University Physics II · unit UPII-07
Kirchhoff's Rules and RC 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.
Kirchhoff's junction rule: infer an input with uncertainty
Kirchhoff's junction rule says that wherever conductors meet, the currents arriving at that point add up to the currents leaving it, because no charge accumulates there. Use the supplied topic model Y_out = i1+i2-i3, where Y_out is I₄ BY THE RULE (A). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. i1 is Current I₁ entering from the left, measured in A. Keep i2 (Current I₂ entering from above) = 0.8 A; i3 (Current I₃ leaving downwards) = 1.4 A 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 = 1.24255 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.
I₄ BY THE RULE against i1
Predictions from the supplied Kirchhoff's junction rule 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): I₄ BY THE RULE / A
Horizontal axis (x): i1: Current I₁ entering from the left / A
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| i1: Current I₁ entering from the left / A | I₄ BY THE RULE / A |
|---|---|
| 0.4975 | -0.10249999999999981 |
| 1.0487499999999998 | 0.44875 |
| 1.5999999999999999 | 1 |
| 2.1512499999999997 | 1.55125 |
| 2.7024999999999997 | 2.1024999999999996 |
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) = -0.1025 A and Y_out(2.7025) = 2.1025 A. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives i1_line = 0.4975 + (1.24255 − -0.1025)(2.7025 − 0.4975)/(2.1025 − -0.1025) = 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 = 1.24255 A.
- The interpolation error in the input is |i1_line − i1| = 4.44089e-16 A. This is within the requested numerical tolerance.
- Invert both limits Y_out = 1.11356 A and 1.37154 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.
Kirchhoff's Loop Rule: infer an input with uncertainty
Kirchhoff's loop rule states that the algebraic sum of the potential changes around any closed path in a circuit is zero, provided no changing magnetic flux passes through that path. Use the supplied topic model Y_out = (e1-e2)/30, where Y_out is LOOP CURRENT I (A). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. e1 is Driving emf ε₁, measured in V. Keep e2 (Opposing emf ε₂) = 6 V; r1 (Resistance R₁ (R₂ = 30 Ω − R₁)) = 10 Ω fixed. For this exercise use only 2.55 V ≤ e1 ≤ 21.45 V 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 = 0.2693 A, with a bounded reading uncertainty ±0.036855 A. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown e1 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00189 V, 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.
LOOP CURRENT I against e1
Predictions from the supplied Kirchhoff's Loop Rule 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): LOOP CURRENT I / A
Horizontal axis (x): e1: Driving emf ε₁ / V
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| e1: Driving emf ε₁ / V | LOOP CURRENT I / A |
|---|---|
| 2.55 | -0.115 |
| 7.2749999999999995 | 0.04249999999999998 |
| 12 | 0.2 |
| 16.724999999999998 | 0.35749999999999993 |
| 21.45 | 0.515 |
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(e1) = 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(2.55) = -0.115 A and Y_out(21.45) = 0.515 A. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives e1_line = 2.55 + (0.2693 − -0.115)(21.45 − 2.55)/(0.515 − -0.115) = 14.079 V. 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 e1 = 14.079 V; substituting back gives Y_out = 0.2693 A.
- The interpolation error in the input is |e1_line − e1| = 0 V. This is within the requested numerical tolerance.
- Invert both limits Y_out = 0.232445 A and 0.306155 A. Sort the roots even if the response decreases: 12.9733 V ≤ e1 ≤ 15.1846 V.
- The lower and upper input deviations are 1.10565 V and 1.10565 V. 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 e1 ≈ 14.079 V; compatible interval [12.9733, 15.1846] V. Endpoint interpolation gives 14.079 V.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Multi-loop circuit strategy: infer an input with uncertainty
Multi-loop analysis turns a network that no series-and-parallel reduction can simplify into a single simultaneous-equation system: one unknown current per branch, and one independent junction or loop equation for each of them. Use the supplied topic model Y_out = e2*(e2*(4+r3)-e1*r3)/(16+8*r3), where Y_out is POWER FROM CELL 2 (W). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. e2 is Second cell emf ε₂, measured in V. Keep e1 (First cell emf ε₁) = 12 V; r3 (Shared resistance R₃) = 6 Ω fixed. For this exercise use only 3.795 V ≤ e2 ≤ 10.725 V 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 = 1.0307402 W, with a bounded reading uncertainty ±0.45747 W. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown e2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000693 V, 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.
