Question

A galvanometer of resistance \( G \) is converted into a voltmeter of range \( (0 - V) \) by using a resistance \( R \). Find the resistance, in terms of \( R \) and \( G \), required to convert it into a voltmeter of range \( (0 - \frac{V}{2}) \).

Hint:

When modifying the range of a voltmeter by changing its series resistance, it’s crucial to ensure that the new resistance provides the correct voltage division to achieve the desired scale without exceeding the galvanometer's maximum current.

Answer 1

Given: Galvanometer resistance \( G \)  Initial voltmeter range \( (0 - V) \) with series resistance \( R \)  Desired voltmeter range \( (0 - \frac{V}{2}) \) The current \( I \) through the galvanometer for full-scale deflection is:
\[ I = \frac{V}{R + G} \] For the new range \( (0 - \frac{V}{2}) \), the current \( I \) should remain the same. Let the new series resistance be \( R' \). Therefore:
\[ I = \frac{\frac{V}{2}}{R' + G} \] Setting the currents equal:
\[ \frac{V}{R + G} = \frac{\frac{V}{2}}{R' + G} \] Simplify and solve for \( R' \):
\[ \frac{1}{R + G} = \frac{1}{2(R' + G)} \] \[ R + G = 2(R' + G) \] \[ R + G = 2R' + 2G \] \[ R - G = 2R' \] \[ R' = \frac{R - G}{2} \] Therefore, the required resistance is:
\[ \boxed{R' = \frac{R - G}{2}} \]