Question

In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is $G$, the resistance of ammeter will be

• A. $\frac{1}{499}G$
• B. $\frac{499}{500}G$
• C. $\frac{1}{500}G$
• D. $\frac{500}{499}G$

Answer 1

The Correct Option is C

In this problem, we are given an ammeter where 0.2% of the main current passes through the galvanometer. We need to find the resistance of the ammeter in terms of the galvanometer resistance $G$.

1. First, understand the current division in the ammeter. Only a small fraction of the total current (main current) goes through the galvanometer, while the rest goes through a shunt resistance, effectively bypassing the galvanometer.
2. Here, the fraction of the main current through the galvanometer is given as 0.2%. This implies that 99.8% of the main current bypasses the galvanometer and flows through the shunt resistance.
3. Assume the total current as $I$. Then, the current through the galvanometer $I_g = 0.002I$.
4. The current through the shunt, $I_s = 0.998I$.
5. Using the relation between currents through parallel circuits: $$ \frac{I_g}{I_s} = \frac{R_s}{G} $$, where $R_s$ is the resistance of the shunt.
6. Substituting the values, we have: $$ \frac{0.002I}{0.998I} = \frac{R_s}{G} $$.
7. Simplify this equation to find $R_s$: $$ R_s = G \cdot \frac{0.002}{0.998} $$.
8. Approximately, $$ R_s = \frac{G}{499} $$.
9. The resistance of the ammeter, which is effectively that of the shunt when combined in parallel with the galvanometer, is essentially negligible because $R_s \ll G$ in magnitude, so the predominant resistance determining factor is this small shunt value: $$ R_{\text{ammeter}} = \frac{1}{500} G $$, which matches our correct option.

Thus, the resistance of the ammeter is $ \frac{1}{500} G $.