POWER FROM CELL 2 against e2
Predictions from the supplied Multi-loop circuit strategy 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): POWER FROM CELL 2 / W
Horizontal axis (x): e2: Second cell emf ε₂ / V
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| e2: Second cell emf ε₂ / V | POWER FROM CELL 2 / W |
|---|---|
| 3.795 | -2.0190585937499996 |
| 5.5275 | -1.4444912109375 |
| 7.26 | 0.06806249999999936 |
| 8.9925 | 2.5186025390624995 |
| 10.725 | 5.90712890625 |
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(e2) = 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(3.795) = -2.01906 W and Y_out(10.725) = 5.90713 W. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives e2_line = 3.795 + (1.03074 − -2.01906)(10.725 − 3.795)/(5.90713 − -2.01906) = 6.46149 V. 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 e2 = 8.0223 V; substituting back gives Y_out = 1.0307402 W.
- The interpolation error in the input is |e2_line − e2| = 1.56081 V. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.5732702 W and 1.4882102 W. Sort the roots even if the response decreases: 7.67786 V ≤ e2 ≤ 8.34179 V.
- The lower and upper input deviations are 0.344442 V and 0.319487 V. 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 e2 ≈ 8.0223 V; compatible interval [7.67786, 8.34179] V. Endpoint interpolation gives 6.46149 V.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Potential Dividers & Measuring Instruments: infer an input with uncertainty
A potential divider is two resistors in series across a supply with the output taken across one of them, so the output is the supply voltage multiplied by that resistor's share of the total resistance. Use the supplied topic model Y_out = 12*f*rl/max(rl+f*(1-f)*rt,0.01), where Y_out is LOADED OUTPUT (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. rl is Load resistance R_L, measured in kΩ. Keep rt (Track resistance R_T) = 10 kΩ; f (Wiper fraction f) = 0.5 fixed. For this exercise use only 6.20625 kΩ ≤ rl ≤ 44.7937 kΩ 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 = 5.5348062 V, with a bounded reading uncertainty ±0.022204 V. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rl by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00385875 kΩ, 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.
LOADED OUTPUT against rl
Predictions from the supplied Potential Dividers & Measuring Instruments 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): LOADED OUTPUT / V
Horizontal axis (x): rl: Load resistance R_L / kΩ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rl: Load resistance R_L / kΩ | LOADED OUTPUT / V |
|---|---|
| 6.20625 | 4.277099784637472 |
| 15.853124999999999 | 5.182700493785118 |
| 25.5 | 5.464285714285714 |
| 35.146874999999994 | 5.6015605544948945 |
| 44.793749999999996 | 5.682833355358795 |
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(rl) = 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(6.20625) = 4.2771 V and Y_out(44.7937) = 5.68283 V. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives rl_line = 6.20625 + (5.53481 − 4.2771)(44.7937 − 6.20625)/(5.68283 − 4.2771) = 40.7304 kΩ. 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 rl = 29.7446 kΩ; substituting back gives Y_out = 5.5348062 V.
- The interpolation error in the input is |rl_line − rl| = 10.9858 kΩ. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 5.5126022 V and 5.5570102 V. Sort the roots even if the response decreases: 28.2757 kΩ ≤ rl ≤ 31.3608 kΩ.
- The lower and upper input deviations are 1.46894 kΩ and 1.6162 kΩ. 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 rl ≈ 29.7446 kΩ; compatible interval [28.2757, 31.3608] kΩ. Endpoint interpolation gives 40.7304 kΩ.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.