电路分析 I（直流）

This is the first course in University Electrical Engineering: direct-current circuit analysis 直流电路分析. A DC circuit is one where currents and voltages settle to steady values — batteries, sensors, logic supplies, and the bias networks inside every amplifier.

The course is a ladder, and every rung is used by the next:

• Units 1.1 and 1.2 need only algebra: quantities, sign conventions, Ohm's law and Kirchhoff's laws.
• Units 1.3 to 1.5 build systematic methods — node and mesh analysis, network theorems, and the ideal op-amp — that turn long circuits into short calculations.
• Units 1.6 and 1.7 add the energy-storing elements and the first transient response 暂态响应, which is where calculus begins to matter.

How to read each unit: a short story says why it matters; a picture or a rule of thumb comes before any formula; then the precise statement; then worked examples that get harder, one step at a time, each with a check at the end; then the mistakes people actually make; then a three-line summary. Every number in the examples is something you can verify a second way — a different loop, an unused node, a limiting case — and you should build that habit here: in later courses the check is the difference between a caught error and a burnt resistor.

A phone charger labelled "20 W" and a kettle labelled "2000 W" both move charge. Why is one a thousand times more powerful? To answer, you need the six quantities every circuit question is asked in — and one convention that keeps their signs honest.

Learn them as a chain, not a list: charge moves; its rate is current; current through a difference in potential transfers energy; the rate of that transfer is power.

• Charge 电荷 $q$ is measured in coulombs 库仑 (C). One coulomb is about $6.24 \times 10^{18}$ elementary charges. Charge is conserved 守恒: it is never created or consumed in a circuit, only moved.
• Current 电流 $i$ is the rate of charge flow, in amperes 安培 (A): $i = dq/dt$, so 1 A = 1 C/s.
• Voltage 电压 $v$ is energy per unit charge, in volts 伏特 (V): 1 V = 1 J/C. It is always measured between two points; "5 V at a node" means 5 V between that node and the reference (ground) node.
• Power 功率 $p$ is energy per unit time, in watts 瓦特 (W): $p = vi$, and 1 W = 1 J/s.
• Energy 能量 $w$ is power integrated over time, $w = \int p \, dt$, in joules 焦耳 (J). Electricity meters sell the kilowatt-hour 千瓦时: $1\ \text{kWh} = 3.6\ \text{MJ}$.

Deriving the power formula. Move a charge $dq$ through a potential difference $v$ and the energy transferred is $dw = v \cdot dq$. Divide by $dt$: $p = dw/dt = v \cdot (dq/dt) = vi$. Power is voltage times current — nothing deeper.

The passive sign convention 无源符号约定 removes all ambiguity. Pick a direction for the current through an element and a polarity for its voltage; when the current arrow enters the $+$ terminal, $p = vi$ is the power the element absorbs. A positive result means the element consumes power; a negative result means it supplies power. Getting this backwards is the most common source of wrong signs in the whole course.

Worked example 1 (power and energy). A 12 V battery drives a lamp that draws 0.25 A, with the current arrow drawn into the lamp's $+$ terminal. For the lamp, $p = vi = 12 \times 0.25 = 3 \text{ W}$ absorbed. Over 2 minutes the lamp takes $w = pt = 3 \times 120 = 360 \text{ J}$. With the same arrow drawn for the battery, the battery's current enters its $-$ terminal, so its absorbed power is $-3$ W: it supplies exactly what the lamp takes, and the bookkeeping closes.

Worked example 2 (the meter reads the wrong number). A 12 V ideal source feeds two 10 M$\Omega$ resistors in series, so each should have 6 V across it. A real voltmeter with 10 M$\Omega$ input resistance is connected across the lower one to "check". The meter is now a second 10 M$\Omega$ resistor in parallel: the lower part combines to 5 M$\Omega$, and the meter reads $\frac{5}{10+5} \times 12 = 4 \text{ V}$ instead of 6 V. The check itself changed the circuit. (Across the ideal source itself the meter would read 12 V correctly: an ideal source holds its voltage whatever is connected.) Rule of thumb: the meter's resistance must be at least a hundred times the resistance it measures. Escalate further: an ammeter 电流表 in current mode is the mirror case — a series meter with non-negligible resistance steals voltage and reads low current.

Common mistakes.

• Reporting "the voltage at point A" without saying what the other point is. Voltage is always between two points; name them.
• Mixing units: minutes with watts, mA with k$\Omega$. Convert first; then the answer's unit falls out by itself ($\text{V}/\Omega = \text{A}$).
• Trusting a meter without asking what resistance it adds.

