ME 123 — Electric Circuits
  • Charge, Current, Voltage, Power & Energy
  • KVL, KCL & Circuit Elements
  • Series/Parallel Networks & Dividers
  • Node-Voltage & Mesh-Current Analysis
  • Thevenin, Norton, Superposition & the Wheatstone Bridge
  • Capacitance & Inductance
  • First-Order Transients & DC Steady State
  • AC Signals & Phasors
  • Complex Impedance & AC Circuit Analysis
  • AC Power
  • Operational Amplifiers
  • Cheat Sheet

Cheat Sheet

Every formula, sign convention, and lookup value from all five tiers — sized for one double-sided letter sheet (the format confirmed on the S25 midterm cover page). Print it, don't read it.
SI Prefixes & Units
T / G / M / k101210^{12}1012 / 10910^{9}109 / 10610^{6}106 / 10310^{3}103
m / μ / n / p10−310^{-3}10−3 / 10−610^{-6}10−6 / 10−910^{-9}10−9 / 10−1210^{-12}10−12
AC/s   ·   V = J/C
WJ/s = V·A
S (siemens)Ω−1\Omega^{-1}Ω−1, conductance G=1/RG=1/RG=1/R
kW·h3.6×1063.6\times10^{6}3.6×106 J
Charge · Current · Voltage
Currenti(t)=dqdti(t)=\dfrac{dq}{dt}i(t)=dtdq​
Chargeq(t1)=∫t0t1 ⁣i dt+q(t0)q(t_1)=\displaystyle\int_{t_0}^{t_1}\! i\,dt + q(t_0)q(t1​)=∫t0​t1​​idt+q(t0​)
Constant iiiq=i Δtq = i\,\Delta tq=iΔt
Double subscriptvab=va−vbv_{ab}=v_a-v_bvab​=va​−vb​,  vba=−vabv_{ba}=-v_{ab}vba​=−vab​
ReferencePolarity/direction is a choice. Negative answer ⇒ true direction is opposite. Never re-draw, just report the sign.
Power & Passive Sign Convention
Powerp=v ip = v\,ip=vi  (W)
Energyw=∫t0t1 ⁣p dtw=\displaystyle\int_{t_0}^{t_1}\! p\,dtw=∫t0​t1​​pdt  (J)
PSCCurrent enters + terminal ⇒ p=vip=vip=vi is power absorbed
Active configCurrent exits + terminal ⇒ p=−vip=-vip=−vi; element supplies
Balance∑p=0\sum p = 0∑p=0 over the whole circuit (supplied = absorbed) — always use as a check
Ohm's Law & Resistors
Ohmv=iRv=iRv=iR,  i=Gvi=Gvi=Gv,  G=1/RG=1/RG=1/R
Dissipationp=vi=i2R=v2R  ≥0p = vi = i^2R = \dfrac{v^2}{R}\;\ge 0p=vi=i2R=Rv2​≥0 always
GeometryR=ρLAR = \dfrac{\rho L}{A}R=AρL​,  ρcopper=1.72×10−8 Ω ⁣⋅ ⁣\rho_{\text{copper}} = 1.72\times10^{-8}\ \Omega\!\cdot\!ρcopper​=1.72×10−8 Ω⋅m
Ideal wireR=0R=0R=0 ⇒ zero drop; both ends are the same node
Open circuiti=0i=0i=0 (any vvv) · Short: v=0v=0v=0 (any iii)
KCL & KVL
KCL∑iin=∑iout\sum i_{\text{in}}=\sum i_{\text{out}}∑iin​=∑iout​ at any node (charge conservation)
KVL∑v=0\sum v = 0∑v=0 around any closed loop (energy conservation)
KVL ruleTraverse the loop; use the first polarity sign encountered ( + ⇒ add, − ⇒ subtract )
Seriessame current through every element
Parallelsame voltage across every element
Series / Parallel — All Elements
R seriesReq=∑RkR_{eq}=\sum R_kReq​=∑Rk​
R parallel1Req=∑1Rk\dfrac{1}{R_{eq}}=\sum\dfrac{1}{R_k}Req​1​=∑Rk​1​; two: R1R2R1+R2\dfrac{R_1R_2}{R_1+R_2}R1​+R2​R1​R2​​
Lseries Leq=∑LkL_{eq}=\sum L_kLeq​=∑Lk​; parallel 1Leq=∑1Lk\frac1{L_{eq}}=\sum\frac1{L_k}Leq​1​=∑Lk​1​  (like R)
Cseries 1Ceq=∑1Ck\frac1{C_{eq}}=\sum\frac1{C_k}Ceq​1​=∑Ck​1​; parallel Ceq=∑CkC_{eq}=\sum C_kCeq​=∑Ck​  (opposite of R)
Zseries Zeq=∑ZkZ_{eq}=\sum Z_kZeq​=∑Zk​; parallel 1Zeq=∑1Zk\frac1{Z_{eq}}=\sum\frac1{Z_k}Zeq​1​=∑Zk​1​
Dividers
Voltage dividerVn=RnR1+⋯+RN VtotalV_n = \dfrac{R_n}{R_1+\cdots+R_N}\,V_{\text{total}}Vn​=R1​+⋯+RN​Rn​​Vtotal​  (series only)
Current divideri1=R2R1+R2 itotali_1 = \dfrac{R_2}{R_1+R_2}\,i_{\text{total}}i1​=R1​+R2​R2​​itotal​  (2 branches, parallel only)
General CDik=Gk∑G itotal=ReqRkitotali_k = \dfrac{G_k}{\sum G}\,i_{\text{total}} = \dfrac{R_{eq}}{R_k}i_{\text{total}}ik​=∑GGk​​itotal​=Rk​Req​​itotal​
WatchVD: bigger RRR gets more vvv. CD: bigger RRR gets less iii — the opposite resistor is on top.
