A galvanometer of $50~\Omega$ resistance has 25 divisions. A current of $4 imes 10^{-4} ext{ A}$ gives a d

Question

A galvanometer of $50~\Omega$ resistance has 25 divisions. A current of $4 	imes 10^{-4} 	ext{ A}$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $25 	ext{ V}$, it should be connected with a resistance of:

Accepted Answer

1. **Concept:** To convert a galvanometer into a voltmeter, a high resistance ($R$) is connected in **series** with the galvanometer coil. This ensures that the total resistance is high enough to not draw significant current from the circuit being measured .
2. **Calculate Full Scale Deflection Current ($I_g$):**
   - Current per division = $4 	imes 10^{-4} 	ext{ A}$.
   - Total divisions = 25.
   - $I_g = 25 	imes 4 	imes 10^{-4} = 100 	imes 10^{-4} = 10^{-2} 	ext{ A} = 0.01 	ext{ A}$.
3. **Formula:** The potential difference ($V$) across the combination is given by Ohm's law:
   $$ V = I_g (G + R) $$
   Where $G$ is the galvanometer resistance and $R$ is the series resistance.
4. **Calculation:**
   - Given $V = 25 	ext{ V}$ and $G = 50~\Omega$.
   - Rearranging the formula for $R$:
     $$ R = \frac{V}{I_g} - G $$
     $$ R = \frac{25}{0.01} - 50 $$
     $$ R = 2500 - 50 = 2450~\Omega $$
5. **Conclusion:** A resistance of $2450~\Omega$ must be connected in series.

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