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
Answer & explanation
Correct answer: option 1
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:
$I = 0.4 \, \text{A}$