Summary: six quantities, one chain (charge to current to energy to power); power is $vi$ under the passive sign convention; meters must be nearly invisible to the circuit (series ammeter = near short, parallel voltmeter = near open).

Two resistors arrive at a node and one leaves. How much current takes each exit? The answer needs the two laws named after Kirchhoff — and they are not new physics. One is "charge does not pile up"; the other is "energy books balance". Both were hiding in unit 1.1.

Ohm's law 欧姆定律 links a resistor 电阻: $v = Ri$, with $R$ in ohms 欧姆 ($\Omega$). Conductance 电导 $G = 1/R$ (siemens) is sometimes handier. Power in a resistor is always absorbed, and combining $v = Ri$ with $p = vi$ gives the three equivalent forms $p = vi = i^{2}R = v^{2}/R$ — use whichever variable you already know.

For series 串联 elements the resistances add and one current is shared; for parallel 并联 elements the conductances add and one voltage is shared. Two resistors in parallel combine as the product over the sum, $R_{1} \parallel R_{2} = \frac{R_{1}R_{2}}{R_{1}+R_{2}}$ — and the result is always smaller than either.

Deriving the divider. Two resistors in series across a source carry one current $i = v_{s}/(R_{1}+R_{2})$. The voltage on $R_{2}$ is that current times $R_{2}$, so $v_{2} = v_{s} \cdot R_{2}/(R_{1}+R_{2})$ — the voltage divider 分压器. The mirror argument on two parallel resistors fed by a current source gives the current divider 分流器: $i_{1} = i_{s} \cdot R_{2}/(R_{1}+R_{2})$. Notice the swap: each branch takes the share proportional to the other resistor.

Kirchhoff's current law 基尔霍夫电流定律 (KCL) restates conservation of charge: the signed currents leaving any closed surface sum to zero. A node 节点 is a junction of two or more elements; every connection of the same wire is one node, however many branches meet there.

Kirchhoff's voltage law 基尔霍夫电压定律 (KVL) restates energy conservation: the signed voltages around any closed loop 回路 sum to zero. Walk a loop and tally the rises and falls: a label that goes up in one traversal is the same label that goes down when you return, so the total must be zero — you cannot gain energy by walking in a circle.

Worked example 1 (divider, checked two ways). A 9 V source feeds $R_{1} = 6$ k$\Omega$ in series with $R_{2} = 3$ k$\Omega$. The current is $i = 9 \text{ V} / 9 \text{ k}\Omega = 1 \text{ mA}$, so $v_{2} = 1 \text{ mA} \times 3 \text{ k}\Omega = 3 \text{ V}$ — the same answer as the divider formula $9 \times 3/9$. Check by KVL: $-9 + 6 + 3 = 0$. Check by power: source $9 \times 1 \text{ mA} = 9$ mW; resistors $6$ mW $+ 3$ mW $= 9$ mW.

Worked example 2 (current divider, then the node it feeds). A 2 A source feeds 6 $\Omega$ in parallel with 3 $\Omega$. The parallel pair is $6 \times 3/9 = 2\ \Omega$, so both see 4 V, and the currents are $4/6 = 0.667$ A and $4/3 = 1.333$ A — the divider formula gives the same shares, and $0.667 + 1.333 = 2$ A at the node, KCL closed. Now escalate: at that node a further branch of unknown current $i_{4}$ also leaves; if another 0.5 A flows in from elsewhere, KCL reads $-0.667 - 1.333 + 0.5 + i_{4} = 0$, so $i_{4} = 1.5$ A out. Charge did not vanish; it re-routed.

Common mistakes.

• Drawing KCL with some arrows in and some out, then adding magnitudes. Write "sum of currents leaving = 0" and let the signs speak.
• Treating two separate blobs of wire as one node (they are one node only if connected).
• Parallel resistors adding to something bigger — the combination is always smaller than the smallest.

Summary: Ohm's law plus two conservation laws solve every DC resistive circuit; dividers come from them, not from memory; label every arrow first, then trust the algebra — a negative answer is information, not a mistake.

A circuit with four nodes and six branches can be solved by clever guesses — once. The same circuit appears a hundred times in a design, and the guess-work has to become a recipe a computer could run. Two recipes exist; both end in a few simultaneous equations.

Nodal analysis 节点法 uses KCL. Choose a reference node 参考节点 (ground), label the remaining $n-1$ node voltages, and write one KCL equation per unknown node, with each branch current written as a voltage difference over a resistance. This method is the workhorse of circuit simulators (SPICE solves exactly these equations, millions of them).

Mesh analysis 网孔法 uses KVL. Assign a circulating mesh current 网孔电流 to each window of a planar circuit and write one KVL equation per mesh; a resistor shared by two meshes carries the difference of the two mesh currents. Pick the method with fewer equations, and remember: a voltage source fixes a node-voltage difference (nodal gets easier), while a current source fixes a branch current (mesh gets easier).

Deriving the nodal equation. At a node $A$ with resistors $R_{1}$ to a source node $V_{s}$ and $R_{2}$ to ground, KCL (sum of currents leaving) gives $\frac{V_{A}-V_{s}}{R_{1}} + \frac{V_{A}-0}{R_{2}} = 0$. Multiply through and collect: the equation is linear in $V_{A}$ — always. Every nodal analysis is this one equation repeated per node.

Worked example 1 (one unknown node). A 12 V source connects through 2 $\Omega$ to node $A$; from $A$, a 6 $\Omega$ resistor goes to ground and a 1 A source pushes current into $A$. KCL at $A$, currents out:

Multiply by 6: $3(V_{A} - 12) + V_{A} - 6 = 0$, so $4V_{A} = 42$ and $V_{A} = 10.5$ V. Check with the source branch: $(12 - 10.5)/2 = 0.75$ A flows from the source, plus 1 A from the current source; the 6 $\Omega$ carries $0.75 + 1 = 1.75$ A, and $10.5/6 = 1.75$ A — the unused route confirms the answer.

Worked example 2 (two meshes, shared resistor). Mesh A: a 12 V source (rising in the clockwise direction) with a 2 $\Omega$ resistor. Mesh B: a 6 $\Omega$ resistor with a 3 V source (opposing). The meshes share a 4 $\Omega$ resistor. KVL with clockwise mesh currents $i_{A}$, $i_{B}$:

Collect: $6i_{A} - 4i_{B} = 12$ and $-4i_{A} + 10i_{B} = -3$. Solving, $i_{A} = 2.455$ A and $i_{B} = 0.682$ A, so the shared resistor carries $i_{A}-i_{B} = 1.773$ A downward in mesh A's sense. Check by substituting back into either original equation: both return zero to within rounding.

Common mistakes.

• Forgetting the shared resistor's current is a difference of mesh currents.
• Choosing nodal for a circuit full of current sources — each source then becomes a constraint you already know; mesh would have been shorter.
• Grounding a node inside a voltage source (that source then fixes two nodes at once — pick its negative terminal as reference instead).

Summary: nodal = KCL per unknown node (simulators' choice); mesh = KVL per window; shared elements see current differences; choose by counting equations.

You replace a sensor board and the lamp it feeds brightens slightly. What did the board look like to the lamp? Not its schematic — one source and one resistance. Three theorems make that replacement exact, and they all say one thing: a linear network is indistinguishable from its behaviour at two terminals.

• Thévenin's theorem 戴维南定理: any linear two-terminal network of sources and resistors behaves, at its terminals, like one voltage source $V_{th}$ in series with one resistance $R_{th}$. $V_{th}$ is the open-circuit voltage 开路电压; $R_{th}$ is the resistance seen with sources zeroed (voltage sources shorted, current sources opened).
• Norton's theorem 诺顿定理 is the same statement as a current source $I_{N} = V_{th}/R_{th}$ in parallel with $R_{th}$. The two are the source transformation 电源变换: swap between them whenever the arithmetic is friendlier.
• Superposition 叠加定理: in a linear circuit with several independent sources, the response is the sum of the responses to each source acting alone (others zeroed). It is a computing strategy, not a property of any one component.

Deriving maximum power transfer. A Thévenin source $V_{th}$, resistance $R_{th}$, feeds $R_{L}$. The load power is $P = \left(\frac{V_{th}}{R_{th}+R_{L}}\right)^{2} R_{L}$. Differentiate with respect to $R_{L}$ and set the result to zero: the maximum sits at $R_{L} = R_{th}$, where $P_{max} = V_{th}^{2}/(4R_{th})$ — and the source resistance then burns exactly as much as the load delivers: 50% efficiency, no more.

