m physical modeling / IICSM

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Electrical engineering models

From charge and circuit conservation to AC power, filters, transformers, and energy: four derivation pathways and twenty solved examples.

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Matter pathway: atom → solid → liquid → gas → plasma. Quantum mechanics and quantum field theory provide foundations across the pathway; they are not additional phases. This is a connected modeling route, not a universal heating curve. Actual phases depend on pressure, composition, and kinetics.

1. Circuit conservation

Definitions & inputs. q charge, i current, v voltage, R resistance; signed currents enter a node.

  1. Define flow of charge and the linear resistor law.

    i=dq/dt,v=Rii=dq/dt,\quad v=Ri

  2. Apply charge conservation and loop energy conservation.

    ∑jij=0,∑loopvj=0\sum_j i_j=0,\quad\sum_{loop}v_j=0

  3. Equal series current adds drops; equal parallel voltage adds currents.

    Rs=∑Rj,Rp−1=∑Rj−1R_s=\sum R_j,\quad R_p^{-1}=\sum R_j^{-1}

Interpretation. Reference directions determine signs; negative results indicate opposite flow.

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2. Storage and transients

Definitions & inputs. C capacitance, L inductance, t time; Vs constant step source.

  1. Constitutive relations connect storage to voltage/current.

    q=Cv,vL=L di/dtq=Cv,\quad v_L=L\,di/dt

  2. Substitute resistor current into the node equation.

    RC dvC/dt+vC=VsRC\,dv_C/dt+v_C=V_s

  3. Integrate the first-order equation; integrate v dq for stored energy.

    vC=Vs(1−e−t/(RC)),EC=CvC2/2v_C=V_s(1-e^{-t/(RC)}),\quad E_C=Cv_C^2/2

Interpretation. An inductor instead stores Li²/2; initial conditions determine transients.

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3. Sinusoidal power

Definitions & inputs. j²=−1, ω angular frequency, RMS voltage V and current I, φ voltage-current phase.

  1. Differentiate sinusoids using complex amplitudes.

    ZR=R,ZL=jωL,ZC=1/(jωC)Z_R=R,\quad Z_L=j\omega L,\quad Z_C=1/(j\omega C)

  2. Average instantaneous power to separate real and reactive terms.

    P=VIcos⁡ϕ,Q=VIsin⁡ϕP=VI\cos\phi,\quad Q=VI\sin\phi

  3. The complex-power magnitude is apparent power in VA.

    S2=P2+Q2,S=VIS^2=P^2+Q^2,\quad S=VI

Interpretation. Power factor is cosφ only for sinusoidal waveforms without distortion.

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4. Filters and conversion

Definitions & inputs. H voltage transfer, f frequency, N turns, subscripts p and s primary and secondary.

  1. Apply a capacitive impedance divider.

    H(jω)=1/(1+jωRC)H(j\omega)=1/(1+j\omega RC)

  2. Half-power cutoff follows |H|²=1/2; equal flux per turn sets voltage ratio.

    fc=(2πRC)−1,Vs/Vp=Ns/Npf_c=(2\pi RC)^{-1},\quad V_s/V_p=N_s/N_p

  3. Add three equal phase real powers and convert to line quantities.

    P3ϕ=3VLLILcos⁡ϕP_{3\phi}=\sqrt3 V_{LL}I_L\cos\phi

Interpretation. Loading, winding loss, saturation, and harmonics require richer models.

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Graphical worked example

5 V step applied to an initially uncharged RC circuit with τ=0.1 s. X axis: Time (s). Y axis: Capacitor voltage (V).
5 V step applied to an initially uncharged RC circuit with τ=0.1 s. Related worked calculation · Download SVG · Plot data

Twenty worked examples

Open a problem to see its defined inputs, assumptions, equation, numerical substitution, result, and interpretation. Values are illustrative analytical exercises.

Example 01. Ohm-law current

Definitions & inputs. V=12 V, R=6 Ω.

  1. Choose the governing model and isolate the requested quantity.

    I=V/RI=V/R

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    I=12/6I=12/6

  3. Evaluate the expression; the result uses the units shown.

    Result=2 A\mathrm{Result}=2\ {\rm A}

Interpretation. Current follows the stated voltage polarity.

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Example 02. Resistor dissipation

Definitions & inputs. I=2 A, R=6 Ω.

  1. Choose the governing model and isolate the requested quantity.

    P=I2RP=I^2R

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    P=22(6)P=2^2(6)

  3. Evaluate the expression; the result uses the units shown.

    Result=24 W\mathrm{Result}=24\ {\rm W}

Interpretation. This heat load must be compared with the resistor rating.

