Circuit Analysis

The mutual inductance M = k√(L₁L₂) that appears when two coils share flux. Sweep the coupling coefficient k yourself and watch M approach the ceiling √(L₁L₂).

The convention that records the sign of the mutual term with a single dot. Toggle aiding and opposing and see how the sign in Lₜ = L₁ + L₂ ± 2M flips.

Any linear two-terminal circuit folds into one source and one resistor. Sweep the load R_L yourself to see the terminal V-I trace a single line, and confirm that the load halving the terminal voltage is R_th.

The response of a linear circuit is the sum of the responses each source makes alone. Toggle two sources and see that the branch currents from each one separately add up to the current with both on.

AC impedance is the complex number Z = R + jX. Sweep the reactance X yourself to watch the impedance triangle tilt up, and confirm that the phase is exactly 45 degrees when the reactance equals the resistance.

Capacitor charging is an exponential curve, and one time constant τ = RC covers 63% of the way. Sweep time along the curve and confirm that the moment it reaches 63% of the final value is τ.

The step-response shape of a series RLC is set by one damping ratio ζ = (R/2)√(C/L). Sweep ζ yourself and, between oscillation (ζ<1) and sluggishness (ζ>1), find critical damping ζ=1, the fastest with no overshoot.

A first-order RC low-pass passes low frequencies and blocks high ones. Sweep the frequency along the magnitude response and find the cutoff ω_c = 1/RC where the output drops to 1/√2 (−3 dB).

Under negative feedback an ideal op-amp drives its inputs equal (a virtual short) with zero input current. Sweep the feedback ratio Rf/Rg yourself to design the gain G = 1 + Rf/Rg to ×3.

In a series RLC the inductive and capacitive reactances cancel at the resonant frequency ω_0 = 1/√(LC), where the impedance is minimal and the current maximal. Sweep the frequency yourself to find the resonant peak where the current spikes.

The power into a load on a Thévenin source is greatest when R_L = R_th (matched). Sweep the load resistance yourself to find the peak of the power curve, and confirm the efficiency there is 50%.

Apparent power S is the right-angle sum of real power P and reactive power Q, and the power factor is cosθ = P/S. Sweep the phase angle θ yourself to tilt the power triangle and find unity power factor (θ=0) where everything becomes work.

The same box can be written as a voltage source in series (Thévenin) or a current source in parallel (Norton). Sweep the Norton current I_N yourself to find where the two V-I lines overlap, and see the transformation V_th = I_N·R_th.

A Laplace pole s = σ + jω sets the time response e^(σt)cos(ωt). Sweep the pole’s real part σ yourself to watch the response fade or blow up, and find the imaginary-axis boundary between stable and unstable.

The same RC as a low-pass, but taking the output across the resistor makes a high-pass. Sweep the frequency along the magnitude response and find the cutoff ω_c = 1/RC where the output rises to 1/√2 (−3 dB).

A bridge of four resistors in a diamond reads zero on the detector when the opposing ratios match. Sweep the standard resistor yourself to find the balance point where the detector nulls, and read off the unknown resistance.

A capacitor stores ½CV², an inductor ½LI², both proportional to the square of voltage or current. Sweep the voltage along the squared curve and set it so the stored energy reaches the target of 18 mJ.

The half-power bandwidth of a bandpass filter is BW = ω_0/Q. Sweep the quality factor Q yourself to watch the passband narrow, and find the Q where the bandwidth reaches the target of 0.25.

Grounding the non-inverting input makes the virtual short pull the inverting input to a 0 V virtual ground. Sweep the feedback ratio Rf/Rin yourself to design the gain −Rf/Rin to −4.

A Bode plot draws gain in decibels (20·log10|H|) against log frequency, approximating a first-order response with two straight lines. Sweep the frequency yourself to find the corner where the gap between the real curve and the asymptotes peaks at 3 dB.

A sinusoid’s RMS value is its peak over √2, the equivalent DC that delivers the same average power (P = V_rms²/R). Sweep the amplitude yourself to find the peak (about 311 V) where the RMS reaches the target of 220 V.

The signed sum of currents at a node is zero (charge conservation). Sweep the unknown branch current yourself to find the point where the node sum is exactly zero and solve for the unknown.

The signed sum of voltage rises and drops once around a closed loop is zero (energy conservation). Sweep the unknown voltage drop yourself to find where the loop sum is exactly zero and solve for the unknown.

Any periodic waveform is a sum of a fundamental and integer-multiple harmonics (Fourier). A square wave is the sum of the odd harmonics; add more harmonics yourself and watch the sum hug the square ever closer.

Replace the inverting amplifier’s feedback resistor with a capacitor and the output becomes the integral of the input, Vout = −(1/RC)∫Vin dt. Sweep the input voltage to find the input where the output ramp reaches the target of −4 V.