AC formSame formulas with ZZZ in place of RRR.
Node-Voltage Analysis
1Pick reference (ground) — usually the − terminal of the largest source / most-connected node.
2Label v1,v2,…v_1, v_2,\dotsv1​,v2​,… at remaining nodes.
3KCL at each unknown node with ALL CURRENTS LEAVING — the course convention, keep it every time.
Branch termvthis−votherR\dfrac{v_{\text{this}}-v_{\text{other}}}{R}Rvthis​−vother​​  (this node's vvv first)
To groundvthisR\dfrac{v_{\text{this}}}{R}Rvthis​​
Source at nodeIf vsv_svs​ ties a node to ground: v1=vsv_1=v_sv1​=vs​, write no KCL there.
Supernodes & Dependent Sources
WhenA voltage source sits between two non-reference nodes (its current is unknown).
DoDraw a bubble around both nodes + the source; write one KCL for everything leaving the bubble.
Plus constraintKVL across the source: v(+ node)−v(− node)=Vsv_{(+\text{ node})}-v_{(-\text{ node})}=V_sv(+ node)​−v(− node)​=Vs​
Dependent srcTreat as an ordinary source, then add the control equation expressing vxv_xvx​ or ixi_xix​ in node voltages.
NeverNever deactivate a dependent source (superposition, RThR_{Th}RTh​ — see those cards).
Mesh-Current Analysis
1Assign a mesh current to every window pane, all clockwise (course convention).
2KVL around each mesh in the direction of its own current.
Own resistordrop =ikR= i_k R=ik​R
Shared resistordrop =(ik−iadj)R=(i_k - i_{\text{adj}})R=(ik​−iadj​)R — own current first
Current src in one meshik=±Isi_k = \pm I_sik​=±Is​ directly; skip that KVL.
Src between meshesSupermesh: KVL around the outside of both, plus ik−ij=Isi_k - i_j = I_sik​−ij​=Is​.
Thévenin & Norton
ThéveninVThV_{Th}VTh​ in series with RThR_{Th}RTh​
NortonINI_NIN​ in parallel with RNR_NRN​
VThV_{Th}VTh​=voc=v_{oc}=voc​ with the load removed
INI_NIN​=isc=i_{sc}=isc​ with the terminals shorted
RThR_{Th}RTh​=RN=vocisc=R_N=\dfrac{v_{oc}}{i_{sc}}=RN​=isc​voc​​
Source-kill RThR_{Th}RTh​Only if no dependent sources: VVV-src → short, III-src → open, then reduce series/parallel.
Superposition
ProcedureLeave one independent source on, deactivate the rest, solve; repeat; add all contributions with sign.
Deactivatevoltage source → short (0 V) · current source → open (0 A)
DependentALWAYS left active in every sub-circuit.
NOT for powerp∝i2p\propto i^2p∝i2 is nonlinear — sum iii first, then compute ppp.
Wheatstone Bridge
LayoutR1,R2R_1,R_2R1​,R2​ one leg; R3,R4R_3,R_4R3​,R4​ other leg; meter across the two midpoints.
