University Physics II · unit UPII-11
Inductance and AC Circuits
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Self-inductance: infer an input with uncertainty
Self-inductance is the flux linkage a circuit's own current produces per unit current, L = NΦ_B/I — a constant fixed by the winding's shape and the material inside it, not by the current. Use the supplied topic model Y_out = 0.00024674*n*n/lc, where Y_out is SELF-INDUCTANCE L (mH). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lc is Winding length ℓ, measured in cm. Keep n (Turns N) = 350; cur (Current I) = 2 A fixed. For this exercise use only 16.85 cm ≤ lc ≤ 23.15 cm 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 = 1.4606703 mH, with a bounded reading uncertainty ±0.023254 mH. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lc by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063 cm, 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.
SELF-INDUCTANCE L against lc
Predictions from the supplied Self-inductance 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): SELF-INDUCTANCE L / mH
Horizontal axis (x): lc: Winding length ℓ / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lc: Winding length ℓ / cm | SELF-INDUCTANCE L / mH |
|---|---|
| 16.85 | 1.793807121661721 |
| 18.425 | 1.64046947082768 |
| 20 | 1.5112825 |
| 21.575 | 1.4009571263035923 |
| 23.15 | 1.3056436285097193 |
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(lc) = 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(16.85) = 1.79381 mH and Y_out(23.15) = 1.30564 mH. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives lc_line = 16.85 + (1.46067 − 1.79381)(23.15 − 16.85)/(1.30564 − 1.79381) = 21.1493 cm. 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 lc = 20.693 cm; substituting back gives Y_out = 1.4606703 mH.
- The interpolation error in the input is |lc_line − lc| = 0.456302 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.4374163 mH and 1.4839243 mH. Sort the roots even if the response decreases: 20.3687 cm ≤ lc ≤ 21.0278 cm.
- The lower and upper input deviations are 0.324272 cm and 0.334764 cm. 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 lc ≈ 20.693 cm; compatible interval [20.3687, 21.0278] cm. Endpoint interpolation gives 21.1493 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Mutual Inductance & Transformers: infer an input with uncertainty
The mutual inductance of two coils is the flux linkage set up in one of them per ampere flowing in the other — a single number, fixed by geometry, that turns a current slope in one circuit into a voltage in the other. Use the supplied topic model Y_out = 0.48*k*f, where Y_out is COIL 2 VOLTAGE |v₂| (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f is Triangle-wave frequency f, measured in Hz. Keep k (Coupling coefficient k) = 0.4; s (Dot sense s: +1 aiding, −1 reversed) = 1 fixed. For this exercise use only 30.8438 Hz ≤ f ≤ 74.1563 Hz 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 = 10.99476 V, with a bounded reading uncertainty ±0.48649 V. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00433125 Hz, 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.
COIL 2 VOLTAGE |v₂| against f
Predictions from the supplied Mutual Inductance & Transformers 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): COIL 2 VOLTAGE |v₂| / V
Horizontal axis (x): f: Triangle-wave frequency f / Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Triangle-wave frequency f / Hz | COIL 2 VOLTAGE |v₂| / V |
|---|---|
| 30.84375 | 5.922 |
| 41.671875 | 8.001 |
| 52.5 | 10.08 |
| 63.328125 | 12.159 |
| 74.15625 | 14.238 |
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(f) = 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(30.8438) = 5.922 V and Y_out(74.1563) = 14.238 V. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives f_line = 30.8438 + (10.9948 − 5.922)(74.1563 − 30.8438)/(14.238 − 5.922) = 57.2644 Hz. 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 f = 57.2644 Hz; substituting back gives Y_out = 10.99476 V.
- The interpolation error in the input is |f_line − f| = 1.42109e-14 Hz. This is within the requested numerical tolerance.
- Invert both limits Y_out = 10.50827 V and 11.48125 V. Sort the roots even if the response decreases: 54.7306 Hz ≤ f ≤ 59.7982 Hz.
