Question

A $50\Omega$ resistance is connected to a battery of $5V$. A galvanometer of resistance $100\Omega$ is to be used as an ammeter to measure current through the resistance, for this a resistance $r_s$ is connected to the galvanometer. Which of the following connections should be employed if the measured current is within $1$% of the current without the ammeter in the circuit?

• A. $r_s=1\Omega$ in parallel with galvanometer
• B. $r_s=0.5\Omega$ in parallel with the galvanometer
• C. $r_s=0.5\Omega$ in series with the galvanometer
• D. $r_s=1\Omega$ in series with galvanometer

Answer 1

The correct option is B $r_s=0.5\Omega$ in parallel with the galvanometer

Current in the circuit without ammeter is $I_{max}=V/R=5/50=0.1A$

Given that current in the circuit with ammeter is within $1\%$ of original current.

Hence, current flowing is $I^{\prime }=0.99I_{max}=0.099A$

Resistance $r_s$ is connected in parallel with the Galvanometer to increase its range.

$50\Omega$ is connected in series to ammeter.Therefore, overall resistance of the circuit is:

$R=50+\frac{100r_s}{100+r_s}$

By Ohm's law:

$V=I^{\prime }R$

$5$ = $0.099\times$ $(50+100r_s)/(100+r_s)$

$⟹$ $r_s$ = $0.5\Omega$