Moving Charges and Magnetism - Result Question 80
82. A galvanometer of resistance $50 \Omega$ is connected to battery of $3 V$ along with a resistance of $2950 \Omega$ 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) $5050 \Omega$
(b) $5550 \Omega$
(c) $6050 \Omega$
(d) $4450 \Omega$
[2008]
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Answer:
Correct Answer: 82. (d)
Solution:
- (d) Total internal resistance $=(50+2950) \Omega$ $=3000 \Omega$
Emf of the cell, $\varepsilon=3 V$
$\therefore$ 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 $1.0 mA$.
$\therefore$ 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{3 V}{(\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
$=(4500-50) \Omega=4450 \Omega$