- The lower and upper input deviations are 2.5338 Hz and 2.5338 Hz. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer f ≈ 57.2644 Hz; compatible interval [54.7306, 59.7982] Hz. Endpoint interpolation gives 57.2644 Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Energy in Inductors & Magnetic Fields: infer an input with uncertainty
The energy stored in an inductor is the work done against its own induced emf while the current is raised from zero to I, and for constant inductance it comes to ½LI². Use the supplied topic model Y_out = 0.5*ind*cur*cur/1000, where Y_out is STORED ENERGY ½ L I² (J). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. cur is Current I, measured in A. Keep ind (Inductance L) = 80 mH fixed. For this exercise use only 1.53125 A ≤ cur ≤ 5.46875 A 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 = 0.61877889 J, with a bounded reading uncertainty ±0.078748 J. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown cur by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00039375 A, 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.
STORED ENERGY ½ L I² against cur
Predictions from the supplied Energy in Inductors & Magnetic Fields 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): STORED ENERGY ½ L I² / J
Horizontal axis (x): cur: Current I / A
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| cur: Current I / A | STORED ENERGY ½ L I² / J |
|---|---|
| 1.53125 | 0.0937890625 |
| 2.515625 | 0.253134765625 |
| 3.5 | 0.49 |
| 4.484375 | 0.804384765625 |
| 5.46875 | 1.1962890625 |
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(cur) = 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(1.53125) = 0.0937891 J and Y_out(5.46875) = 1.19629 J. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives cur_line = 1.53125 + (0.618779 − 0.0937891)(5.46875 − 1.53125)/(1.19629 − 0.0937891) = 3.40621 A. 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 cur = 3.93312 A; substituting back gives Y_out = 0.61877889 J.
- The interpolation error in the input is |cur_line − cur| = 0.526911 A. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.54003089 J and 0.69752689 J. Sort the roots even if the response decreases: 3.67434 A ≤ cur ≤ 4.1759 A.
- The lower and upper input deviations are 0.258785 A and 0.242779 A. 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 cur ≈ 3.93312 A; compatible interval [3.67434, 4.1759] A. Endpoint interpolation gives 3.40621 A.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
LR Growth & Decay: infer an input with uncertainty
In an LR circuit the current grows or decays exponentially towards the value the resistance alone would allow, with a time constant τ = L/R built from the inductance and the total resistance of the loop. Use the supplied topic model Y_out = 12/r*(1-exp(-t*r/(1000*l))), where Y_out is CURRENT AT THE CURSOR (A). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. t is Cursor time t, measured in ms. Keep l (Inductance L) = 0.4 H; r (Total loop resistance R) = 20 Ω fixed. For this exercise use only 12.75 ms ≤ t ≤ 107.25 ms 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 = 0.5822359 A, with a bounded reading uncertainty ±0.0022426 A. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00945 ms, 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.
CURRENT AT THE CURSOR against t
Predictions from the supplied LR Growth & Decay 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): CURRENT AT THE CURSOR / A
Horizontal axis (x): t: Cursor time t / ms
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Cursor time t / ms | CURRENT AT THE CURSOR / A |
|---|---|
| 12.75 | 0.28283261744040156 |
| 36.375 | 0.5026629541452448 |
| 59.99999999999999 | 0.5701277589792816 |
| 83.62499999999999 | 0.5908323621724514 |
| 107.24999999999999 | 0.597186498887753 |
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(t) = 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(12.75) = 0.282833 A and Y_out(107.25) = 0.597186 A. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 12.75 + (0.582236 − 0.282833)(107.25 − 12.75)/(0.597186 − 0.282833) = 102.756 ms. 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 t = 70.395 ms; substituting back gives Y_out = 0.5822359 A.
- The interpolation error in the input is |t_line − t| = 32.3606 ms. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.5799933 A and 0.5844785 A. Sort the roots even if the response decreases: 68.0172 ms ≤ t ≤ 73.0941 ms.
- The lower and upper input deviations are 2.37775 ms and 2.69907 ms. 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 t ≈ 70.395 ms; compatible interval [68.0172, 73.0941] ms. Endpoint interpolation gives 102.756 ms.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.