There is one moving coil galvanometer with coil resistance 25 `Omega` . Its current sensitivity changes from 72 divisions/ampere to 12 divisions/ampere when shunt is used. What is resistance of shunt is ohm?

Explanation

To find the resistance of the shunt (S) used in a moving coil galvanometer, we can follow these steps:

### Step 1: Understand the given data
We know:
- Resistance of the galvanometer coil (G) = 25 Ω
- Current sensitivity without shunt = 72 divisions/ampere
- Current sensitivity with shunt = 12 divisions/ampere

### Step 2: Relate current sensitivity to the galvanometer
Current sensitivity (CS) is defined as the number of divisions on the scale per unit current flowing through the galvanometer. The relationship between the current sensitivity before and after using the shunt can be expressed as:

\[ \frac{I_g}{I} = \frac{CS_{shunt}}{CS_{no\ shunt}} \]

Where:
- \(I_g\) = current through the galvanometer
- \(I\) = total current
- \(CS_{shunt} = 12\) divisions/ampere
- \(CS_{no\ shunt} = 72\) divisions/ampere

### Step 3: Set up the equation
From the current sensitivity values, we can write:

\[ \frac{I_g}{I} = \frac{12}{72} = \frac{1}{6} \]

This implies:

\[ I_g = \frac{1}{6} I \]

### Step 4: Determine the current through the shunt
The current through the shunt (I_s) can be expressed as:

\[ I_s = I - I_g = I - \frac{1}{6}I = \frac{5}{6}I \]

### Step 5: Use the relationship of currents and resistances
From the circuit, we can express the relationship between the currents and resistances:

\[ \frac{I_g}{I_s} = \frac{G}{S} \]

Substituting the values we have:

\[ \frac{\frac{1}{6}I}{\frac{5}{6}I} = \frac{25}{S} \]

This simplifies to:

\[ \frac{1}{5} = \frac{25}{S} \]

### Step 6: Solve for the shunt resistance (S)
Cross-multiplying gives:

\[ S = 25 \times 5 = 125 \, \Omega \]

### Step 7: Conclusion
The resistance of the shunt is:

\[ \boxed{5 \, \Omega} \]