Question

A $50 \, \Omega$ resistance is connected to a battery of $5\, V$. 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. $r_s = 0.5 \, \Omega$ in parallel with the galvanometer
• B. $r_s = 0.5 \, \Omega$ in series with the galvanometer
• C. $r_s = 1 \, \Omega$ in series with galvanometer
• D. $r_s = 1 \Omega$ in parallel with galvanometer

Answer 1

Answer: (A)

$I=\frac{5}{50}=0.1$
$I '=0.099$
$R_{e}=50+\frac{100S}{100+S}=\frac{V}{I}$
$\frac{100S}{100+S}=\frac{5}{0.099}-50$
$\frac{100S}{100+S}=50\cdot50-50$
$\frac{100\,S}{100+S}=0.5$
$100 S=50+0.55$
$99.5S=50$
$S=\frac{50}{99.05}=0.5\,\Omega$