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Example 03. Series resistance

Definitions & inputs. R1=100 Ω, R2=220 Ω.

  1. Choose the governing model and isolate the requested quantity.

    Rs=R1+R2R_s=R_1+R_2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Rs=100+220R_s=100+220

  3. Evaluate the expression; the result uses the units shown.

    Result=320 Ω\mathrm{Result}=320\ \Omega

Interpretation. Both resistors carry the same current.

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Example 04. Parallel resistance

Definitions & inputs. R1=100 Ω, R2=300 Ω.

  1. Choose the governing model and isolate the requested quantity.

    Rp=R1R2/(R1+R2)R_p=R_1R_2/(R_1+R_2)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Rp=100(300)/400R_p=100(300)/400

  3. Evaluate the expression; the result uses the units shown.

    Result=75 Ω\mathrm{Result}=75\ \Omega

Interpretation. Equivalent resistance is less than either branch.

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Example 05. Unloaded divider

Definitions & inputs. Vin=10 V; upper R1=1 kΩ, lower R2=4 kΩ.

  1. Choose the governing model and isolate the requested quantity.

    Vo=VinR2/(R1+R2)V_o=V_{in}R_2/(R_1+R_2)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Vo=10(4)/5V_o=10(4)/5

  3. Evaluate the expression; the result uses the units shown.

    Result=8 V\mathrm{Result}=8\ {\rm V}

Interpretation. A connected load changes the lower effective resistance.

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Example 06. Node balance

Definitions & inputs. Currents entering are 3 A and 2 A; one current leaves.

  1. Choose the governing model and isolate the requested quantity.

    Io=I1+I2I_o=I_1+I_2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Io=3+2I_o=3+2

  3. Evaluate the expression; the result uses the units shown.

    Result=5 A\mathrm{Result}=5\ {\rm A}

Interpretation. This assumes no net charge accumulation at the node.

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Example 07. Capacitor charge

Definitions & inputs. C=100 μF, V=5 V.

  1. Choose the governing model and isolate the requested quantity.

    q=CVq=CV

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    q=100×10−6(5)q=100\times10^{-6}(5)

  3. Evaluate the expression; the result uses the units shown.

    Result=0.0005 C\mathrm{Result}=0.0005\ {\rm C}

Interpretation. Stored charge magnitude is equal on the two plates.

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Example 08. Capacitor energy

Definitions & inputs. C=100 μF, V=5 V.

  1. Choose the governing model and isolate the requested quantity.

    E=CV2/2E=CV^2/2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    E=(100×10−6)52/2E=(100\times10^{-6})5^2/2

  3. Evaluate the expression; the result uses the units shown.

    Result=0.00125 J\mathrm{Result}=0.00125\ {\rm J}

Interpretation. Charging losses depend on the source circuit.

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Example 09. Inductor energy

Definitions & inputs. L=20 mH, I=2 A.

  1. Choose the governing model and isolate the requested quantity.

    E=LI2/2E=LI^2/2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    E=0.02(22)/2E=0.02(2^2)/2

  3. Evaluate the expression; the result uses the units shown.

    Result=0.04 J\mathrm{Result}=0.04\ {\rm J}

Interpretation. Use a nonsaturating linear inductance.

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Example 10. RC time constant

Definitions & inputs. R=10 kΩ, C=10 μF.

  1. Choose the governing model and isolate the requested quantity.

    τ=RC\tau=RC

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    τ=104(10−5)\tau=10^4(10^{-5})

  3. Evaluate the expression; the result uses the units shown.

    Result=0.1 s\mathrm{Result}=0.1\ {\rm s}

Interpretation. About 63.2% of a step is reached after one time constant.

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Example 11. RC charging at one time constant

Definitions & inputs. Vs=5 V, initially zero, t=τ.

  1. Choose the governing model and isolate the requested quantity.

    v=Vs(1−e−t/τ)v=V_s(1-e^{-t/\tau})

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    v=5(1−e−1)v=5(1-e^{-1})

  3. Evaluate the expression; the result uses the units shown.

    Result=3.160603 V\mathrm{Result}=3.160603\ {\rm V}

Interpretation. The final voltage is 5 V only for negligible leakage and loading.

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Example 12. RC cutoff

Definitions & inputs. R=1 kΩ, C=100 nF.

  1. Choose the governing model and isolate the requested quantity.

    fc=(2πRC)−1f_c=(2\pi RC)^{-1}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    fc=[2π(103)(10−7)]−1f_c=[2\pi(10^3)(10^{-7})]^{-1}

  3. Evaluate the expression; the result uses the units shown.

    Result=1591.549 Hz\mathrm{Result}=1591.549\ {\rm Hz}

Interpretation. The output amplitude is 1/√2 at cutoff.

