Question

The resistance of a moving coil galvanometer is $5\Omega$. The maximum current it can measure is $0.015\text{A}$. To convert it into an ammeter to measure $1.5\text{A}$:

• A. connect $\frac{5}{99}\Omega$ in series
• B. connect $\frac{99}{5}\Omega$ in parallel
• C. connect $\frac{5}{99}\Omega$ in parallel
• D. connect $\frac{99}{5}\Omega$ in series

Answer 1

The correct option is C connect $\frac{5}{99}\Omega$ in parallel

To convert galvanometer into ammeter, resistance is connected in parallel with the galvanometer as shown below.


Given:
$G=5\Omega ,I=1.5\text{A},I_g=0.015\text{A}$

Since both are in parallel, potential across both will be equal.

$\therefore I_g\times G=(I-I_g)\times r(\because V=IR)$

$0.015\times 5=(1.5-0.015)\times r$

$\therefore r=0.0505\Omega \approx \frac{5}{99}\Omega$

Therefore, option $\left(C\right)$ is correct.

Why this question ? Concept: To convert galvanometer into ammeter, resistance is connected in parallel to the galvanometer.