To solve the problem, we need to determine the resistance required in series with the galvanometer to reduce the deflection from 20 divisions to 10 divisions. Let's break down the solution step by step.
### Step 1: Calculate the initial current (I_g) for full-scale deflection
The total resistance in the circuit when the galvanometer shows full-scale deflection (20 divisions) is the sum of the galvanometer resistance and the series resistance.
Given:
- Resistance of galvanometer (R_g) = 30 Ω
- Series resistance (R_s) = 1970 Ω
- EMF of the battery (V) = 2 V
The total resistance (R_total) is:
\[ R_{total} = R_g + R_s = 30 \, \Omega + 1970 \, \Omega = 2000 \, \Omega \]
Using Ohm's law, the current (I_g) through the circuit can be calculated as:
\[ I_g = \frac{V}{R_{total}} = \frac{2 \, V}{2000 \, \Omega} = 0.001 \, A = 1 \, mA \]
### Step 2: Calculate the current (I') for reduced deflection
To reduce the deflection to 10 divisions, we need to find the new current (I') flowing through the galvanometer. Since the deflection is directly proportional to the current, we can use the ratio of the deflections to find I'.
The relationship is:
\[ I' = I_g \times \frac{10}{20} = \frac{1 \, mA}{2} = 0.5 \, mA \]
### Step 3: Calculate the new equivalent resistance (R'_total)
Now we need to find the new equivalent resistance when the current is I' (0.5 mA). Using Ohm's law again:
\[ R'_{total} = \frac{V}{I'} = \frac{2 \, V}{0.5 \, mA} = \frac{2 \, V}{0.0005 \, A} = 4000 \, \Omega \]
### Step 4: Calculate the new series resistance (R_s')
The new total resistance (R'_{total}) is the sum of the galvanometer resistance and the new series resistance (R_s'):
\[ R'_{total} = R_g + R_s' \]
\[ 4000 \, \Omega = 30 \, \Omega + R_s' \]
\[ R_s' = 4000 \, \Omega - 30 \, \Omega = 3970 \, \Omega \]
### Conclusion
The resistance required in series to reduce the deflection to 10 divisions is:
\[ R_s' = 3970 \, \Omega \]
