A galvanometer has a coil of resistance `100 Omega` and gives a full-scale deflection for `30 mA` current. If it is to work as a voltmeter of `30 V` range, the resistance required to be added will be

Options

A. `900 Omega`
B. `1800 Omega`
C. `500 Omega`
D. `1000 Omega`

Explanation

To solve the problem of converting a galvanometer into a voltmeter, we need to determine the resistance that must be added in series with the galvanometer. Here’s a step-by-step solution:

### Step 1: Understand the given values
- Resistance of the galvanometer coil, \( R_G = 100 \, \Omega \)
- Full-scale deflection current, \( I_G = 30 \, \text{mA} = 0.030 \, \text{A} \)
- Desired voltage range of the voltmeter, \( V = 30 \, \text{V} \)

### Step 2: Calculate the voltage across the galvanometer at full-scale deflection
The voltage across the galvanometer when it shows full-scale deflection can be calculated using Ohm's law:
\[ V_G = I_G \times R_G \]
Substituting the known values:
\[ V_G = 0.030 \, \text{A} \times 100 \, \Omega = 3 \, \text{V} \]

### Step 3: Determine the total resistance needed for the voltmeter
To convert the galvanometer into a voltmeter that can measure up to 30 V, we need to find the total resistance \( R_T \) that will allow a current of \( I_G \) to flow when a voltage of 30 V is applied:
\[ R_T = \frac{V}{I_G} \]
Substituting the known values:
\[ R_T = \frac{30 \, \text{V}}{0.030 \, \text{A}} = 1000 \, \Omega \]

### Step 4: Calculate the resistance to be added
The resistance \( R_V \) that needs to be added in series with the galvanometer is given by:
\[ R_V = R_T - R_G \]
Substituting the values we calculated:
\[ R_V = 1000 \, \Omega - 100 \, \Omega = 900 \, \Omega \]

### Final Answer
The resistance required to be added is \( 900 \, \Omega \).

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