Maximum power transfer 最大功率传输 is a signal rule, not an energy rule: audio and RF systems match impedances to keep signals strong; power grids never do (50% efficiency would double every bill). A Thévenin equivalent does not preserve the internal losses of the original network; its efficiency must be calculated from the actual sources.

Worked example 1 (find the equivalent, then load it). A 10 V source and a 4 $\Omega$ resistor form one series branch across a 6 $\Omega$ resistor. Find the equivalent seen by an external load connected across that 6 $\Omega$ resistor. With the external load disconnected, the terminal voltage is $V_{th} = 10 \times 6/10 = 6$ V (a divider). Zeroing the source leaves 4 $\Omega$ parallel with 6 $\Omega$: $R_{th} = 2.4\ \Omega$. With $R_{L} = R_{th} = 2.4\ \Omega$, the loop current is $6/(2.4+2.4) = 1.25$ A and the load power is $1.25^{2} \times 2.4 = 3.75$ W. Check against the formula: $V_{th}^{2}/(4R_{th}) = 36/9.6 = 3.75$ W.

Worked example 2 (superposition, checked head-on). A 12 V source through 2 $\Omega$ feeds node $A$; a 3 A source also injects into $A$; 4 $\Omega$ runs from $A$ to ground. By superposition: the 12 V source alone (3 A opened) drives the 4 $\Omega$ through a $2+4$ divider: $V_{A1} = 12 \times 4/6 = 8$ V. The 3 A source alone (12 V shorted) sees $2\ \Omega \parallel 4\ \Omega = 1.333\ \Omega$: $V_{A2} = 3 \times 1.333 = 4$ V. Total $V_{A} = 8 + 4 = 12$ V. Direct check by one nodal equation: $(V_{A}-12)/2 + V_{A}/4 - 3 = 0$ gives $V_{A} = 12$ V — the two methods agree exactly.

Common mistakes.

• Computing $R_{th}$ with sources still alive. Zero them: shorts for voltage sources, opens for current sources.
• Applying superposition to power (it is linear in voltage and current, not in their squares — compute responses first, then square).
• Believing the 50%-efficiency point is a design target for power delivery.

Summary: measure two things ($V_{th}$ open-circuit, $R_{th}$ zeroed) and any linear one-port becomes a battery and a resistor; swap series↔parallel forms freely; add responses, never powers; match resistance for maximum signal power, not for economy.

A microphone produces 10 mV. A speaker wants 2 V. Something must multiply by 200 — with a dial to set exactly how much. That something is the operational amplifier, and the whole unit is about one trick: feedback makes an absurdly high-gain device behave like an exact, mild-gain one.

An operational amplifier 运算放大器 is a differential voltage amplifier with very large gain, drawn as a triangle with $+$ and $-$ inputs and one output. Real op-amps have a gain-bandwidth product 增益带宽积 and finite input currents; the ideal op-amp 理想运算放大器 model used in analysis assumes infinite open-loop gain, infinite input resistance and zero output resistance.

Why the golden rules follow from feedback. With negative feedback 负反馈 (a path from output back to the $-$ input), the output does whatever is needed to make the input difference nearly zero — with gain above $10^{5}$, an output of volts needs an input difference of microvolts, which rounds to zero. Two golden rules 黄金法则 hold:

1. No current flows into either input (infinite input resistance).
2. The voltage difference between the inputs is zero — a virtual short 虚短路. The inputs track each other without being connected.

Deriving the inverting gain. The workhorse is the inverting amplifier 反相放大器: input through $R_{in}$ to the $-$ node, feedback through $R_{f}$, $+$ input grounded. Rule 2 pins the $-$ node at 0 V (a virtual ground 虚地). Rule 1 says the input current $v_{in}/R_{in}$ cannot enter the chip, so it all flows through $R_{f}$:

The gain is set by two resistors — the chip's enormous gain cancels out of the answer. The non-inverting amplifier 同相放大器 puts the signal at $+$ and a divider at $-$: $v_{out}/v_{in} = 1 + R_{f}/R_{in}$, always at least unity (at unity it is the voltage follower 电压跟随器, a buffer).

Worked example 1 (inverting stage, fully accounted). $R_{in} = 10$ k$\Omega$, $R_{f} = 100$ k$\Omega$, $v_{in} = +0.2$ V. Gain is $-100/10 = -10$, so $v_{out} = -2$ V, comfortably inside $\pm 15$ V supplies. The feedback current is $0.2/10\text{k} = 20\ \mu\text{A}$ and flows into the output; the ideal input carries no current. Check by the virtual ground: the $-$ node sits at $0$ V, so the drop on $R_{f}$ is $0 - (-2) = 2$ V $= 20\ \mu\text{A} \times 100$ k$\Omega$.

