A galvanometer can withstand safely a maximum current of 5 mA. If is converted into voltmeter readding upto 20 V by connecting in series an external resistance of `3960 Omega`. The resistance of galvanometer is

Options

A. `48 Omega`
B. `44 Omega`
C. `36 Omega`
D. `40 Omega`

Answer

To find the resistance of the galvanometer (G), we can use the relationship between the total resistance in the circuit, the voltage across the voltmeter (V), and the maximum current (I_G) that the galvanometer can handle. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the given values:**
- Maximum current (I_G) = 5 mA = \(5 \times 10^{-3}\) A
- Voltage (V) = 20 V
- External resistance (R) = 3960 Ω

2. **Use the formula for total resistance in a voltmeter setup:**
The total resistance (R_total) in the circuit when the galvanometer is converted into a voltmeter is given by:
\[ R_{total} = R + G \]
where G is the resistance of the galvanometer.

3. **Relate voltage, current, and resistance:**
From Ohm's law, we know that:
\[ V = I_G \times R_{total} \]
Rearranging gives:
\[ R_{total} = \frac{V}{I_G} \]

4. **Substitute the values into the equation:**
Substitute V and I_G into the equation:
\[ R_{total} = \frac{20 \text{ V}}{5 \times 10^{-3} \text{ A}} = \frac{20}{0.005} = 4000 \, \Omega \]

5. **Set up the equation for G:**
Now we can substitute R_total back into the equation for total resistance:
\[ R_{total} = R + G \implies 4000 = 3960 + G \]

6. **Solve for G:**
Rearranging the equation gives:
\[ G = 4000 - 3960 = 40 \, \Omega \]

### Final Answer:
The resistance of the galvanometer (G) is **40 Ω**.

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