BalancedR1R2=R3R4\dfrac{R_1}{R_2}=\dfrac{R_3}{R_4}R2​R1​​=R4​R3​​  ⟺  R1R4=R2R3R_1R_4=R_2R_3R1​R4​=R2​R3​
At balancevbridge=0v_{\text{bridge}}=0vbridge​=0, no current through the meter branch — it can be removed or shorted freely.
UnknownRx=RknownR2R1R_x = R_{\text{known}}\dfrac{R_2}{R_1}Rx​=Rknown​R1​R2​​
Capacitors
i–vi=Cdvdti = C\dfrac{dv}{dt}i=Cdtdv​
v from iv(t)=1C∫0t ⁣i dτ+v(0)v(t)=\dfrac1C\displaystyle\int_{0}^{t}\! i\,d\tau + v(0)v(t)=C1​∫0t​idτ+v(0)
Energyw=12Cv2w=\tfrac12 C v^2w=21​Cv2  (stored, never dissipated)
ContinuityvCv_CvC​ cannot jump: vC(0+)=vC(0−)v_C(0^+)=v_C(0^-)vC​(0+)=vC​(0−) (a jump needs infinite iii); iCi_CiC​ can jump freely.
DC steady stateopen circuit (dv/dt=0dv/dt=0dv/dt=0 ⇒ i=0i=0i=0)
Inductors
v–iv=Ldidtv = L\dfrac{di}{dt}v=Ldtdi​  (H)
i from vi(t)=1L∫0t ⁣v dτ+i(0)i(t)=\dfrac1L\displaystyle\int_{0}^{t}\! v\,d\tau + i(0)i(t)=L1​∫0t​vdτ+i(0)
Energyw=12Li2w=\tfrac12 L i^2w=21​Li2
ContinuityiLi_LiL​ cannot jump: iL(0+)=iL(0−)i_L(0^+)=i_L(0^-)iL​(0+)=iL​(0−) (a jump needs infinite vvv)
DC steady stateshort circuit (di/dt=0di/dt=0di/dt=0 ⇒ v=0v=0v=0)
First-Order Transients
Universalx(t)=x(∞)+[x(0+)−x(∞)]e−t/τx(t)=x(\infty)+\big[x(0^{+})-x(\infty)\big]e^{-t/\tau}x(t)=x(∞)+[x(0+)−x(∞)]e−t/τ
τ\tauτRC: τ=RThC\tau=R_{Th}Cτ=RTh​C  · RL: τ=L/RTh\tau=L/R_{Th}τ=L/RTh​
RThR_{Th}RTh​resistance seen by the C or L with sources deactivated, after the switch moves
Charge/dischargeCharging: vC=Vs(1−e−t/τ)v_C=V_s(1-e^{-t/\tau})vC​=Vs​(1−e−t/τ).  Discharging: vC=V0e−t/τv_C=V_0e^{-t/\tau}vC​=V0​e−t/τ.  Charging current: i=VsRe−t/τi=\frac{V_s}{R}e^{-t/\tau}i=RVs​​e−t/τ.
Transient Procedure & τ Table
1t<0t<0t<0: old circuit at DC steady state (C open, L short) ⇒ find vC(0−)v_C(0^-)vC​(0−) / iL(0−)i_L(0^-)iL​(0−).
2Continuity: vC(0+)=vC(0−)v_C(0^+)=v_C(0^-)vC​(0+)=vC​(0−), iL(0+)=iL(0−)i_L(0^+)=i_L(0^-)iL​(0+)=iL​(0−).
3t→∞t\to\inftyt→∞: new circuit at DC steady state ⇒ x(∞)x(\infty)x(∞).
4τ\tauτ from the new circuit's RThR_{Th}RTh​; substitute into the universal formula. Other quantities: get from vCv_CvC​/iLi_LiL​ via Ohm/KVL, not another exponential.
Solve for tt=−τln⁡ ⁣x(t)−x(∞)x(0+)−x(∞)t=-\tau\ln\!\dfrac{x(t)-x(\infty)}{x(0^{+})-x(\infty)}t=−τlnx(0+)−x(∞)x(t)−x(∞)​
AC Sinusoids & RMS
Course formv(t)=Vmsin⁡(ωt+θ)v(t)=V_m\sin(\omega t+\theta)v(t)=Vm​sin(ωt+θ)
ω\omegaω=2πf=2πT=2\pi f = \dfrac{2\pi}{T}=2πf=T2π​  (rad/s)
RMSVrms=Vm2=0.707 VmV_{rms}=\dfrac{V_m}{\sqrt2}=0.707\,V_mVrms​=2​Vm​​=0.707Vm​  (sinusoids only)
Phase lead/lagϕ=θv−θi\phi = \theta_v-\theta_iϕ=θv​−θi​.  ϕ>0\phi>0ϕ>0: vvv leads iii (inductive). ϕ<0\phi<0ϕ<0: iii leads vvv (capacitive).  Δϕ=ωΔt\Delta\phi=\omega\Delta tΔϕ=ωΔt.
