Moving Charges And Magnetism Ques 81
81. 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
[2008]
(a) $5050 $ $\Omega$
(b) $5550 $ $\Omega$
(c) $6050 $ $\Omega$
(d) $4450 $ $\Omega$
Show Answer
Answer:
Correct Answer: 81.(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$
BETA
