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