Phasors
PhasorV=V∠θ\mathbf{V}=V\angle\thetaV=V∠θ — magnitude + phase at a fixed ω\omegaω; time is dropped.
MUST DO FIRSTPut every source in one reference form (all sine or all cosine) before reading off angles.
Identitysin⁡(ωt+θ)=cos⁡(ωt+θ−90∘)\sin(\omega t+\theta)=\cos(\omega t+\theta-90^\circ)sin(ωt+θ)=cos(ωt+θ−90∘); −sin⁡(x)=sin⁡(x±180∘)-\sin(x)=\sin(x\pm180^\circ)−sin(x)=sin(x±180∘) — a minus sign is a 180∘180^\circ180∘ shift.
MagnitudeState whether you're using peak or RMS phasors and stay consistent; AC power formulas below assume RMS.
Complex Arithmetic
Engineeringj=−1j=\sqrt{-1}j=−1​ (never iii — that's current)
Rect → polar∣Z∣=R2+X2|Z|=\sqrt{R^2+X^2}∣Z∣=R2+X2​, ∠Z=arctan⁡ ⁣XR\angle Z=\arctan\!\dfrac{X}{R}∠Z=arctanRX​ (if R<0R<0R<0, add 180∘180^\circ180∘ — calculators only return ±90∘\pm90^\circ±90∘)
Polar → rectZ=∣Z∣cos⁡θ+j∣Z∣sin⁡θZ=|Z|\cos\theta + j|Z|\sin\thetaZ=∣Z∣cosθ+j∣Z∣sinθ
+/− vs ×/÷+/− use rectangular. ×/÷ use polar: multiply/divide magnitudes, add/subtract angles.
jjj powersj2=−1j^2=-1j2=−1,  1j=−j\dfrac1j=-jj1​=−j,  j=1∠90∘j=1\angle 90^\circj=1∠90∘
Complex Impedance
Phasor OhmV=ZI\mathbf{V}=\mathbf{Z}\mathbf{I}V=ZI
ResistorZR=R=R∠0∘Z_R=R=R\angle 0^\circZR​=R=R∠0∘ — vvv, iii in phase
InductorZL=jωL=ωL∠90∘Z_L=j\omega L=\omega L\angle 90^\circZL​=jωL=ωL∠90∘ — vvv leads iii by 90∘90^\circ90∘
CapacitorZC=−jωC=1ωC∠ ⁣− ⁣90∘Z_C=\dfrac{-j}{\omega C}=\dfrac{1}{\omega C}\angle\!-\!90^\circZC​=ωC−j​=ωC1​∠−90∘ — iii leads vvv by 90∘90^\circ90∘
ReactanceXL=ωLX_L=\omega LXL​=ωL,  XC=−1ωCX_C=-\dfrac{1}{\omega C}XC​=−ωC1​;  Z=R+jXZ=R+jXZ=R+jX;  Y=1/Z=G+jBY=1/Z=G+jBY=1/Z=G+jB (S). ω ⁣→ ⁣0\omega\!\to\!0ω→0: ZL ⁣→ ⁣0Z_L\!\to\!0ZL​→0, ∣ZC∣ ⁣→ ⁣∞|Z_C|\!\to\!\infty∣ZC​∣→∞ (matches DC). ω ⁣→ ⁣∞\omega\!\to\!\inftyω→∞: reverse.
AC Circuit Analysis
1Fix ω\omegaω; convert every source to a phasor and every R/L/C to an impedance.
2Every DC technique applies unchanged with ZZZ for RRR: series/parallel, dividers, node, mesh, Thévenin, superposition.
3Do the complex algebra (polar for ×/÷, rectangular for +/−).
ZThZ_{Th}ZTh​same rules as RThR_{Th}RTh​, complex-valued. Series RLC: Z=R+j(ωL−1ωC)Z=R+j(\omega L-\frac1{\omega C})Z=R+j(ωL−ωC1​).
Resonanceω0=1LC\omega_0=\dfrac{1}{\sqrt{LC}}ω0​=LC​1​: XXX cancels, Z=RZ=RZ=R (purely real, PF=1=1=1). ∣VL∣,∣VC∣|\mathbf V_L|,|\mathbf V_C|∣VL​∣,∣VC​∣ can each exceed the source — normal, not an error.
