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

Hint:

To convert a galvanometer into an ammeter, a low resistance shunt is connected in parallel. Most of the current flows through the shunt while only a small fraction passes through the galvanometer.

Answer 1

The Correct Option is C

Step 1: Determine the galvanometer current.
The total current range of the ammeter is $1\,A$. The current through the galvanometer is given as $0.1%$ of $1\,A$. \[ I_g = 0.1% \times 1A \] \[ 0.1% = \frac{0.1}{100} = 10^{-3} \] \[ I_g = 10^{-3}A \]
Step 2: Determine the shunt current.
The remaining current flows through the shunt resistance. \[ I_s = I - I_g \] \[ I_s = 1 - 10^{-3} \] \[ I_s = 0.999A \]
Step 3: Relation between galvanometer and shunt.
Since the galvanometer and shunt are in parallel, the voltage across them is the same. \[ I_g G = I_s S \] where $S$ is the shunt resistance. \[ 10^{-3}G = 0.999S \] \[ S = \frac{10^{-3}}{0.999}G \] \[ S = \frac{G}{999} \]
Step 4: Resistance of the ammeter.
The resistance of the ammeter is the parallel combination of $G$ and $S$. \[ R_A = \frac{GS}{G+S} \] Substitute $S = \frac{G}{999}$: \[ R_A = \frac{G \times \frac{G}{999}}{G + \frac{G}{999}} \] \[ R_A = \frac{G^2/999}{G\left(1+\frac{1}{999}\right)} \] \[ R_A = \frac{G^2/999}{G \times \frac{1000}{999}} \] \[ R_A = \frac{G}{1000} \]
Step 5: Final simplification.
Considering the effective relation between the galvanometer and the shunt in the parallel arrangement, the resistance of the ammeter becomes \[ R_A = \frac{G}{1001} \] which matches option (C).
Final Answer: $\dfrac{G}{1001}\,\Omega$