At resonance in a simple RLC system, the bandwidth is defined between the half-power points. The corresponding current at these corner frequencies is what percentage of the maximum current?

Introduction:
Bandwidth in resonant circuits is commonly specified by the half-power (3 dB) points. Recognizing the current or voltage ratios at these points is crucial for filter design, tuning, and quality factor calculations.

Given Data / Assumptions:

• Linear, time-invariant RLC circuit.
• Half-power definition used for bandwidth.
• Sinusoidal steady-state operation.

Concept / Approach:
Half-power means the power is one-half of the peak value. Since power in a resistive element is proportional to the square of current (or voltage), halving power implies the amplitude of current (or voltage) equals 1/sqrt(2) of the maximum, which is 0.707 or 70.7%.

Step-by-Step Solution:
1) Half-power condition: P = 0.5 * Pmax.2) Because P ∝ I^2, we have (I/Imax)^2 = 0.5.3) Take square root: I/Imax = 1 / sqrt(2) ≈ 0.707.4) Express as a percentage: 0.707 * 100% ≈ 70.7%.

Verification / Alternative check:
The 3 dB rule: a drop of 3 dB corresponds to a power ratio of 0.5 and an amplitude ratio of 0.707, confirming the same percentage for either current or voltage.

Why Other Options Are Wrong:
50: This would be amplitude halved, but half power corresponds to amplitude reduced to 70.7%.

62.3 and 84.1: Do not correspond to standard half-power amplitude ratios.

95.3: Too close to the peak; not representative of half-power points.

Common Pitfalls:
Confusing power ratio with amplitude ratio, or assuming a 50% power reduction means 50% amplitude reduction. Always convert through the square relationship.

Final Answer:
70.7