When 0.005A current flows through a moving coil galvanometer, it gives fullscale deflection. It is converted into a voltmeter to read 5 Volt, using an external resistance of `975Omega`. The resistance of galvanometer in ohms is

Options

A. 5
B. 10
C. 15
D. 25

Answer

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

### Step 1: Understand the relationship between voltage, current, and resistance
The voltage across the galvanometer and the external resistance can be expressed using Ohm's law:
\[ V = I \times R_{\text{eq}} \]
where \( V \) is the total voltage, \( I \) is the current, and \( R_{\text{eq}} \) is the equivalent resistance of the circuit.

### Step 2: Identify the known values
From the problem, we know:
- The current \( I = 0.005 \, \text{A} \)
- The voltage \( V = 5 \, \text{V} \)
- The external resistance \( R = 975 \, \Omega \)

### Step 3: Write the equation for equivalent resistance
The equivalent resistance \( R_{\text{eq}} \) in this case is the sum of the galvanometer resistance \( R_g \) and the external resistance \( R \):
\[ R_{\text{eq}} = R_g + 975 \]

### Step 4: Substitute into the voltage equation
Now, substituting \( R_{\text{eq}} \) into the voltage equation:
\[ V = I \times (R_g + 975) \]
Substituting the known values:
\[ 5 = 0.005 \times (R_g + 975) \]

### Step 5: Solve for \( R_g \)
First, isolate \( R_g + 975 \):
\[ R_g + 975 = \frac{5}{0.005} \]
Calculating the right side:
\[ R_g + 975 = 1000 \]

Now, solve for \( R_g \):
\[ R_g = 1000 - 975 \]
\[ R_g = 25 \, \Omega \]

### Final Answer
The resistance of the galvanometer \( R_g \) is \( 25 \, \Omega \).

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