Question

The scale of a galvanometer is divided into 150 equal division. The galvanometer has the current sensitivity of 10 divisions per mA and the voltage sensitivity of 2 divisions per mV . How the galvanometer can be designed to read (a) 6A / division and (b) 1V /division ?

• A. $S=8.3\times 10^{-5}\Omega ,R=9995\Omega$
• B. $S=8.3\times 10^{-2}\Omega ,R=995\Omega$
• C. $S=4.3\times 10^{-5}\Omega ,R=9950\Omega$
• D. $R=8.3\times 10^{-5}\Omega ,S=995\Omega$

Answer 1

Answer: (A)

Here $I_g=\left(\frac{1mA}{10dv}\right)(150div)=15mA$
$V_g=\left(\frac{1mA}{2div}\right)(150div)=75mV$
$G=\frac{V_g}{I_g}=\frac{75mV}{15mA}=5\Omega$
$(a)I$ (current to be measured)
$=\left(\frac{6A}{div}\right)(150div)=900A$
$Asn=\frac{I}{I_s}=\frac{900A}{15mA}=\frac{900A}{15\times 10^{-3}}=6\times 10^4$
$S=\frac{G}{(n-1)}=\frac{5\Omega }{(6\times 10^4-1)}=\frac{5\Omega }{6\times 10^4}=8.3\times 10^{-5}\Omega$
(b) V (voltage to be measured)
$=\left(\frac{1V}{div}\right)(150div)=150V$
$Asn=\frac{V}{V_g}=\frac{150V}{75mV}=\frac{150V}{75\times 10^{-3}V}=2\times 10^3$
$R=G(n-1)=5\Omega (2\times 10^3-1)$
$=5\Omega (2000-1)=9995\Omega$