In a galvanometer, there is a deflection of 10 divisions per mA. The resistance of the galvanometer is 60 Ω. If a shunt of 1 Ω is connected to the galvanometer and there are 100 divisions in all, on the scale of the galvanometer, the maximum current that this galvanometer can read is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 610 mA
Given:
Deflection sensitivity: 10 divisions/mA → Current per division $i_d = 0.1$ mA
Galvanometer resistance: $R_g = 60$ Ω
Shunt resistance: $R_s = 1$ Ω
Total divisions: 100
Maximum galvanometer current: $I_g^\text{max} = i_d \times 100 = 0.1 \times 100 = 10$ mA
Maximum current through galvanometer with shunt: $I_\text{max} = I_g^\text{max} \left(1 + \frac{R_g}{R_s}\right)$
$I_\text{max} = 10 \text{ mA} \left(1 + \frac{60}{1}\right) = 10 \text{ mA} \times 61 = 610$ mA
Answer: $I_\text{max} = 610$ mA