Start Learning Free

Before You Begin

Question #43336

Difficulty: 8/10 | 🟢 Free

The Question

A student constructs an ideal solenoid of length l=50.0 mml = 50.0 \text{ mm} and radius r=10.0 mmr = 10.0 \text{ mm} with N=1000N = 1000 turns. The measurements have maximum uncertainties Δl=0.5 mm\Delta l = 0.5 \text{ mm} and Δr=0.1 mm\Delta r = 0.1 \text{ mm}. The solenoid is connected in series with an exact resistor R=10 ΩR = 10 \ \Omega and an ideal AC voltage source V(t)=V0sin⁡(ωt)V(t) = V_0 \sin(\omega t), where V0V_0 and ω=2500 rad/s\omega = 2500 \text{ rad/s} are exact. Assume π2=10\pi^2 = 10 and μ0=4π×10−7 T⋅m/A\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} are exact. The student calculates the wattless component of the RMS current (IwI_w) in the circuit. What is the maximum percentage error in the calculated value of IwI_w?

Options

  • A.

    1.8%1.8\%

  • B.

    3.0%3.0\%

  • C.

    2.4%2.4\%

  • D.

    1.2%1.2\%

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Wattless Current|Relative Error Propagation|Self-Inductance|Small Angle Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the propagation of errors in AC circuits, specifically targeting the trap of treating dependent variables as independent.

Students often calculate the wattless current as Iw=Irmssin⁡ϕI_w = I_{rms} \sin\phi.

Finding the relative error in IrmsI_{rms} (2.4%2.4\%) and sin⁡ϕ\sin\phi (0.6%0.6\%) separately, they mistakenly add them to get 3.0%3.0\%.

This linear error propagation fails because both IrmsI_{rms} and sin⁡ϕ\sin\phi depend on the same physical quantity (inductance LL).

An increase in LL decreases IrmsI_{rms} but increases sin⁡ϕ\sin\phi, meaning their errors are anti-correlated and partially cancel.

The correct approach requires expressing IwI_w entirely in terms of LL before applying calculus, revealing a sensitivity factor cos⁡(2ϕ)\cos(2\phi) that scales the relative error.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAlternating Current - Class 12
ConceptWattless Current
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree