Question

A.galvanometer of resistance 30 Ω is connected to a battery of emf 2 V with 1970 Ω resistance in series. A full scale deflection of 20 divisions is obtained in the galvanometer. To reduce the deflection to 10 divisions, the resistance in series required is

• A. 4030 Ω
• B. 4000 Ω
• C. 3970 Ω
• D. 2000 Ω

Answer 1

Answer: (C)

3970 Ω

Answer 2

1. Determine full‐scale current:

Total resistance with the given series resistor is

$$R_{\text{total}} = 1970\,\Omega + 30\,\Omega = 2000\,\Omega.$$

Thus, the full-scale current (yielding 20 divisions) is

$$I_{\text{fs}} = \frac{2\,\text{V}}{2000\,\Omega} = 1\,\text{mA}.$$

2. Find current for 10 divisions:

Since 20 divisions correspond to 1 mA, 10 divisions correspond to

$$I' = \frac{1}{2}\, \text{mA} = 0.5\,\text{mA}.$$

3. Calculate the new series resistor:

For the current through the galvanometer (with internal resistance $30\,\Omega$) to be $0.5\,\text{mA}$, the total required resistance is

$$R'_{\text{total}} = \frac{2\,\text{V}}{0.0005\,\text{A}} = 4000\,\Omega.$$

Therefore, the required external resistance is

$$R' = R'_{\text{total}} - 30\,\Omega = 4000\,\Omega - 30\,\Omega = 3970\,\Omega.$$