A galvanometer of `50 Omega` resistance when connected across the terminals of a battery of emf 2V along with the resistance `200 Omega` the deflection produced in the galvanometer is 10 divisions. If the total number of divisions on the galvanometer ·scale on either side of central zero is 30, then the maximum current that can pass through the galvanometer is

Options

A. `0.24 A`
B. `0.24 m A`
C. `0.024 m A`
D. `2.4 mA`

Answer

To solve the problem, we need to find the maximum current that can pass through the galvanometer. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the given values
- Resistance of the galvanometer (G) = 50 Ω
- External resistance (R) = 200 Ω
- EMF of the battery (V) = 2 V
- Deflection produced in the galvanometer = 10 divisions
- Total divisions on the galvanometer scale = 30 divisions

### Step 2: Determine the fraction of maximum current
The galvanometer shows a deflection of 10 divisions out of a maximum of 30 divisions. This means the current passing through the galvanometer (Ig) is:
\[ \frac{10}{30} \text{ of the maximum current (Imax)} \]
Thus,
\[ Ig = \frac{1}{3} I_{max} \]

### Step 3: Apply Kirchhoff's Law
According to Kirchhoff's law, the total voltage (V) across the circuit is equal to the sum of the voltage drops across the external resistance (R) and the galvanometer (G):
\[ V = I \cdot R + I_g \cdot G \]
Substituting the values we have:
\[ 2 = I \cdot 200 + Ig \cdot 50 \]

### Step 4: Substitute Ig in terms of Imax
From Step 2, we substitute \(Ig\) in the equation:
\[ 2 = I \cdot 200 + \left(\frac{1}{3} I_{max}\right) \cdot 50 \]
Let’s express \(I\) in terms of \(I_{max}\):
\[ I = \frac{2}{3} I_{max} \]
Substituting this into the equation gives:
\[ 2 = \left(\frac{2}{3} I_{max}\right) \cdot 200 + \left(\frac{1}{3} I_{max}\right) \cdot 50 \]

### Step 5: Solve for Imax
Now we simplify the equation:
\[ 2 = \frac{400}{3} I_{max} + \frac{50}{3} I_{max} \]
Combine the terms:
\[ 2 = \frac{450}{3} I_{max} \]
Multiply both sides by 3:
\[ 6 = 450 I_{max} \]
Now, divide by 450:
\[ I_{max} = \frac{6}{450} = \frac{1}{75} \text{ A} \]
Calculating this gives:
\[ I_{max} = 0.01333 \text{ A} \]

### Step 6: Convert to milliAmperes
To express this in milliAmperes:
\[ I_{max} = 0.01333 \times 1000 = 13.33 \text{ mA} \]

### Final Answer
The maximum current that can pass through the galvanometer is approximately **13.33 mA**.

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