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Step 1: Understand the Problem
We have a galvanometer with an internal resistance of $20\,\Omega$ and a figure of merit (FOM) of $0.005\,\text{A/div}$. The galvanometer scale is marked with $30$ divisions on each side (making full-scale deflection $30$ divisions). We wish to convert this galvanometer into a voltmeter capable of reading up to $15\,\text{V}$. We need to determine the external (series) resistance $R$ required.
Step 2: Note the Known Quantities
Galvanometer resistance, $R_g = 20\,\Omega$
Figure of merit, FOM $= 0.005\,\text{A/div}$
Full-scale deflection divisions, $N = 30$
Desired full-scale voltage, $V_{\text{full-scale}} = 15\,\text{V}$
Step 3: Calculate the Maximum Current for Full-Scale Deflection
The figure of merit tells us how much current per division is needed. Hence, the current for one division = $0.005\,\text{A}$. For $30$ divisions (full-scale deflection), the current is:
$I_g = 0.005 \times 30 = 0.15\,\text{A}
Step 4: Relate Voltage to the Desired Current and Total Series Resistance
For this galvanometer to measure up to $15\,\text{V}$ at full-scale deflection, the total current through the galvanometer circuit is $I_g = 0.15\,\text{A}$. The total resistance required in the circuit (galvanometer + series resistor) is given by Ohm’s law:
$V_{\text{full-scale}} = I_g \times (R_g + R)$
Substitute the known values:
$15 = 0.15 \times (20 + R)$
Step 5: Solve for the Series Resistance $R$
Rearranging the equation:
$20 + R = \frac{15}{0.15} = 100
Hence,
$R = 100 - 20 = 80\,\Omega
Step 6: Final Answer
The required series resistance is $80\,\Omega.$
