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A millivoltmeter of 25 millivolt range is to be converted into an ammeter of 25 ampere range. The value (in ohm) of necessary shunt will be
A
0.05
B
0.001
C
0.01
D
1
Detailed Solution
Full-scale voltage across the meter: $i_gR_g = 25\times10^{-3}$ V
The shunt is in parallel, so $S(i - i_g) = i_gR_g$
$S = \frac{i_gR_g}{i - i_g} \approx \frac{25\times10^{-3}}{25}$ (since $i_g \ll i$)
$S = 10^{-3}\ \Omega = 0.001\ \Omega$
The shunt is in parallel, so $S(i - i_g) = i_gR_g$
$S = \frac{i_gR_g}{i - i_g} \approx \frac{25\times10^{-3}}{25}$ (since $i_g \ll i$)
$S = 10^{-3}\ \Omega = 0.001\ \Omega$