Question

A galvanometer of resistance 8 $\Omega$ gives full scale deflection for a current of 4 mA. The resistance to be connected in series to the galvanometer to convert it into a voltmeter to measure a maximum potential difference of 20 V is

• A. 4992 $\Omega$
• B. 5008 $\Omega$
• C. 3992 $\Omega$
• D. 4008 $\Omega$

Hint:

- Always convert mA to A before calculations.
- Formula: $R_s = V/I_g - R_g$.
- The series resistance ensures the galvanometer does not get damaged at full voltage.
- To verify, calculate total resistance: $R_\text{total} = R_g + R_s = 8 + 4992 = 5000~\Omega$, then $V = I_g R_\text{total} = 0.004 \cdot 5000 = 20$ V.
- Practicing a few examples helps remember series/parallel conversions for meters.

Answer 1

The Correct Option is A

1. To convert a galvanometer into a voltmeter: $V = I_g (R_g + R_s)$, where $I_g$ = full scale deflection current, $R_g$ = galvanometer resistance, $R_s$ = series resistance.
2. Given: $V = 20$ V, $I_g = 4$ mA = 0.004 A, $R_g = 8~\Omega$.
3. Series resistance required: $R_s = \frac{V}{I_g} - R_g = \frac{20}{0.004} - 8 = 5000 - 8 = 4992~\Omega$