Question

The galvanometer shown in the figure has a resistance $50\Omega$ and full-scale deflection current 1 mA. What are the the resistances $R_1,R_2$ and $R_3$ required to convert it into ammeter having ranges as indicated?

• A. $\frac{1}{22}\Omega ,\frac{1}{33}\Omega ,\frac{1}{44}\Omega$
• B. $\frac{1}{22}\Omega ,\frac{15}{33}\Omega ,\frac{1}{100}\Omega$
• C. $\frac{1}{198}\Omega ,\frac{1}{22}\Omega ,\frac{15}{33}\Omega$
• D. $\frac{1}{22}\Omega ,\frac{15}{198}\Omega ,\frac{5}{44}\Omega$

Answer 1

The correct option is C

$\frac{1}{198}\Omega ,\frac{1}{22}\Omega ,\frac{15}{33}\Omega$

The maximum current in the galvonometer $I_g$, the resistance of the galvonometer $r$, the shunt resistance $R$, and the maximum current of the ammeter $I$ are related as
$R=\frac{r}{\frac{I}{I_g}-1}$

For the range I = 0.1 A,
$R_1+R_2+R_3=\frac{r}{\frac{I}{I_g}-1}=\frac{50}{100-1}=0.5\Omega$

For the range I = 1 A,
$R_1+R_2=\frac{r+R_3}{\frac{I}{I_g}-1}=\frac{50+R_3}{1000-1}$
$R_1+R_2+R_3=0.5$. Hence, $R_3<0.5\Rightarrow 50+R_3\simeq 50$
$\therefore R_1+R_2=\frac{50+R_3}{1000-1}\simeq 0.05$

For the range I = 10 A,
$R_1=\frac{r+R_2+R_3}{\frac{I}{I_g}-1}=\frac{50+R_2+R_3}{10000-1}\simeq 0.005\Omega$
$R_2=0.05-R_1=0.05-0.005\simeq 0.05\Omega$
$R_3=0.5-R_1-R_2=0.5-0.005-0.05\simeq 0.5\Omega$