Question:
A galvanometer has current range of $15\, mA$ and voltage range $750\, mV$. To convert this galvanometer into an ammeter of range $25\, A$, the required shunt is
A galvanometer has current range of $15\, mA$ and voltage range $750\, mV$. To convert this galvanometer into an ammeter of range $25\, A$, the required shunt is
Updated On: Apr 26, 2024
- $0.8 \, \Omega$
- $0.93 \, \Omega$
- $0.03 \, \Omega$
- $2.0 \, \Omega$
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The Correct Option is C
Solution and Explanation
Given : $V = 750 \times 10^{-3} V;$
$ I_{g} = 15 \times 10^{-3} A $
and $ I = 25\, A $
Using therelation
$a = \frac{Y}{I_{g}}$
$ = \frac{750\times 10^{-3}}{15\times 10^{-3}}$
$= 50\, \Omega $
$I_{g} = \frac{S}{S + a}\times I $
$ 15\times 106-3 = \left(\frac{S}{S + 50}\right) \times 25 $
$\therefore \, \, S = 0.03\, \Omega$
$ I_{g} = 15 \times 10^{-3} A $
and $ I = 25\, A $
Using therelation
$a = \frac{Y}{I_{g}}$
$ = \frac{750\times 10^{-3}}{15\times 10^{-3}}$
$= 50\, \Omega $
$I_{g} = \frac{S}{S + a}\times I $
$ 15\times 106-3 = \left(\frac{S}{S + 50}\right) \times 25 $
$\therefore \, \, S = 0.03\, \Omega$
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Magnetic Field:
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