In an LCR circuit, at half-power frequencies (f 1 and f 2 ), the current is I max /√2. The bandwidth is:

f 2 - f 1 = R/(2πL) = R/ω_0/Q

f 2 - f 1 = f 0 during normal conditions

f 2 - f 1 = 1/(RC) as generally observed

f 2 - f 1 = 1/Q in typical laboratory settings

⚛️ Physics · Alternating Current · NEET & JEE

In an LCR circuit, at half-power frequencies (f₁ and f₂), the current is I_max/√2. The bandwidth is:

Answer: f₂ - f₁ = R/(2πL) = R/ω_0/Q.

A f₂ - f₁ = R/(2πL) = R/ω_0/Q
B f₂ - f₁ = f₀ during normal conditions
C f₂ - f₁ = 1/(RC) as generally observed
D f₂ - f₁ = 1/Q in typical laboratory settings

Correct answer: A. f₂ - f₁ = R/(2πL) = R/ω_0/Q

Explanation: At half-power points, I = I_max/√2. Bandwidth Δf = f₂ - f₁ = R/(2πL) = f₀/Q. These are the -3 dB frequencies on either side of resonance.

In a pure inductor the current lags the voltage by 90° (energy is briefly stored in the magnetic field before current responds); in a pure capacitor the current leads the voltage by 90° (current must flow to build up charge before voltage rises) - the mnemonic "ELI the ICE man" keeps these straight.

Concept context

AC circuits, impedance, LCR series circuit, resonance, transformers, and power factor.

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In an LCR circuit, at half-power frequencies (f1 and f2), the current is Imax/√2. The bandwidth is:

Concept context

More Alternating Current questions

In an LCR circuit, at half-power frequencies (f₁ and f₂), the current is I_max/√2. The bandwidth is: