An electrical circuit contains a battery of 4 V, and a resistance of 2 Ω. A galvanometer of resistance 40 Ω shunted by 10 Ω is used to measure current in the circuit. The value of current in the circuit will be

0.4 A

The correct answer is Option (1) → 0.4 A

Given:

Battery emf, $E = 4 \, \text{V}$

External resistance, $R = 2 \, \Omega$

Galvanometer resistance, $G = 40 \, \Omega$

Shunt resistance, $S = 10 \, \Omega$

Effective resistance of galvanometer and shunt in parallel:

$R_g = \frac{G \times S}{G + S} = \frac{40 \times 10}{40 + 10} = \frac{400}{50} = 8 \, \Omega$

Total circuit resistance:

$R_{total} = R + R_g = 2 + 8 = 10 \, \Omega$

Total current in circuit:

$I = \frac{E}{R_{total}} = \frac{4}{10} = 0.4 \, \text{A}$

Final Answer:

$I = 0.4 \, \text{A}$