Question

A galvanometer of resistance $ 50\,\Omega $ is connected to a 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 $

Answer 1

The Correct Option is D

Current through the galvanometer

$I=\frac{3}{(50+2950)} =10^{-3} A $
Current for 30 divisions $=10^{-3} A$
Current for 20 divisions $=\frac{10^{-3}}{30} \times 20$
$=\frac{2}{3} \times 10^{-3} A$
For the same deflection to obtain for 20 divisions, let resistance added be $R$
$\therefore \frac{2}{3} \times 10^{-3} =\frac{3}{(50+1 R)}$
or $R =4450 \Omega$

Answer 2

To solve this problem, we need to understand the relationship between the current through the galvanometer and the resistance in the circuit. Let's go step by step:
 Given Data:
1. Galvanometer resistance, \( R_g = 50 \Omega \)
2. Battery voltage, \( V = 3 \text{ V} \)
3. Initial series resistance, \( R_s = 2950 \Omega \)
4. Full-scale deflection current corresponds to 30 divisions.
Step-by-Step Solution:
1. Determine the current for full-scale deflection:
The total resistance in the circuit when the galvanometer shows full-scale deflection (30 divisions) is:
 \[ R_{\text{total}} = R_g + R_s = 50 \Omega + 2950 \Omega = 3000 \Omega \]
  The current through the circuit is given by Ohm's law:
 \[ I = \frac{V}{R_{\text{total}}} = \frac{3 \text{ V}}{3000 \Omega} = 0.001 \text{ A} = 1 \text{ mA} \]
So, a full-scale deflection (30 divisions) corresponds to a current of 1 mA.
2. Current for 20 divisions:
 Since full-scale deflection is 30 divisions for 1 mA, the current for 20 divisions can be determined using direct proportion:
  \[ I_{20} = \frac{20}{30} \times 1 \text{ mA} = \frac{2}{3} \text{ mA} = \frac{2}{3} \times 0.001 \text{ A} = \frac{2}{3000} \text{ A} = \frac{1}{1500} \text{ A} \]
3. Determine the total resistance needed for 20 divisions:
Let the new series resistance be \( R'_s \). The total resistance in the circuit for 20 divisions is:
\[ R'_{\text{total}} = R_g + R'_s \]
 Using Ohm's law, we can set up the equation:
  \[ \frac{V}{R'_{\text{total}}} = I_{20} \]
  \[ \frac{3 \text{ V}}{R_g + R'_s} = \frac{1}{1500} \text{ A} \]
  \[ 3 = \frac{R_g + R'_s}{1500} \]
  \[ 3 \times 1500 = R_g + R'_s \]
  \[ 4500 = 50 + R'_s \]
  \[ R'_s = 4500 - 50 \]
  \[ R'_s = 4450 \Omega \]
 Conclusion:
To reduce the deflection from 30 divisions to 20 divisions, the resistance in series should be option (D) \( \boxed{4450 \Omega} \).