Worked example 2 (a mixer, and a saturation trap). Extend the inverting stage to two inputs ($R_{1} = R_{2} = 10$ k$\Omega$, same $R_{f} = 100$ k$\Omega$): the virtual ground collects both currents, so $v_{out} = -\left(\frac{R_f}{R_1}v_{1} + \frac{R_f}{R_2}v_{2}\right)$ — a summing amplifier 求和放大器. With $v_{1} = 0.2$ V and $v_{2} = -0.05$ V: $v_{out} = -(10 \times 0.2 + 10 \times (-0.05)) = -(2 - 0.5) = -1.5$ V. Now the trap: swap $R_{f}$ for 1 M$\Omega$ (gain $-100$ per input) and the same inputs ask for $-(100 \times 0.2 + 100 \times (-0.05)) = -15$ V. The saturation 饱和 rule stops it: the output cannot leave its supply rails, and a $\pm 15$ V op-amp rests near $-13.5$ V. The golden rules have silently stopped applying — report the clamp, not the impossible voltage.

Common mistakes.

• Forgetting the inverting stage's minus sign (or "correcting" it away).
• Applying rule 2 when there is no negative feedback path — no feedback, no virtual short.
• Pushing the answer past the rails and reporting it anyway.

Summary: infinite gain tamed by feedback gives two golden rules; the inverting stage turns them into $-R_{f}/R_{in}$, the non-inverting into $1 + R_{f}/R_{in}$; virtual ground makes mixers; saturation is a real output state, not a rounding error.

Every flash in a phone camera collects energy slowly and releases it in a millisecond. No arrangement of resistors can hold anything — resistors burn energy; the two new elements store it, and that single change is what makes circuits dynamic and worth a calculus course.

A capacitor 电容器 stores charge separated by an insulator: two plates with electric field between them. Its capacitance 电容 $C$, in farads 法拉 (F), is charge per volt, $q = Cv$. Differentiate: current is the rate of change of voltage,

so a capacitor carries current only while its voltage changes: steady DC is an open circuit to it. Integrating the same law gives the stored energy $w = \frac{1}{2}Cv^{2}$ — a state memory that depends only on the present voltage, not on how it got there.

An inductor 电感器 stores energy in a magnetic field around a coil. Its inductance 电感 $L$, in henrys 亨利 (H), gives the mirror law,

so an inductor develops voltage only while its current changes: steady DC is a short circuit to it. Stored energy is $w = \frac{1}{2}Li^{2}$.

Why the combinations mirror the resistor rules. Parallel capacitors share one voltage, so charges add: $C_{eq} = C_{1}+C_{2}$ (areas add). Series inductors share one current, so flux linkages add: $L_{eq} = L_{1}+L_{2}$ (lengths add). Series capacitors and parallel inductors combine by the reciprocal rule — the exact mirror of resistor arithmetic, because the storage laws swap the roles of $v$ and $i$.

Worked example 1 (two energies, two units). A 100 µF capacitor charged to 12 V holds $w = \frac{1}{2} \times 100\times10^{-6} \times 12^{2} = 7.2$ mJ. A 10 mH inductor carrying 2 A holds $w = \frac{1}{2} \times 0.01 \times 2^{2} = 20$ mJ. Both are small by kettle standards, but the capacitor's 7.2 mJ delivered in 1 ms is 7.2 W — and in 1 µs it would be 7.2 kW. Stored energy is modest; rate makes it dramatic, which is why large capacitors are discharged before you touch them.

Worked example 2 (slope law and combinations). A 10 µF capacitor's voltage rises at 100 V/s; the current is $i = C\,dv/dt = 10\times10^{-6} \times 100 = 1$ mA — constant, because the slope is constant. Now combine: a 2 µF and a 3 µF capacitor in series give $2 \times 3/5 = 1.2$ µF (each holds the same charge, so the "capacity per volt" shrinks); in parallel they give 5 µF. A 10 mH and 20 mH inductor in series give 30 mH; the same two in parallel give $10 \times 20/30 = 6.67$ mH.

⚠ An inductor's current is continuous: opening a switch in an inductor's path forces $di/dt$ to be huge, and the induced voltage $L\,di/dt$ can arc across the opening contacts. This is real, not theoretical — it is how old ignition coils made their spark.

