A galvanometer can be converted into a voltmeter of certain range by connecting a resistance of `880Omega` in series with it. When the resistance of `420Omega` is connected in series, the range becomes half. Find the resistance of galvanometer.

Explanation

To solve the problem, we need to find the resistance of the galvanometer (let's denote it as \( G \)). We are given two scenarios with different series resistances and their effects on the voltage range of the galvanometer.

### Step-by-Step Solution:

1. **Define the Variables:**
- Let \( G \) be the resistance of the galvanometer.
- The first series resistance \( R_1 = 880 \, \Omega \).
- The second series resistance \( R_2 = 420 \, \Omega \).
- The full-scale deflection current of the galvanometer is denoted as \( I_g \).
- The voltage across the galvanometer when using \( R_1 \) is \( V \), and when using \( R_2 \), it becomes \( \frac{V}{2} \).

2. **Write the Current Equations:**
- For the first case (using \( R_1 \)):
\[ I_g = \frac{V}{R_1 + G} = \frac{V}{880 + G} \]
- For the second case (using \( R_2 \)):
\[ I_g = \frac{V/2}{R_2 + G} = \frac{V/2}{420 + G} \]

3. **Set the Two Equations Equal:**
Since both expressions represent the same current \( I_g \), we can set them equal to each other:
\[ \frac{V}{880 + G} = \frac{V/2}{420 + G} \]

4. **Eliminate \( V \):**
We can eliminate \( V \) from the equation by multiplying both sides by \( 2(880 + G)(420 + G) \):
\[ 2V(420 + G) = V(880 + G) \]
Simplifying gives:
\[ 2(420 + G) = 880 + G \]

5. **Expand and Rearrange:**
Expanding the left side:
\[ 840 + 2G = 880 + G \]
Rearranging gives:
\[ 2G - G = 880 - 840 \]
Simplifying further:
\[ G = 40 \, \Omega \]

### Final Answer:
The resistance of the galvanometer \( G \) is \( 40 \, \Omega \).