University Physics II · unit UPII-06
Current, Resistance, and DC 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.
Electric Current & Charge Flow: infer an input with uncertainty
The current through a surface is the net charge crossing it per unit time, counted positive in whichever direction has been chosen as positive: I = ΔQ/Δt for a steady flow, and I = dq/dt when the rate is changing. Use the supplied topic model Y_out = i1-i2, where Y_out is BRANCH 3 CURRENT I₃ (A). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. i1 is Branch 1 current I₁ (into node), measured in A. Keep i2 (Branch 2 current I₂ (out of node)) = 0.7 A; dt (Counting window Δt) = 20 s fixed. For this exercise use only 0.4975 A ≤ i1 ≤ 2.7025 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 = 1.14255 A, with a bounded reading uncertainty ±0.12899 A. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown i1 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0002205 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.
BRANCH 3 CURRENT I₃ against i1
Predictions from the supplied Electric Current & Charge Flow 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): BRANCH 3 CURRENT I₃ / A
Horizontal axis (x): i1: Branch 1 current I₁ (into node) / A
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| i1: Branch 1 current I₁ (into node) / A | BRANCH 3 CURRENT I₃ / A |
|---|---|
| 0.4975 | -0.20249999999999996 |
| 1.0487499999999998 | 0.3487499999999999 |
| 1.5999999999999999 | 0.8999999999999999 |
| 2.1512499999999997 | 1.4512499999999997 |
| 2.7024999999999997 | 2.0024999999999995 |
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(i1) = 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(0.4975) = -0.2025 A and Y_out(2.7025) = 2.0025 A. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives i1_line = 0.4975 + (1.14255 − -0.2025)(2.7025 − 0.4975)/(2.0025 − -0.2025) = 1.84255 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 i1 = 1.84255 A; substituting back gives Y_out = 1.14255 A.
- The interpolation error in the input is |i1_line − i1| = 2.22045e-16 A. This is within the requested numerical tolerance.
- Invert both limits Y_out = 1.01356 A and 1.27154 A. Sort the roots even if the response decreases: 1.71356 A ≤ i1 ≤ 1.97154 A.
- The lower and upper input deviations are 0.12899 A and 0.12899 A. 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 i1 ≈ 1.84255 A; compatible interval [1.71356, 1.97154] A. Endpoint interpolation gives 1.84255 A.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Current with calculus: infer an input with uncertainty
The instantaneous current through a chosen surface is the time derivative of the net charge that has crossed it, i = dq/dt, with the sign fixed by which direction through the surface you called positive. Use the supplied topic model Y_out = i0*exp(-tc/tau), where Y_out is CURRENT i(t) (mA). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. tc is Cursor time t, measured in ms. Keep i0 (Initial current I₀) = 3 mA; tau (Time constant τ) = 2 ms fixed. For this exercise use only 1.50937 ms ≤ tc ≤ 8.99062 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 = 0.14401232 mA, with a bounded reading uncertainty ±0.016579 mA. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown tc by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000748125 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 i(t) against tc
Predictions from the supplied Current with calculus 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 i(t) / mA
Horizontal axis (x): tc: Cursor time t / ms
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tc: Cursor time t / ms | CURRENT i(t) / mA |
|---|---|
| 1.509375 | 1.4104725479993074 |
| 3.3796874999999997 | 0.5536450722634982 |
| 5.25 | 0.21731927110275437 |
| 7.120312499999999 | 0.08530314448487542 |
| 8.990625 | 0.033483576592556026 |
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(tc) = 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(1.50937) = 1.41047 mA and Y_out(8.99062) = 0.0334836 mA. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives tc_line = 1.50937 + (0.144012 − 1.41047)(8.99062 − 1.50937)/(0.0334836 − 1.41047) = 8.39012 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 tc = 6.07294 ms; substituting back gives Y_out = 0.14401232 mA.
- The interpolation error in the input is |tc_line − tc| = 2.31718 ms. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.12743332 mA and 0.16059132 mA. Sort the roots even if the response decreases: 5.85501 ms ≤ tc ≤ 6.31755 ms.
