Question

It is required to convert a galvanometer of current range $15$mA and voltage range $750$mV into an ammeter of range $25$A. The value of necessary resistance that must be connected in parallel with the galvanometer is?

• A. $0.4\Omega$
• B. $0.03\Omega$
• C. $1\Omega$
• D. $0.5\Omega$

Answer 1

The correct option is B $0.03\Omega$

Given, $I_g=15mA$
$=15\times {10}^{-3}A$
$V_g=750mV$
$=7.5\times {10}^{-1}V$
$I=25A$
$S=?$
$\therefore S=G\frac{I_g}{I-I_g}$
$=\left(\frac{V_g}{I_g}\right)\cdot \frac{I_g}{I-I_g}$
$=\frac{7.5\times {10}^{-1}}{25-0.015}$
$=\frac{0.75}{24.985}\approx 0.03\Omega$.