Passive components: frequency and time
The same inductor is almost a short circuit at 50 Hz and almost an open circuit at 1 MHz. Complex impedance explains why, and the RC time constant explains everything that happens at the instant a switch is closed.
01The voltage-current relationship of each component
Three passive elements are enough to build almost any linear circuit. Each one relates voltage and current in a different way, and that difference is what makes this whole course interesting.
| Component | v-i relationship | Does it have memory? | Energy it stores |
|---|---|---|---|
| Resistor | \( v = Ri \) | No: the voltage depends only on the current at that instant | None; it dissipates as heat |
| Inductor | \( v = L\dfrac{di}{dt} \) | Yes: the voltage depends on how the current changes | \( \tfrac12 Li^2 \), in the magnetic field |
| Capacitor | \( i = C\dfrac{dv}{dt} \) | Yes: the current depends on how the voltage changes | \( \tfrac12 Cv^2 \), in the electric field |
“Having memory” is the key idea: the voltage of an inductor and the current of a capacitor cannot be calculated by looking only at the present instant; you need the history of how they got there. That is, exactly, what turns a circuit with \( L \) or \( C \) into a problem of differential equations.
02Phasors: turning a differential equation into algebra
Solving differential equations every time a circuit has a sinusoidal source would be exhausting. The phasor trick avoids it: in steady state, every voltage and every current of a linear circuit excited at a single frequency \( \omega \) is a sinusoid of that same frequency, and a sinusoid is completely determined by two numbers, its amplitude and its phase. A phasor is, precisely, those two numbers packed into a single complex number.
That correspondence —differentiating is multiplying by \( j\omega \)— is what turns the circuit’s whole differential equation into an algebraic equation with complex numbers, as mechanical to solve as a purely resistive circuit.
03Complex impedance: the generalized Ohm’s law
With phasors, each passive component behaves like a “complex resistance,” the impedance \( Z = \mathbf{V}/\mathbf{I} \), and Ohm’s law and Kirchhoff’s laws apply exactly the same as with pure resistances, now with complex numbers.
| Component | At 50 Hz (mains) | At 1 MHz (AM radio) |
|---|---|---|
| 10 mH inductor | \( |Z_L| = 3.14 \) Ω | \( |Z_L| = 62{,}832 \) Ω |
| 100 nF capacitor | \( |Z_C| = 31{,}831 \) Ω | \( |Z_C| = 1.59 \) Ω |
The same pair of components behaves in almost opposite ways depending on the working frequency: at 50 Hz the inductor is almost a short circuit and the capacitor almost an open circuit; at 1 MHz, the other way around. Designing a filter is, in essence, exploiting that crossover.
04The time domain: the transient response of an RC circuit
Phasors only describe the steady state: what happens long after any switching event —closing a switch, connecting a source— has stopped having an effect. The moment of switching itself demands going back to the time domain and solving the differential equation. The simplest and most important case in the whole course is the first-order RC circuit.
You choose \( R \) and \( C \); the plot traces the charge from 0 V to 5 V and marks \( \tau \), \( 3\tau \) and \( 5\tau \).
05The two domains are the same physics, seen in two ways
The complete solution of any linear circuit with a sinusoidal source is always the sum of two parts: the steady-state response —the sinusoid given by phasor analysis, which lasts forever— and the transient response —a decaying exponential that describes how the circuit gets from its initial state to that steady state—. Phasors compute the first part effortlessly, at the cost of not being able to see the second; the differential equation in time computes both, at the cost of much more work. Choosing the right tool depends on which question needs to be answered.
06Where it shows up in electronics
| Application | What it uses |
|---|---|
| Switch-bounce suppression (debounce) for a pushbutton | An RC with a \( \tau \) of a few milliseconds filters out the mechanical bounce of the contact |
| Power-up circuit (power-on reset) | An RC holds a pin low for a few milliseconds after power-up, so that the microcontroller starts in a known state |
| RC low-pass / high-pass filter | The same impedance \( Z_C \) that falls with frequency, used to separate bands |
| RC snubber on a relay or a thyristor | It absorbs the switching transient so as not to damage the semiconductor |
07In the lab
Using this page’s lab, choose \( R \) and \( C \) so that \( 5\tau \) is between 5 and 10 ms, a typical debounce time. Verify with at least two different combinations of \( R \) and \( C \) that they give the same \( \tau \).
Calculate \( |Z_L| \) and \( |Z_C| \) for a 1 mH inductor and a 1 µF capacitor at 100 Hz, 10 kHz and 1 MHz. Identify in which range each one behaves most like a short circuit and in which like an open circuit.
For an RC circuit driven by a sinusoidal source that is connected at \( t=0 \), identify which part of the response a phasor computes and which it does not, and explain why the complete differential equation is needed only at the instant of connection.
08Common mistakes
- Applying phasors during a transient. Phasors only describe the steady state; the moment of switching needs the differential equation in the time domain.
- Forgetting that \( Z_L \) and \( Z_C \) depend on frequency. They are not fixed numbers like \( R \): they change with \( \omega \), and have to be recalculated for each frequency of interest.
- Confusing the time constant \( \tau \) with the time it takes for the transient to “finish.” \( \tau \) is 63.2 %; the practical convention for finished is \( 5\tau \).
- Thinking an inductor or a capacitor “has” a fixed resistance. Its impedance is reactive —purely imaginary, in the ideal case— and depends on \( \omega \); it is not a constant real resistance.
- Ignoring the initial condition of the capacitor or the inductor. \( V_0 \) in the charging formula is not always zero; with a nonzero initial charge, the curve is different.
09Self-assessment
Why do an inductor or a capacitor “have memory” and a resistor does not?
Because their v-i relationship involves a derivative or an integral: their value at an instant depends on how they got there, not only on the present value of the other variable.
What does a phasor represent?
The amplitude and the phase of a steady-state sinusoid, packed into a complex number, without the explicit dependence on time.
How is the impedance of a capacitor calculated?
\( Z_C = 1/(j\omega C) \): it falls with frequency and is inversely proportional to the capacitance.
What does the time constant \( \tau = RC \) mean?
The time it takes the capacitor voltage to cover 63.2 % of the distance it still has to go to its final value.
Why are phasors not enough to describe the instant a switch closes?
Because they describe only the steady state; the transient that follows the closing needs the complete differential equation solved in the time domain.
Which combination of \( R \) and \( C \) would you use for a debounce circuit with a \( \tau \) of 1 ms?
Any whose product \( RC \) gives 1 ms, for example 10 kΩ with 100 nF.
10Further reading
- Charles K. Alexander and Matthew N. O. Sadiku. Fundamentals of Electric Circuits. 7th ed., McGraw-Hill, 2021. The chapters on phasors, impedance and first-order response, which develop in detail everything this page summarizes.
- James W. Nilsson and Susan A. Riedel. Electric Circuits. 11th ed., Pearson, 2019. A very gradual treatment of the RC and RL transient response, with dozens of examples worked step by step.
- Horowitz and Hill. The Art of Electronics. 3rd ed., Cambridge University Press, 2015. Chapter 1 connects this theory with real everyday RC circuits: debouncing, filters and snubbers.