Replace the inverting amplifier’s input resistor with a capacitor and the output becomes the derivative of the input, Vout = −RC·dVin/dt. Sweep the slope of the input ramp to find the slope where the output sits on the target of −4 V.

Capacitors are the opposite of resistors: in parallel the capacitances add (C_p = C1 + C2) and in series the reciprocals add (1/C_s = 1/C1 + 1/C2). Sweep C2 to find the value where the parallel capacitance reaches the target of 10 μF.

Inductors are on the same side as resistors: in series the inductances add (L_s = L1 + L2) and in parallel the reciprocals add (1/L_p = 1/L1 + 1/L2). Sweep L2 to find the value where the series inductance reaches the target of 10 mH.

Two resistors in series pass the same current and split the source voltage by resistance ratio. The output is V_out = V_in·R2/(R1+R2). Sweep R2 to find the value where the output reaches the target of 8 V.

The mirror of the voltage divider. Two parallel branches share the same voltage and split the source current by conductance ratio. The R2-branch current is I_2 = I_in·R1/(R1+R2), with the opposite resistor on top. Sweep R2 to find the value where the branch current reaches the target of 4 mA.

The mirror of the RC transient. The inductor current climbs along an exponential, and one time constant τ = L/R covers 63% of the remaining distance to the final value I = V/R. Sweep the time t to find the instant τ where the current reaches 63%.

The mirror of the inverting amplifier. Feed the input into the non-inverting terminal and the virtual short holds V− at Vin, while the Rf-Rg feedback divider sets the gain to 1+Rf/Rg. The sign is preserved and the gain is always at least one. Sweep the ratio Rf/Rg to find the value where the gain reaches the target of 4.

A tool for circuits that are neither series nor parallel. Swap a triangle (Δ) and a star (Y) sharing three terminals so they look identical at the terminals. When balanced, a star arm is one-third of a triangle side, R_Y = R_Δ/3. Sweep R_Y to find the value where the two networks are equivalent.

The foundation of AC steady state. Write a same-frequency sinusoid A cos(ωt+φ) as a rotating arrow A∠φ and adding sinusoids becomes the vector sum of arrows. Lengths 3 and 4 at a right angle make 5. Sweep the phase φ to find the value where the summed amplitude reaches the target of 5.

The capstone of mutual inductance. Two perfectly coupled coils step the voltage up by as much as the current down through the turns ratio n = N2/N1: V2 = n·V1, I2 = I1/n, and impedance transforms by the square, Z_in = Z_L/n². Sweep the ratio to find the value where the secondary voltage reaches the target of 25 V.

Feed one signal into the non-inverting (+) input and the other into the inverting (−) to amplify only their difference. With matched resistors Vout = (Rf/R1)(V2 − V1), and the common mode (shared noise) cancels and vanishes. Sweep V2 to find the value where the output reaches the target of 6 V.

Widen jω into the complex frequency s and an inductor is sL, a capacitor 1/sC. The circuit becomes voltage division in s, giving the transfer function H(s). The pole −1/RC of the RC low-pass H(s)=1/(1+sRC) fixes the transient, cutoff, and DC gain at once. Sweep the time constant τ to place the pole at the target −2.

The AC extension of DC maximum power transfer (CR-MP). In AC, matching the resistance alone is not enough. Cancel the source reactance with the load reactance (X_L=−X_s) and match the resistance, and the load becomes the source’s conjugate Z_L=Z_s*=R_s−jX_s, where the power is maximal. Sweep the load reactance X_L to find the power peak.

The dual mirror of series resonance (CR-RE). In parallel the susceptances ωC and 1/ωL cancel at ω_0=1/√(LC), so the impedance is maximal (=R) and the source current minimal. Yet inside the tank, L and C exchange Q times the source current (current magnification). Sweep the frequency to find resonance.

The root of transients and reactance. Unlike a resistor, a capacitor and inductor respond not to the value but to the rate of change. From the plate charge q=Cv comes i=C·dv/dt, with the inductor as its mirror v=L·di/dt. Sweep the rate of change dv/dt to find the slope where the current reaches the target of 4.

The complement of the bandpass (CR-BW), completing the four basic filters. Take the output across the series L and C of a series RLC, and at resonance ω_0=1/√(LC) the LC becomes a short (Z=0), so the output sinks to zero. A band-stop that discards a single frequency. Sweep the frequency to find the notch.

The basis of every op-amp circuit. Under negative feedback the closed-loop gain is G=A/(1+Aβ). With large loop gain Aβ the A cancels and G≈1/β, so the gain is set by the precise feedback β (gain desensitization). Sweep the raw gain A to watch G lock onto the ideal value 1/β=10.