Step 1: Understanding the Concept:
To convert a galvanometer into a voltmeter of a higher range, a high resistance must be connected in series with it.
The series resistance limits the current to the galvanometer's full-scale value when the target voltage is applied.
Step 2: Key Formula or Approach:
Ohm's Law for the whole device: \(V = I_g (G + R)\).
Resistance of galvanometer: \(G = V_g / I_g\).
Step 3: Detailed Explanation:
Given:
Full scale current \(I_g = 16\text{ mA} = 0.016\text{ A}\).
Full scale voltage drop \(V_g = 80\text{ mV} = 0.08\text{ V}\).
First, calculate internal resistance \(G\):
\[ G = \frac{V_g}{I_g} = \frac{0.08}{0.016} = 5\Omega \]
Now, target voltage range \(V = 160\text{ V}\).
Apply the voltmeter formula:
\[ 160 = 0.016 \times (5 + R) \]
Divide by \(0.016\):
\[ \frac{160}{0.016} = 5 + R \implies 10000 = 5 + R \]
Solve for \(R\):
\[ R = 10000 - 5 = 9995\Omega \]
Since it's a voltmeter conversion, the resistance is in series.
Step 4: Final Answer:
A resistance of \(9995\Omega\) must be connected in series.