To convert a galvanometer into a voltmeter, we need to determine the resistance that must be connected in series with the galvanometer. Here’s how to solve the problem step by step:
### Step 1: Determine the total current the galvanometer can measure
The sensitivity of the galvanometer is given as \( 16 \, \mu A/\text{div} \) and it has 30 divisions. To find the maximum current that the galvanometer can measure, we multiply the sensitivity by the number of divisions:
\[
I_G = \text{Sensitivity} \times \text{Number of Divisions} = 16 \, \mu A/\text{div} \times 30 \text{ div} = 480 \, \mu A
\]
### Step 2: Convert the current to amperes
Since \( 1 \, \mu A = 10^{-6} \, A \), we convert the current:
\[
I_G = 480 \, \mu A = 480 \times 10^{-6} \, A = 0.00048 \, A
\]
### Step 3: Use the formula for the resistance needed to convert the galvanometer to a voltmeter
To convert the galvanometer into a voltmeter that can read up to \( 3 \, V \), we use the formula:
\[
R = \frac{V}{I_G} - G
\]
Where:
- \( R \) is the resistance to be connected in series,
- \( V \) is the voltage range (3 V),
- \( I_G \) is the maximum current (0.00048 A),
- \( G \) is the internal resistance of the galvanometer (not given, so we will leave it as \( G \)).
### Step 4: Substitute the values into the formula
Substituting the known values into the formula:
\[
R = \frac{3 \, V}{0.00048 \, A} - G
\]
Calculating \( \frac{3}{0.00048} \):
\[
R = 6250 - G
\]
### Step 5: Approximate the internal resistance
If we assume a typical value for the internal resistance \( G \) of the galvanometer, we can estimate \( R \). For example, if \( G \) is negligible compared to 6250 ohms, we can say:
\[
R \approx 6250 \, \Omega \text{ (or about 6.25 kΩ)}
\]
### Conclusion
Thus, the resistance that needs to be connected in series to convert the galvanometer into a voltmeter capable of reading 3 V is approximately \( 6250 \, \Omega \) or about \( 6 \, k\Omega \).
