Diagnosing circuit behavior from data

Diagnosing a circuit from data means comparing measured voltages and currents with the values predicted by Ohm's law and Kirchhoff's rules. For each element compute the implied resistance $R = U/I$ and compare it with the nominal value: a large deviation signals a fault. In a series branch an open element carries no current and takes the whole source voltage, while a shorted element carries current with almost zero voltage across it. In a parallel branch a short lowers the node voltage and raises the total current, whereas an open branch removes only its share of the current and leaves the node voltage near nominal. Because the data come from a simulation, faults (open, short, resistance drift) can be created without risk of overheating, fire, or damaged components. The workflow is: record data, compute implied resistance, classify each element, and state the single fault that explains all readings.

In a simulated series circuit the source provides $U = 9.0\ \text{V}$, with nominal resistors $R_1 = 100\ \Omega$ and $R_2 = 200\ \Omega$. Measurements: $I = 30.0\ \text{mA}$, $U_1 = 3.0\ \text{V}$, $U_2 = 6.0\ \text{V}$. Is the circuit healthy?

The same model with $U = 9.0\ \text{V}$, $R_1 = 100\ \Omega$, $R_2 = 200\ \Omega$, but now the measurements are: $I = 90.0\ \text{mA}$, $U_1 = 9.0\ \text{V}$, $U_2 = 0.0\ \text{V}$. Identify the fault and explain why the data are incompatible with an open circuit.

A parallel model: $U = 6.0\ \text{V}$, nominal values $R_1 = 300\ \Omega$, $R_2 = 150\ \Omega$. Measurements: total current $I = 20.0\ \text{mA}$, branch currents $I_1 = 20.0\ \text{mA}$, $I_2 = 0.0\ \text{mA}$. Find the fault and predict what the instruments would read if instead $R_2$ were shorted.

A safe self-study plan in a simulator (for example Falstad Circuit Simulator or PhET): build a series circuit with a source $U = 9.0\ \text{V}$ and two resistors $R_1 = 100\ \Omega$, $R_2 = 200\ \Omega$. Measure the circuit current and the voltage across each resistor in three states: normal, $R_2$ open (broken branch), and $R_2$ shorted (zero-resistance jumper). Fill in a table "state — $I$ — $U_1$ — $U_2$ — $R_1^{\text{implied}}$ — $R_2^{\text{implied}}$ — diagnosis", compute the implied resistances with $R = U/I$, and check that each row is explained by one fault. Work only in the simulator: a real short can overheat the source and wiring, so direct measurement is unsafe here.

Show how a measured pair (element voltage and branch current) corresponds to a specific fault type, so the diagnosis can be formed before calculation.

A sketch of a series circuit with a source and two resistors, alongside a signature table: "normal" ($U_1 = 3.0\ \text{V}$, $U_2 = 6.0\ \text{V}$, $I = 30.0\ \text{mA}$), "$R_2$ open" ($U_1 = 0.0\ \text{V}$, $U_2 = 9.0\ \text{V}$, $I = 0.0\ \text{mA}$), "$R_2$ shorted" ($U_1 = 9.0\ \text{V}$, $U_2 = 0.0\ \text{V}$, $I = 90.0\ \text{mA}$), with hint arrows pointing to the corresponding elements of the diagram.

A power bank gives $5.10\ \text{V}$ at its socket with nothing connected. With a phone drawing $1.0\ \text{A}$, the voltage at the phone's socket drops to $4.55\ \text{V}$; with a short thick cable at the same current it is $4.95\ \text{V}$. The difference is $0.55\ \text{V}$ against $0.15\ \text{V}$.

This is recovering a parameter from data: two pairs $(I, U)$ define the straight line $U = U_0 - I\,R$, and its slope gives the resistance of the cable together with its connectors — $0.55\ \Omega$ for the thin one and $0.15\ \Omega$ for the thick one. The verdict "the cable is at fault" comes from comparing the measured line with the ideal-source model $U = U_0$.

The battery is $6.0\ \text{V}$, the lamp is marked $0.60\ \text{W}$, so the model predicts a current $I = P/U = 0.10\ \text{A}$. A voltmeter across the lamp itself reads $2.4\ \text{V}$ and an ammeter reads $0.040\ \text{A}$; the switch with corroded contacts drops $3.6\ \text{V}$ at the same current.

The data confirm the lamp itself is fine: $R = U/I = 2.4\ \text{V}/0.040\ \text{A} = 60\ \Omega$, which matches the rated value $U^2/P = 60\ \Omega$. The deviation from the model — the missing $3.6\ \text{V}$ — is assigned to another element of the circuit, and its resistance is found from the same data: $3.6\ \text{V}/0.040\ \text{A} = 90\ \Omega$.

The strip is rated $12\ \text{V}$. A $2\ \text{m}$ section draws $0.80\ \text{A}$, two sections draw $1.60\ \text{A}$ at $12.0\ \text{V}$. With a third section the linear model gives $2.40\ \text{A}$, but the measurement gives $1.90\ \text{A}$ while the supply voltage falls to $10.5\ \text{V}$.

The model $I = U/R$ with constant $R$ predicts $2.40\ \text{A}$, and the disagreement with the measurement shows the supply has hit its current limit and no longer holds $12\ \text{V}$. Diagnosing from data means locating the point where the model stops working and naming the element that breaks it.