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. $r_s=0.5\Omega$ 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)

$I=\frac{5}{50}=0.1$
$I^{\prime }=0.099$
$R_e=50+\frac{100S}{100+S}=\frac{V}{I}$
$\frac{100S}{100+S}=\frac{5}{0.099}-50$
$\frac{100S}{100+S}=50\cdot 50-50$
$\frac{100S}{100+S}=0.5$
$100S=50+0.55$
$99.5S=50$
$S=\frac{50}{99.05}=0.5\Omega$