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A galvanometer having a resistance of $4 \Omega$ is shunted by a wire of resistance $2 \Omega$. If the total current is $1.5 \mathrm{~A}$, the current passing through shunt is
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The correct answer is:
$1 \mathrm{~A}$
From figure,
Shunt current $=i-i_g$
galvanometer current $=i_{\mathrm{g}}$.
$i=$ total current
$\therefore \quad \quad i=1.5 \mathrm{~A}$
By KVL,
Putting $G=4 \Omega$ and $S=2 \Omega$ and $i=1.5 \mathrm{~A}$
$\therefore \quad i_g \times 4=\left(1.5-i_g\right) \times 2$
$4 i_g=3-2 i_g$
$6 i_g=3 \Rightarrow i_g=0.5 \mathrm{~A}$
$\therefore$ Current passing through shunt $=i-i_g$
$=1.5-0.5=1 \mathrm{~A}$
Shunt current $=i-i_g$
galvanometer current $=i_{\mathrm{g}}$.
$i=$ total current
$\therefore \quad \quad i=1.5 \mathrm{~A}$
By KVL,
Putting $G=4 \Omega$ and $S=2 \Omega$ and $i=1.5 \mathrm{~A}$
$\therefore \quad i_g \times 4=\left(1.5-i_g\right) \times 2$
$4 i_g=3-2 i_g$
$6 i_g=3 \Rightarrow i_g=0.5 \mathrm{~A}$
$\therefore$ Current passing through shunt $=i-i_g$
$=1.5-0.5=1 \mathrm{~A}$
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