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The resistance of an ammeter is $13\,\Omega$ and its scale is graduated for a current upto 100 amps. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto 750 amperes by this meter. The value of shunt-resistance is
A
$2\,\Omega$
B
$0.2\,\Omega$
C
$2\,k\Omega$
D
$20\,\Omega$
Detailed Solution
The shunt S is connected in parallel with the ammeter of resistance $G = 13\,\Omega$.
At full scale the ammeter carries $I_g = 100$ A and the shunt carries the remaining current $I - I_g = 750 - 100 = 650$ A.
The potential difference across both is the same: $I_g G = (I - I_g)S$
Equivalently, $\dfrac{I}{I_g} = 1 + \dfrac{G}{S}$
$\dfrac{750}{100} = 1 + \dfrac{13}{S}$
$7.5 - 1 = \dfrac{13}{S} \Rightarrow \dfrac{13}{S} = 6.5$
$S = \dfrac{13}{6.5} = 2\,\Omega$
At full scale the ammeter carries $I_g = 100$ A and the shunt carries the remaining current $I - I_g = 750 - 100 = 650$ A.
The potential difference across both is the same: $I_g G = (I - I_g)S$
Equivalently, $\dfrac{I}{I_g} = 1 + \dfrac{G}{S}$
$\dfrac{750}{100} = 1 + \dfrac{13}{S}$
$7.5 - 1 = \dfrac{13}{S} \Rightarrow \dfrac{13}{S} = 6.5$
$S = \dfrac{13}{6.5} = 2\,\Omega$