In a galvanometer there is a deflection of 10 divisions per `50 mA`. The internal resistance of the galvanometer is `60Omega`. If a shunt of `2*5Omega` is connected to the galvanometer and there are 50 divisions in all on the scale of galvanometer what maximum current can this galvanometer read?

Explanation

To solve the problem step-by-step, we will follow these steps:

### Step 1: Determine the current through the galvanometer (I_G)
We know that the galvanometer deflects 10 divisions for 50 mA. Thus, the current corresponding to 1 division is given by:

\[ I_G = \frac{50 \, \text{mA}}{10} = 5 \, \text{mA} = 5 \times 10^{-3} \, \text{A} \]

### Step 2: Identify the internal resistance of the galvanometer (R_G) and the shunt resistance (R_S)
The internal resistance of the galvanometer is given as:

\[ R_G = 60 \, \Omega \]

The shunt resistance connected to the galvanometer is:

\[ R_S = 2.5 \, \Omega \]

### Step 3: Use the relationship between the total current (I), the galvanometer current (I_G), and the shunt current (I_S)
The total current flowing through the circuit can be expressed as:

\[ I = I_G + I_S \]

Where \(I_S\) is the current through the shunt. The relationship between these currents can also be expressed as:

\[ I_G = \frac{I \cdot R_S}{R_G + R_S} \]

### Step 4: Rearranging the equation to find the maximum current (I)
From the above relationship, we can rearrange it to find \(I\):

\[ I = I_G \cdot \frac{R_G + R_S}{R_S} \]

### Step 5: Substitute the known values into the equation
Substituting the values we have:

\[ I = 5 \times 10^{-3} \, \text{A} \cdot \frac{60 \, \Omega + 2.5 \, \Omega}{2.5 \, \Omega} \]

Calculating the numerator:

\[ 60 + 2.5 = 62.5 \, \Omega \]

Thus, we have:

\[ I = 5 \times 10^{-3} \cdot \frac{62.5}{2.5} \]

Calculating the fraction:

\[ \frac{62.5}{2.5} = 25 \]

So now we can calculate \(I\):

\[ I = 5 \times 10^{-3} \cdot 25 = 125 \times 10^{-3} \, \text{A} = 0.125 \, \text{A} \]

### Step 6: Convert the current to milliamperes
To express the maximum current in milliamperes:

\[ I = 0.125 \, \text{A} = 125 \, \text{mA} \]

### Final Answer
The maximum current that the galvanometer can read is **125 mA**.

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