Common mistakes.

• Using $w = \frac{1}{2}Cv^{2}$ with volts swapped for something else — check the units: farads times volts squared is joules.
• Adding parallel capacitances as reciprocals (that is the series rule).
• Believing a capacitor "blocks" DC instantly — it blocks steady DC; during changes it carries current gladly.

Summary: capacitors remember voltage, inductors remember current; $i = C\,dv/dt$ and $v = L\,di/dt$; energies $\frac{1}{2}Cv^{2}$, $\frac{1}{2}Li^{2}$; combinations mirror resistor arithmetic with $v$ and $i$ swapped; never open an inductor's path suddenly.

You press a button and the light fades up over a second — or the sensor reading creeps to its new value over milliseconds. Switch a source, an energy-storing element and a resistor into a new arrangement, and the stored quantity cannot jump: it slides to its new steady value. A circuit with one storage element is first order, and every such circuit has the same shape of answer — learn the shape once.

Define the natural response 自然响应 as the decay from an initial state with the source removed, and the step response 阶跃响应 as the move from an initial state to a new forced value $V_{F}$. For a capacitor voltage,

Deriving it. After the switch, a capacitor $C$ discharges through $R$: KCL gives $C\,dv/dt + v/R = 0$. Separate variables: $dv/v = -dt/(RC)$, integrate: $\ln v = -t/(RC) + \text{const}$, so $v = V_{0}e^{-t/RC}$ — an exponential with time constant 时间常数 $\tau = RC$. Add a source and the same equation gains a constant solution $V_{F}$; the general answer is the constant plus a decaying difference: exponential to the new value, always. The RL mirror replaces $\tau = RC$ with $\tau = L/R$.

At $t = \tau$ the gap has closed to 37% of its starting size; after $5\tau$ the circuit is within 1% and is treated as settled. The initial value $V_{0}$ comes from the stored energy (voltage continuity for C, current continuity for L), and $V_{F}$ from the circuit the switch left behind, with the storage element treated as its steady-state open (C) or short (L).

Worked example 1 (RC discharge). A 100 µF capacitor charged to 12 V discharges through 10 k$\Omega$. $\tau = RC = 10\,000 \times 100\times10^{-6} = 1$ s, so $v(t) = 12e^{-t}$ volts. After 2 s, $v = 12 \times e^{-2} = 1.62$ V. After 5 s ($5\tau$) it is $12e^{-5} = 0.08$ V — settled. Check the energy story: the initial 7.2 mJ has all gone into the resistor by then.

Worked example 2 (charging between two non-zero values). The same RC network, but the switch moves the resistor from a 2 V supply to a 9 V supply with the capacitor already at 2 V. $V_{0} = 2$, $V_{F} = 9$, so $v(t) = 9 + (2-9)e^{-t} = 9 - 7e^{-t}$ volts. At $t = 1$ s ($= \tau$) it reaches $9 - 7 \times 0.368 = 6.42$ V — the gap (7 V) has closed by 63.2%, exactly the definition. Endpoint checks: $v(0) = 2$ V and $v(5) = 9 - 0.047 \approx 9$ V.

Worked example 3 (the RL mirror, fast). A 20 mH inductor in series with 100 $\Omega$ switches onto a 10 V supply. $\tau = L/R = 0.02/100 = 0.2$ ms, final current $I_{F} = 10/100 = 0.1$ A, so $i(t) = 0.1(1 - e^{-t/0.2\text{ms}})$. At $t = 0.6$ ms (three time constants) it has reached 95%: $0.095$ A. The same shape as the capacitor voltage — one law, two elements.

Method for every switching problem: (1) find $V_{0}$ from the old circuit; (2) find $V_{F}$ from the new circuit; (3) compute $\tau$ from the new circuit with the source zeroed; (4) write the one-line answer and check it against both endpoints.

Common mistakes.

• Using the old circuit for $V_{F}$ or the new one for $V_{0}$ — the values come from different snapshots, deliberately.
• Computing $\tau$ with the source still present — zero the source to see the true resistance the storage element faces.
• Believing $5\tau$ is a physical constant — it is a 1% engineering convention.

Summary: one storage element means one exponential; $V_{0}$ from continuity, $V_{F}$ from the final circuit, $\tau$ from the zeroed-source resistance; $\tau = RC$ or $L/R$; $5\tau$ is settled for practical purposes, and every answer checks at both ends.