Moving-coil conversion to ammeter: A 0.5 Ω, 0.1 A full-scale movement is to be converted into a 0–10 A ammeter. Calculate the required shunt resistance that must be connected in parallel with the coil.

Introduction / Context:
Converting a sensitive moving-coil instrument into a high-range ammeter is a classic instrumentation task. The idea is to bypass most of the load current through a low-value shunt so that the delicate movement always carries only its rated full-scale current. This problem checks shunt sizing using basic parallel-circuit relationships and voltage equality across parallel branches.

Given Data / Assumptions:

• Coil (movement) resistance, Rm = 0.5 Ω.
• Full-scale current of movement, Im = 0.1 A.
• Desired ammeter range: 0–10 A, so maximum total current I = 10 A.
• Shunt is ideal (negligible inductance) and connected in parallel with the movement.

Concept / Approach:

At full-scale, the coil must carry Im and the shunt carries the remaining current Is = I − Im. Since the coil and shunt are in parallel, the voltage drop across each is equal. Therefore, Rs is determined by equating the coil's full-scale voltage to the shunt voltage at Is. This ensures protection of the movement and correct overall range.

Step-by-Step Solution:

Verification / Alternative check:

Check power in the movement at FS: P = Im^2 * Rm = 0.01 * 0.5 = 0.005 W = 5 mW, a reasonable level. The shunt voltage equals 0.05 V; multiplying by Is gives 0.495 W dissipated in the shunt—consistent with a small, low-ohm, higher-wattage resistor selection.

Why Other Options Are Wrong:

• 0.004 Ω / 0.008 Ω: would not maintain the correct coil current at 10 A total; coil would be under/over-driven.
• 0.05 Ω / 0.1 Ω: far too large; the coil would be forced to carry more than 0.1 A before reaching 10 A total.

Common Pitfalls:

• Forgetting that the coil's voltage at full scale sets the shunt voltage.
• Using series-resistor formulas instead of parallel-voltage equality.

Final Answer:

0.005 Ω