Question

An electric meter of internal resistance $20\Omega$ gives a full scale deflection when one milliampere current flows through it. The maximum current, that can be measured by using three resistors of resistance $12\Omega$ each, in milliampere is :

• A. $10$
• B. $8$
• C. $6$
• D. $4$

Answer 1

Answer: (C)

$6$

Given,
Meter has a resistance G = 20 $\Omega$
Current through meter for full deflections
$t_g=1mA=1\times {10}^{-3}A$
Now, for stunt three resistors each have resistance 12 $\Omega$ are connected in parallel.
Hence,
$\frac{1}{R_p}=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}\Rightarrow R_p=4\Omega$
Now, from the formula
$R=\frac{I_g}{I-I_g}$ ...(i)
( where, R is shunt resistance R = 4 $\Omega$ )
Now, putting the values in Eq. (i)
$4=\frac{1\times {10}^{-3}\times 20}{I-1\times {10}^{-3}}$
$4I-4\times {10}^{-3}=20\times {10}^{-3}$
$4I=20\times {10}^{-3}+4\times {10}^{-3}=24\times {10}^{-3}$
$I=\frac{24\times {10}^{-3}}{4}=6\times {10}^{-3}A=6mA$