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

Answer

To find the resistance of the galvanometer (Rg), we can follow these steps:

### Step 1: Identify the given values
- Maximum current (I_max) = 5 mA = 5 × 10^(-3) A
- Voltage (V) = 20 V
- External resistance (R_ext) = 3960 Ω

### Step 2: Use the formula for maximum current in terms of voltage and resistance
The formula relating current, voltage, and resistance in a circuit is given by Ohm's Law:
\[ I = \frac{V}{R} \]
In this case, the total resistance (R_total) in the circuit is the sum of the galvanometer resistance (Rg) and the external resistance (R_ext):
\[ I_{max} = \frac{V}{R_g + R_{ext}} \]

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation:
\[ 5 \times 10^{-3} = \frac{20}{R_g + 3960} \]

### Step 4: Rearrange the equation to solve for Rg
Rearranging the equation gives:
\[ R_g + 3960 = \frac{20}{5 \times 10^{-3}} \]
Calculating the right-hand side:
\[ R_g + 3960 = \frac{20}{0.005} = 4000 \]

### Step 5: Solve for Rg
Now, isolate Rg:
\[ R_g = 4000 - 3960 \]
\[ R_g = 40 \, \Omega \]

### Conclusion
The resistance of the galvanometer (Rg) is 40 Ω.

---