A moving coil galvanometer of resistance `20 Omega` gives a full scale deflection when a current of `1mA` is passed through it. It is to be converted into an ammeter reading `20 A` on full scale. But the shunt of `0.005 Omega` only is available. What resistance should be connected in series with the galvanometer coil?

Explanation

To solve the problem, we need to determine the resistance that should be connected in series with the galvanometer to convert it into an ammeter that reads up to 20 A. We will use the given values and the concept of current division in parallel circuits.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance of the galvanometer, \( R_g = 20 \, \Omega \)
- Full-scale deflection current of the galvanometer, \( I_g = 1 \, \text{mA} = 0.001 \, A \)
- Desired full-scale current for the ammeter, \( I = 20 \, A \)
- Shunt resistance available, \( R_s = 0.005 \, \Omega \)

2. **Calculate the Current through the Shunt:**
- The current through the galvanometer when the total current is 20 A can be found using:
\[ I_s = I - I_g = 20 \, A - 0.001 \, A = 19.999 \, A \]

3. **Apply Kirchhoff’s Voltage Law:**
- The potential drop across the galvanometer and the series resistance \( R \) must equal the potential drop across the shunt resistance.
- The voltage across the galvanometer is given by:
\[ V_{g} = I_g \cdot (R_g + R) = 0.001 \cdot (20 + R) \]
- The voltage across the shunt is given by:
\[ V_{s} = I_s \cdot R_s = 19.999 \cdot 0.005 \]

4. **Set the Voltages Equal:**
- According to Kirchhoff's law:
\[ V_g = V_s \]
- Therefore:
\[ 0.001 \cdot (20 + R) = 19.999 \cdot 0.005 \]

5. **Calculate the Right Side:**
- Calculate \( 19.999 \cdot 0.005 \):
\[ 19.999 \cdot 0.005 = 0.099995 \, V \]

6. **Set Up the Equation:**
- Now we have:
\[ 0.001 \cdot (20 + R) = 0.099995 \]

7. **Solve for R:**
- Divide both sides by 0.001:
\[ 20 + R = \frac{0.099995}{0.001} = 99.995 \]
- Now, isolate \( R \):
\[ R = 99.995 - 20 = 79.995 \, \Omega \]

8. **Round the Result:**
- The resistance \( R \) can be approximated to:
\[ R \approx 80 \, \Omega \]

### Final Answer:
The resistance that should be connected in series with the galvanometer is approximately \( 80 \, \Omega \).

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