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Example 13. Inductive reactance

Definitions & inputs. f=50 Hz, L=0.1 H.

  1. Choose the governing model and isolate the requested quantity.

    XL=2πfLX_L=2\pi fL

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    XL=2π(50)(0.1)X_L=2\pi(50)(0.1)

  3. Evaluate the expression; the result uses the units shown.

    Result=31.41593 Ω\mathrm{Result}=31.41593\ \Omega

Interpretation. Inductive voltage leads current by 90°.

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Example 14. Capacitive reactance magnitude

Definitions & inputs. f=50 Hz, C=100 μF.

  1. Choose the governing model and isolate the requested quantity.

    ∣XC∣=(2πfC)−1|X_C|=(2\pi fC)^{-1}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    ∣XC∣=[2π(50)(10−4)]−1|X_C|=[2\pi(50)(10^{-4})]^{-1}

  3. Evaluate the expression; the result uses the units shown.

    Result=31.83099 Ω\mathrm{Result}=31.83099\ \Omega

Interpretation. Capacitive impedance has a negative imaginary part.

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Example 15. Series R-L impedance magnitude

Definitions & inputs. R=30 Ω, XL=40 Ω.

  1. Choose the governing model and isolate the requested quantity.

    ∣Z∣=R2+XL2|Z|=\sqrt{R^2+X_L^2}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    ∣Z∣=302+402|Z|=\sqrt{30^2+40^2}

  3. Evaluate the expression; the result uses the units shown.

    Result=50 Ω\mathrm{Result}=50\ \Omega

Interpretation. A 100 V RMS source would deliver 2 A RMS.

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Example 16. Single-phase real power

Definitions & inputs. V=230 V RMS, I=10 A RMS, power factor 0.8.

  1. Choose the governing model and isolate the requested quantity.

    P=VIcos⁡ϕP=VI\cos\phi

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    P=230(10)(0.8)P=230(10)(0.8)

  3. Evaluate the expression; the result uses the units shown.

    Result=1840 W\mathrm{Result}=1840\ {\rm W}

Interpretation. Apparent power is 2300 VA.

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Example 17. Ideal transformer output

Definitions & inputs. Vp=120 V RMS, Np=1000, Ns=100 turns.

  1. Choose the governing model and isolate the requested quantity.

    Vs=VpNs/NpV_s=V_pN_s/N_p

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Vs=120(100)/1000V_s=120(100)/1000

  3. Evaluate the expression; the result uses the units shown.

    Result=12 V\mathrm{Result}=12\ {\rm V}

Interpretation. Loaded regulation and winding losses are neglected.

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Example 18. Balanced three-phase power

Definitions & inputs. Line voltage 400 V RMS, line current 10 A RMS, power factor 0.9.

  1. Choose the governing model and isolate the requested quantity.

    P=3VLLILcos⁡ϕP=\sqrt3 V_{LL}I_L\cos\phi

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    P=3(400)(10)(0.9)P=\sqrt3(400)(10)(0.9)

  3. Evaluate the expression; the result uses the units shown.

    Result=6235.383 W\mathrm{Result}=6235.383\ {\rm W}

Interpretation. Line voltage differs from phase voltage in a star connection.

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Example 19. LC resonance

Definitions & inputs. L=10 mH, C=1 μF.

  1. Choose the governing model and isolate the requested quantity.

    f0=(2πLC)−1f_0=(2\pi\sqrt{LC})^{-1}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    f0=[2π10−210−6]−1f_0=[2\pi\sqrt{10^{-2}10^{-6}}]^{-1}

  3. Evaluate the expression; the result uses the units shown.

    Result=1591.549 Hz\mathrm{Result}=1591.549\ {\rm Hz}

Interpretation. Resistance sets resonance width and amplitude.

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Example 20. Battery ideal energy

Definitions & inputs. Nominal voltage 12 V, capacity 10 Ah.

  1. Choose the governing model and isolate the requested quantity.

    E=VQAhE=VQ_{Ah}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    E=12(10)E=12(10)

  3. Evaluate the expression; the result uses the units shown.

    Result=120 Wh\mathrm{Result}=120\ {\rm Wh}

Interpretation. Usable energy depends on discharge rate, temperature, and cutoff.

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Symbols and units

Each derivation and problem defines its own symbols and inputs. Symbols may be reused with different meanings in other subjects. Keep units consistent, retain sufficient precision during calculation, and apply the stated validity limits.