University Physics II · topic UPII-11.07
LC oscillations and series RLC resonance
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
LC Oscillations & Series RLC Resonance: predict a finite change
An LC loop oscillates as energy trades between magnetic and electric storage at ω₀ = 1/√(LC). Resistance damps the free oscillation and, when the loop is driven, sets the sharpness Q = ω₀L/R of the current peak at ω₀. Use the supplied topic model Y_out = sqrt(L*0.001/0.000001)/R/pi, where Y_out is FREE RING TO 1/e (cycles). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. R is Total series resistance R, measured in Ω. Keep L (Inductance L) = 10 mH fixed. For this exercise use only 11.4 Ω ≤ R ≤ 36.6 Ω on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at R = 16.944 Ω and 20.472 Ω. Find the signed output change ΔY_out and the average sensitivity ΔY_out/ΔR across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FREE RING TO 1/e against R
Predictions from the supplied LC Oscillations & Series RLC Resonance model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): FREE RING TO 1/e / cycles
Horizontal axis (x): R: Total series resistance R / Ω
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| R: Total series resistance R / Ω | FREE RING TO 1/e / cycles |
|---|---|
| 11.4 | 2.792191984068339 |
| 17.7 | 1.798360938891473 |
| 24 | 1.3262911924324612 |
| 30.299999999999997 | 1.050527677174227 |
| 36.599999999999994 | 0.8696991425786632 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of R separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At R = 16.944 Ω, substitute into Y_out = sqrt(L*0.001/0.000001)/R/pi: Y₀ = 1.8786 cycles.
- At R = 20.472 Ω, the same fixed-parameter model gives Y₁ = 1.55485 cycles.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.323745 cycles; the input change is ΔR = 3.528 Ω.
- Average sensitivity is ΔY_out/ΔR = -0.0917643 cycles per Ω. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 1.8786 cycles, Y₁ = 1.55485 cycles; ΔY_out = -0.323745 cycles; average sensitivity = -0.0917643 cycles per Ω.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
LC Oscillations & Series RLC Resonance: infer an input with uncertainty
An LC loop oscillates as energy trades between magnetic and electric storage at ω₀ = 1/√(LC). Resistance damps the free oscillation and, when the loop is driven, sets the sharpness Q = ω₀L/R of the current peak at ω₀. Use the supplied topic model Y_out = sqrt(L*0.001/0.000001)/R/pi, where Y_out is FREE RING TO 1/e (cycles). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. R is Total series resistance R, measured in Ω. Keep L (Inductance L) = 10 mH fixed. For this exercise use only 11.4 Ω ≤ R ≤ 36.6 Ω on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 1.1889657 cycles, with a bounded reading uncertainty ±0.04789 cycles. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown R by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00252 Ω, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
FREE RING TO 1/e against R
Predictions from the supplied LC Oscillations & Series RLC Resonance model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): FREE RING TO 1/e / cycles
Horizontal axis (x): R: Total series resistance R / Ω
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| R: Total series resistance R / Ω | FREE RING TO 1/e / cycles |
|---|---|
| 11.4 | 2.792191984068339 |
| 17.7 | 1.798360938891473 |
| 24 | 1.3262911924324612 |
| 30.299999999999997 | 1.050527677174227 |
| 36.599999999999994 | 0.8696991425786632 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(R) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(11.4) = 2.79219 cycles and Y_out(36.6) = 0.869699 cycles. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives R_line = 11.4 + (1.18897 − 2.79219)(36.6 − 11.4)/(0.869699 − 2.79219) = 32.4151 Ω. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives R = 26.772 Ω; substituting back gives Y_out = 1.1889657 cycles.
- The interpolation error in the input is |R_line − R| = 5.64306 Ω. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.1410757 cycles and 1.2368557 cycles. Sort the roots even if the response decreases: 25.7354 Ω ≤ R ≤ 27.8956 Ω.
- The lower and upper input deviations are 1.03659 Ω and 1.1236 Ω. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer R ≈ 26.772 Ω; compatible interval [25.7354, 27.8956] Ω. Endpoint interpolation gives 32.4151 Ω.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.