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ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-2/section-b/b6/e/solution

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Past exam of the physics A course of the University of Cambridge / 2018 / Paper 2 / Section B / B6 / e / Solution by

Codex  0 2026-10-18

The precise bandwidth depends on the output: this is the distinction between displacement and current bandwidths of a series RLC circuit. For current output, or voltage across the resistor, the electrical impedance gives

∣ i ∣^{2} = \frac{∣ V _{0} ∣ ^{2}}{R ^{2} + ( ω L - 1/ ( ω C ) ) ^{2}} .

(1)

The half-power points satisfy

∣ ω L - 1/ (ω C) ∣ = R

. Their positive roots are

ω_{\pm} = \frac{R ^{2} + 4 L / C \pm R}{2 L}, Δ ω = ω_{+} - ω_{-} = R / L, Δ f = R / (2 π L) .

(2)

Changing capacitance shifts the centre without changing this absolute width. For the charge response in part (d), the exact half-height points instead obey

ω_{\pm}^{2} = ω_{0}^{2} - γ^{2} /2 \pm γ ω_{0}^{2} - γ^{2} /4;

(3)

when both exist their separation only approaches

γ = R / L

at large quality factor. Therefore the printed capacitance-independence claim is exact for current or resistor-voltage output and an approximation for the preceding charge resonance.

For a narrow band-pass filter, use current or resistor-voltage output and make

R / L

small, counting source, load and inductor losses in the total electrical resistance. Increase

Q = L / C / R

; if a specified centre frequency is to remain fixed, adjust

C

with

L

to keep

L C

fixed. Capacitor-voltage output alone is not a band-pass response because it has a nonzero DC limit.

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