\therefore \]Resistance to be added \[=(4500-50)\Omega \] \[=4450\Omega \]
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