A galvanometer of 25 ohm resistance can read a maximum current of 6mA. It can be used as a voltmeter to measure maximum potential difference of 6V by connecting a resistance to galvanometer. Identify the correct choice from the following

Options

A. `1025Omega` in series
B. `1025Omega` is parallel
C. `975Omega` is series
D. `975Omega` in parallel

Answer

To solve the problem of converting a galvanometer into a voltmeter, we will follow these steps:

### Step 1: Understand the given values
We have a galvanometer with:
- Internal resistance \( R_g = 25 \, \Omega \)
- Maximum current \( I_g = 6 \, \text{mA} = 6 \times 10^{-3} \, \text{A} \)
- Maximum potential difference \( V = 6 \, \text{V} \)

### Step 2: Determine the total resistance needed
When converting a galvanometer into a voltmeter, we need to add a resistance \( R \) in series with the galvanometer. The total resistance in the circuit will be:
\[ R_{\text{total}} = R_g + R \]

### Step 3: Use Ohm's Law to relate voltage, current, and resistance
According to Ohm's Law, the voltage across the total resistance can be expressed as:
\[ V = I \times R_{\text{total}} \]
Substituting the known values:
\[ 6 \, \text{V} = 6 \times 10^{-3} \, \text{A} \times (25 \, \Omega + R) \]

### Step 4: Solve for \( R \)
Rearranging the equation:
\[ 6 = 6 \times 10^{-3} \times (25 + R) \]
Dividing both sides by \( 6 \times 10^{-3} \):
\[ 1000 = 25 + R \]
Now, isolating \( R \):
\[ R = 1000 - 25 \]
\[ R = 975 \, \Omega \]

### Conclusion
To convert the galvanometer into a voltmeter capable of measuring up to 6 V, we need to add a resistance of \( 975 \, \Omega \) in series with the galvanometer.