A moving coil galvanometer is converted into an ammeter reads upto `0.03 A` by connecting a shunt of resistance `4r` across it and ammeter reads up `0.06 A`, when a shunt of resistance `r` is used. What is the maximum current which can be sent through this galvanometer if no shunt is used ?

Options

A. `0.01A`
B. `0.02A`
C. `0.03A`
D. `0.04A`

Answer

To solve the problem, we need to determine the maximum current that can be sent through the galvanometer without using a shunt. We will use the information given about the shunt resistances and the corresponding currents.

### Step 1: Understand the relationship between the galvanometer current and the total current
When a shunt resistance \( S \) is connected in parallel with the galvanometer, the total current \( I \) splits into two parts: the current through the galvanometer \( I_G \) and the current through the shunt \( I - I_G \). The voltages across both components are equal.

### Step 2: Write the equations for the two cases
1. For the first case with shunt \( S = 4R \) and total current \( I = 0.03 \, A \):
\[ I_G = \frac{S}{G + S} \cdot I \]
Substituting \( S = 4R \) and \( I = 0.03 \):
\[ I_G = \frac{4R}{G + 4R} \cdot 0.03 \quad \text{(Equation 1)} \]

2. For the second case with shunt \( S = R \) and total current \( I = 0.06 \, A \):
\[ I_G = \frac{S}{G + S} \cdot I \]
Substituting \( S = R \) and \( I = 0.06 \):
\[ I_G = \frac{R}{G + R} \cdot 0.06 \quad \text{(Equation 2)} \]

### Step 3: Equate the two expressions for \( I_G \)
From Equation 1 and Equation 2, we can set them equal to each other:
\[ \frac{4R}{G + 4R} \cdot 0.03 = \frac{R}{G + R} \cdot 0.06 \]

### Step 4: Simplify the equation
Cross-multiplying gives:
\[ 4R \cdot 0.03 (G + R) = R \cdot 0.06 (G + 4R) \]
Dividing both sides by \( R \) (assuming \( R \neq 0 \)):
\[ 4 \cdot 0.03 (G + R) = 0.06 (G + 4R) \]
Expanding both sides:
\[ 0.12G + 0.12R = 0.06G + 0.24R \]

### Step 5: Rearranging the equation
Bringing all terms involving \( G \) to one side and \( R \) to the other:
\[ 0.12G - 0.06G = 0.24R - 0.12R \]
\[ 0.06G = 0.12R \]
Thus, we find:
\[ G = 2R \]

### Step 6: Substitute \( G \) back into one of the equations to find \( I_G \)
Using Equation 1:
\[ I_G = \frac{4R}{2R + 4R} \cdot 0.03 \]
This simplifies to:
\[ I_G = \frac{4R}{6R} \cdot 0.03 = \frac{2}{3} \cdot 0.03 = 0.02 \, A \]

### Conclusion
The maximum current that can be sent through the galvanometer without using a shunt is:
\[ \boxed{0.02 \, A} \]