Question

A galvanometer of resistance $G \, \Omega$ is converted into an ammeter of range 0 to 1 A. If the current through the galvanometer is 0.1\% of 1 A, the resistance of the ammeter is:

• A. $\frac{G}{999} \, \Omega$
• B. $\frac{G}{1000} \, \Omega$
• C. $\frac{G}{1001} \, \Omega$
• D. $\frac{G}{100.1} \, \Omega$

Hint:

To convert a galvanometer into an ammeter, always calculate the shunt resistance $S$ using $S = \frac{I_g G}{I_s}$. The total resistance of the ammeter is the effective parallel resistance of $G$ and $S$.

Answer 1

The Correct Option is B

To convert a galvanometer into an ammeter, a shunt resistance $S$ is connected in parallel with the galvanometer. The current through the galvanometer is given as $I_g = 0.001 \, \text{A}$ (0.1\% of 1 A). The total current through the ammeter is $I = 1 \, \text{A}$. The current through the shunt is: \[ I_s = I - I_g = 1 - 0.001 = 0.999 \, \text{A}. \] The potential difference across the galvanometer and the shunt is the same: \[ I_g G = I_s S. \] Substitute $I_s$ and rearrange for $S$: \[ S = \frac{I_g G}{I_s} = \frac{(0.001) G}{0.999}. \] Simplify: \[ S = \frac{G}{999}. \] The total resistance of the ammeter is the parallel combination of $G$ and $S$: \[ R_{\text{ammeter}} = \frac{G S}{G + S}. \] Substitute $S = \frac{G}{999}$: \[ R_{\text{ammeter}} = \frac{G \cdot \frac{G}{999}}{G + \frac{G}{999}} = \frac{G^2}{999G + G} = \frac{G}{1000}. \] Thus, the resistance of the ammeter is: \[ \boxed{\frac{G}{1000} \, \Omega}. \] \vspace{1em}