Question

An ammeter has a current range of 0-5 A, and its internal resistance is $0.2\Omega$. In order to change the range to $0-25A$ , we need to add a resistance of

• A. $0.8\Omega$ in series with the meter
• B. $1.0\Omega$ in series with the meter
• C. $0.04\Omega$ in parallel with the meter
• D. $0.05\Omega$ in parallel with the meter

Answer 1

The correct option is D $0.05\Omega$ in parallel with the meter

To extend the range of the ammeter, a resistance $R_{sh}$ is connected across the meter.


$I_m=fullscaledeflectioncurrent=5A$

Multiplying power,

$m=\frac{I}{I_m}=\frac{25}{5}=5$

$I_m=\frac{I{R}_{sh}}{R_{sh}+R_m}$

$m=1+\frac{R_m}{R_{sh}}$

$m=1+\frac{R_m}{R_{sh}}$

$R_{sh}=\frac{R_m}{m-1}=\frac{0.2}{5-1}=0.05$