ELECTENG291 · Fundamentals of Electrical Engineering
Capacitors, Inductors & Energy Storage
Module 2 of University of Auckland ELECTENG 291 introduces the two energy-storage elements and the rules that make transients possible: the capacitor law i = C dv/dt and inductor law v = L di/dt (both under the passive sign convention), the energy stored (½Cv² and ½Li²), and the crucial continuity rules — a capacitor's voltage and an inductor's current cannot change instantly. Setting correct initial conditions from these rules is the first step of every first- and second-order transient problem in the tests and the exam.
What this chapter covers
- 01Capacitor terminal law (PSC): i = C dv/dt — current is proportional to the rate of change of voltage
- 02Inductor terminal law (PSC): v = L di/dt — voltage is proportional to the rate of change of current
- 03DC steady state: a capacitor becomes an open circuit (i_C = 0) and an inductor becomes a short circuit (v_L = 0)
- 04Continuity: v_C(t) and i_L(t) are continuous — they cannot jump, so v_C(0⁺) = v_C(0⁻) and i_L(0⁺) = i_L(0⁻)
- 05Energy stored: capacitor ½C v_C², inductor ½L i_L² [J]
- 06Rate/state relations at a switching instant: dv_C/dt = i_C/C and di_L/dt = v_L/L
- 07Reading initial conditions from the pre-switch DC steady state; final conditions from the post-switch steady state
Initial conditions and stored energy across a switching event
- +1Use continuity for the capacitor voltage: v_C cannot jump, so v_C(0⁺) = v_C(0⁻) = 12 V.
- +1Use continuity for the inductor current: i_L cannot jump, so i_L(0⁺) = i_L(0⁻) = 3 A. (These two continuity facts are what set every transient's initial conditions.)
- +1Energy in the capacitor: W_C = ½ C v_C² = ½ × 4 µF × (12)² = ½ × 4×10⁻⁶ × 144 = 288 µJ.
- +1Energy in the inductor: W_L = ½ L i_L² = ½ × 2 H × (3)² = ½ × 2 × 9 = 9 J. Both are finite and continuous through the switching instant, consistent with the continuity rules.
Key terms
- Capacitor terminal law
- i = C dv/dt under the passive sign convention: the current equals the capacitance times the rate of change of voltage. It follows that a capacitor passes no current at DC (an open circuit) and its voltage cannot change instantly.
- Inductor terminal law
- v = L di/dt under the passive sign convention: the voltage equals the inductance times the rate of change of current. At DC an inductor is a short circuit (no voltage), and its current cannot change instantly.
- Continuity of v_C and i_L
- A capacitor's voltage and an inductor's current are continuous functions of time — they cannot jump. Hence v_C(0⁺) = v_C(0⁻) and i_L(0⁺) = i_L(0⁻), which fix the initial conditions of a transient.
- DC steady state
- The condition after a circuit has settled with constant sources: all derivatives are zero, so a capacitor behaves as an open circuit (i_C = 0) and an inductor as a short circuit (v_L = 0). Used to read both initial and final conditions.
- Stored energy
- The energy held in an energy-storage element: ½C v_C² in a capacitor and ½L i_L² in an inductor, in joules [J]. Because these depend on v_C and i_L, and those are continuous, the stored energy cannot change instantaneously.
- Initial condition (0⁺)
- The value of a circuit variable just after a switching event. For v_C and i_L it equals the value just before (continuity); other variables (like i_C or v_L) may jump and are found from KVL/KCL at t = 0⁺.
Capacitors, Inductors & Energy Storage FAQ
Why can't a capacitor voltage or inductor current change instantly?
Because an instantaneous jump would require infinite power. The stored energy is ½Cv² for a capacitor and ½Li² for an inductor; changing v_C or i_L in zero time would change that energy in zero time, which is physically impossible. So v_C and i_L are continuous, and their pre-switch values carry straight across as the initial conditions for a transient.
How do capacitors and inductors behave at DC steady state?
Once everything is constant, all derivatives vanish. In i = C dv/dt, a constant v_C gives i_C = 0, so a capacitor looks like an open circuit. In v = L di/dt, a constant i_L gives v_L = 0, so an inductor looks like a short circuit. You use exactly these facts to read the pre-switch initial conditions and the post-switch final conditions.
How do I find the initial conditions for a transient?
First analyse the circuit in its pre-switch DC steady state (capacitor open, inductor short) to read v_C(0⁻) and i_L(0⁻). Then apply continuity: v_C(0⁺) = v_C(0⁻) and i_L(0⁺) = i_L(0⁻). Any other variable that is allowed to jump — such as the capacitor current or inductor voltage at 0⁺ — is then found from KVL/KCL using those known values.
Can Sia help me with capacitors, inductors and initial conditions?
Yes, as a study aid. Sia can apply the terminal laws, work out v_C(0⁺), i_L(0⁺), final values and stored energy, and explain why continuity holds. It explains and drills you on fresh numbers; it does not do graded assessment for you, and University of Auckland academic-integrity rules apply — confirm what is permitted on Canvas.
Exam move
Anchor everything on two continuity facts — v_C and i_L cannot jump — because they set the initial conditions for every transient in Module 2. Practise the three-snapshot habit: analyse the pre-switch DC steady state (capacitor open, inductor short) to get v_C(0⁻) and i_L(0⁻); carry them across by continuity to 0⁺; then analyse the post-switch DC steady state to get the final values. Keep the terminal laws i = C dv/dt and v = L di/dt under the passive sign convention so the signs stay consistent. Be able to compute stored energy (½Cv², ½Li²) and to find the variables that can jump (i_C, v_L at 0⁺) from KVL/KCL. This groundwork is what makes the first- and second-order chapters quick. Confirm assessment details on Canvas.
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