A galvanometer having a resistance of 7 Ω is shunted by a resistance of 3 Ω. If the total current is 1 A, the value of current passing through the shunt is

0.7 A

The correct answer is Option (2) → 0.7 A

Let current through galvanometer = $I_g$

Resistance of galvanometer = $R_g = 7 \, \Omega$

Resistance of shunt = $R_s = 3 \, \Omega$

Total current = $I = 1 \, A$

Since voltage across galvanometer and shunt is same,

$I_g R_g = I_s R_s$

$\Rightarrow I_s = \frac{R_g}{R_s} I_g = \frac{7}{3} I_g$

Total current, $I = I_g + I_s = I_g + \frac{7}{3} I_g = \frac{10}{3} I_g$

$\Rightarrow I_g = \frac{3}{10} = 0.3 \, A$

$\Rightarrow I_s = \frac{7}{3} \times 0.3 = 0.7 \, A$

∴ Current through shunt = 0.7 A