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A moving-coil instrument has internal resistance 0.6 Ω and gives full-scale deflection at 0.1 A. To convert it into an ammeter of 0–15 A range, what shunt resistance should be connected?

Difficulty: Medium

Correct Answer: 0.004 Ω

Explanation:


Introduction / Context:
Moving-coil meters measure small currents. To extend their range, a low-value shunt resistor bypasses most of the current, leaving only the full-scale meter current through the movement. Correct calculation ensures accurate division of current and prevents damage to the meter.


Given Data / Assumptions:

  • Meter resistance R_m = 0.6 Ω.
  • Meter full-scale current I_m = 0.1 A.
  • Desired total range I_total = 15 A → shunt current I_s = 15 − 0.1 = 14.9 A.
  • Meter and shunt are in parallel at the terminals.


Concept / Approach:
At full-scale: the same voltage appears across the meter and the shunt. Hence V_m = I_m * R_m = I_s * R_s. Solve for the shunt resistance R_s = (I_m * R_m) / I_s.


Step-by-Step Solution:

Compute V_m at full-scale: V_m = 0.1 * 0.6 = 0.06 V.Compute shunt resistance: R_s = 0.06 / 14.9 ≈ 0.0040268 Ω.Round to a practical value: R_s ≈ 0.004 Ω.


Verification / Alternative check:

Check current split: I_s = V_m / R_s ≈ 0.06 / 0.004 = 15 A; with 0.1 A through the meter, total ≈ 15.1 A (rounding explains the small difference).


Why Other Options Are Wrong:

0.6 Ω, 0.06 Ω, and 0.1 Ω are far too large; they would route excessive current through the meter.


Common Pitfalls:

Using series instead of parallel formulas; forgetting that the voltage across meter and shunt is equal at the design point.


Final Answer:

0.004 Ω
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