In the circuit, shown the galvanometer $G$ of resistance $60\, \Omega$ is shunted by a resistance $r=0.02\, \Omega$. The current through $R$ (in $ohm$) is nearly $1\, A$. The value of resistance $R$ (in $ohm$) is nearly (in $\Omega$)
AIIMS 2018, Medium
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Here, resistance of the galvanometer
$R_{G}=60 \Omega$
When the galvanometer is shunted by a resistance $r$, its effective resistance
$R_{P}=\frac{R_{G} r}{R_{G}+r}=\frac{60 \times 0.02}{60+0.02} \approx 0.02 \Omega$
Total resistance of the circuit $=R+R_{P}=R+0.02$
Current, $I=\frac{5}{R+0.02} \Rightarrow 1=\frac{5}{R+0.02}$
$R+0.02=5 \Rightarrow R=5-0.02=4.98 \approx 5 \Omega$
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