Question

In a circuit with a 3 V source and a 3 Ω resistor, the galvanometer \((R_G = 60\ \Omega)\) is converted into an ammeter using a shunt resistance \(r_s = 0.02\ \Omega\). The measured current is approximately:

पर्याय

• 0.99 A

• 0.048 A

• 0.02 A

• 1.00 A

Answer 1

The galvanometer–shunt parallel combination has \[R_{eq} = \frac{60 \times 0.02}{60 + 0.02} \approx 0.02\ \Omega\] Total resistance \(= 3 + 0.02 = 3.02\ \Omega\), so \[I = \frac{3}{3.02} \approx 0.99\ \text{A}\] — very close to the ideal 1.00 A.