- The lower and upper input deviations are 0.217928 ms and 0.244611 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 tc ≈ 6.07294 ms; compatible interval [5.85501, 6.31755] ms. Endpoint interpolation gives 8.39012 ms.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Current density and drift speed: infer an input with uncertainty
Current density J = n q v(d) is current per unit cross-sectional area, a vector along the conventional current. Drift velocity v(d) is the slow average velocity the field adds to the carriers' fast random motion. Use the supplied topic model Y_out = i/max(a2,0.05), where Y_out is J, NARROW SECTION (MA m⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. a2 is Narrow cross-section A₂, measured in mm². Keep i (Current I) = 3 A fixed. For this exercise use only 0.765625 mm² ≤ a2 ≤ 2.73438 mm² 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 = 1.5255045 MA m⁻², with a bounded reading uncertainty ±0.064254 MA m⁻². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000196875 mm², 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.
J, NARROW SECTION against a2
Predictions from the supplied Current density and drift speed 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): J, NARROW SECTION / MA m⁻²
Horizontal axis (x): a2: Narrow cross-section A₂ / mm²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a2: Narrow cross-section A₂ / mm² | J, NARROW SECTION / MA m⁻² |
|---|---|
| 0.765625 | 3.9183673469387754 |
| 1.2578125 | 2.3850931677018634 |
| 1.75 | 1.7142857142857142 |
| 2.2421875 | 1.3379790940766552 |
| 2.734375 | 1.0971428571428572 |
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(a2) = 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(0.765625) = 3.91837 MA m⁻² and Y_out(2.73438) = 1.09714 MA m⁻². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a2_line = 0.765625 + (1.5255 − 3.91837)(2.73438 − 0.765625)/(1.09714 − 3.91837) = 2.43545 mm². 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 a2 = 1.96656 mm²; substituting back gives Y_out = 1.5255045 MA m⁻².
- The interpolation error in the input is |a2_line − a2| = 0.468887 mm². This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.4612505 MA m⁻² and 1.5897585 MA m⁻². Sort the roots even if the response decreases: 1.88708 mm² ≤ a2 ≤ 2.05304 mm².
- The lower and upper input deviations are 0.0794835 mm² and 0.0864735 mm². 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 a2 ≈ 1.96656 mm²; compatible interval [1.88708, 2.05304] mm². Endpoint interpolation gives 2.43545 mm².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Resistance and Ohm's law: infer an input with uncertainty
The resistance of a device is the potential difference across it divided by the current through it, R = V/I, measured in ohms; the device is ohmic when that ratio stays the same as V and I change. Use the supplied topic model Y_out = v/(0.4*max(n,0.05)*(v/6)^n), where Y_out is SLOPE r_d = dV/dI (Ω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. n is Exponent n in I ∝ Vⁿ, dimensionless. Keep v (Operating voltage V) = 4 V fixed. For this exercise use only 0.57 ≤ n ≤ 1.83 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 = 12.854789 Ω, with a bounded reading uncertainty ±0.2068 Ω. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown n by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000126, 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.
SLOPE r_d = dV/dI against n
Predictions from the supplied Resistance and Ohm's law 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): SLOPE r_d = dV/dI / Ω
Horizontal axis (x): n: Exponent n in I ∝ Vⁿ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Exponent n in I ∝ Vⁿ | SLOPE r_d = dV/dI / Ω |
|---|---|
| 0.5700000000000001 | 22.105338109415076 |
| 0.885 | 16.17698167626459 |
| 1.2000000000000002 | 13.555897139971233 |
| 1.5150000000000001 | 12.20016276222964 |
| 1.83 | 11.476142026242911 |
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(n) = 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(0.57) = 22.1053 Ω and Y_out(1.83) = 11.4761 Ω. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives n_line = 0.57 + (12.8548 − 22.1053)(1.83 − 0.57)/(11.4761 − 22.1053) = 1.66657. 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 n = 1.3386; substituting back gives Y_out = 12.854789 Ω.
- The interpolation error in the input is |n_line − n| = 0.327973. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 12.647989 Ω and 13.061589 Ω. Sort the roots even if the response decreases: 1.29357 ≤ n ≤ 1.38803.
- The lower and upper input deviations are 0.0450273 and 0.0494273. 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 n ≈ 1.3386; compatible interval [1.29357, 1.38803]. Endpoint interpolation gives 1.66657.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.