Difficulty: Easy
Correct Answer: Incorrect
Explanation:
Introduction / Context:Series R–L–C circuits are used to build simple frequency-selective filters. Whether the network is band-pass or band-stop depends on where the output is measured. This item asks you to judge the claim that taking the output across the series resistor makes the circuit a band-stop (notch) filter.
Given Data / Assumptions:
Concept / Approach:In a series R–L–C, the current peaks at resonance because the net reactance cancels (XL = XC), leaving the loop impedance approximately R. The resistor's voltage is v_R = i * R, so when current peaks at resonance, the resistor's voltage also peaks. A response that peaks at a center frequency and falls off on both sides is the hallmark of a band-pass, not a band-stop. By contrast, taking the output across L or across C in the same series loop yields a deep minimum at resonance (because the reactive element's voltage tends toward zero there), which is a notch or band-stop behavior.
Step-by-Step Solution:
Write impedance: Z = R + j(XL − XC). At resonance: XL = XC ⇒ Z ≈ R ⇒ i = V_in / R (maximum). Output across R: v_out = i * R ≈ V_in at resonance (peak output). Hence the R output is band-pass, not band-stop.Verification / Alternative check:Plotting |v_R/V_in| vs frequency shows a peak near f_0; plotting |v_L/V_in| or |v_C/V_in| shows a dip at f_0. This matches standard lab measurements and textbook Bode plots.
Why Other Options Are Wrong:Correct: contradicts series-R output behavior (band-pass). Applies only with an ideal current source: the statement concerns a voltage source; with a current source the mapping of outputs changes but does not make the R output a notch under a voltage-source assumption. Cannot be determined: with the stated series topology and output node, it is determinable.
Common Pitfalls:Confusing the R output (band-pass) with L or C outputs (notch). Mixing up series vs parallel tuned circuits.
Final Answer:Incorrect
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