AC Power (RMS values)
RealP=VrmsIrmscos⁡ϕP=V_{rms}I_{rms}\cos\phiP=Vrms​Irms​cosϕ  (W) — only RRR dissipates
ReactiveQ=VrmsIrmssin⁡ϕQ=V_{rms}I_{rms}\sin\phiQ=Vrms​Irms​sinϕ  (VAR) — stored/returned by L and C
ApparentS=VrmsIrms=P2+Q2S=V_{rms}I_{rms}=\sqrt{P^2+Q^2}S=Vrms​Irms​=P2+Q2​  (VA)
⚠ Peak vs RMSThese need RMS. With peak phasors, P=12VmImcos⁡ϕP=\tfrac12 V_mI_m\cos\phiP=21​Vm​Im​cosϕ.
Power factorPF=cos⁡ϕ=PS\mathrm{PF}=\cos\phi=\dfrac{P}{S}PF=cosϕ=SP​,  0≤PF≤10\le \mathrm{PF}\le 10≤PF≤1. Lagging = inductive (iii lags vvv); leading = capacitive (iii leads vvv).
Complex Power & Triangle
Complex powerS=VrmsIrms∗=P+jQ=S∠ϕ\mathbf{S}=\mathbf{V}_{rms}\mathbf{I}^{*}_{rms}=P+jQ = S\angle\phiS=Vrms​Irms∗​=P+jQ=S∠ϕ
NoteThe conjugate on I\mathbf{I}I is what makes ∠S=θv−θi\angle\mathbf{S}=\theta_v-\theta_i∠S=θv​−θi​. Forgetting it flips the sign of QQQ.
ConservationPPP and QQQ each add over all elements: ∑P=0\sum P=0∑P=0, ∑Q=0\sum Q=0∑Q=0.  SSS does not add arithmetically.
⚠ Q signWith ϕ=θv−θi\phi=\theta_v-\theta_iϕ=θv​−θi​ and S=VrmsIrms∗\mathbf{S}=\mathbf{V}_{rms}\mathbf{I}^{*}_{rms}S=Vrms​Irms∗​: Q>0Q>0Q>0 for an inductive (lagging) load, Q<0Q<0Q<0 for capacitive (leading) — matches the course's own worked example (V=120∠30°\mathbf{V}=120\angle30°V=120∠30°, I=10∠0°\mathbf{I}=10\angle0°I=10∠0°, lagging, Q=+600Q=+600Q=+600 VAR). One slide bullet states the opposite rule in words; trust this formula and the worked numbers over that bullet.
PF correctionAdd C in parallel to cancel inductive QQQ; PPP unchanged, SSS drops, PF→1\mathrm{PF}\to1PF→1.
Ideal Op-Amp
RinR_{in}Rin​∞\infty∞ ⇒ no current into either input
RoutR_{out}Rout​000 ⇒ ideal voltage source at the output
Golden rulesWith negative feedback: v+=v−v_+=v_-v+​=v−​ (virtual short) and i+=i−=0i_+=i_-=0i+​=i−​=0.
Virtual groundIf +++ is tied to ground, then v−=0v_-=0v−​=0 V — but it is not a real ground; no current sinks into it.
SaturationOutput clips at the supply rails ±VCC\pm V_{CC}±VCC​; the golden rules stop holding there.
Inverting Amplifier
Circuit+++ to ground; vinv_{in}vin​ through RinR_{in}Rin​ to the −-− node; RfR_fRf​ from output back to −-−.
GainAv=voutvin=−RfRinA_v=\dfrac{v_{out}}{v_{in}}=-\dfrac{R_f}{R_{in}}Av​=vin​vout​​=−Rin​Rf​​
Minus sign180∘180^\circ180∘ inversion — positive in, negative out. Do not drop it.
ExampleRf=10R_f=10Rf​=10 kΩ\OmegaΩ, Rin=1R_{in}=1Rin​=1 kΩ\OmegaΩ ⇒ Av=−10A_v=-10Av​=−10; 1 V in gives −10-10−10 V out. Always verify ∣vout∣≤VCC|v_{out}|\le V_{CC}∣vout​∣≤VCC​, else it saturates.
Exam Gotchas — Final Check
SignsPSC: current into +++ ⇒ absorbing. Node KCL: all currents leaving. Mesh: all clockwise. Deactivating: V-src→short, I-src→open, dependent sources never.
Units & modeConvert kΩ/μF/mH before substituting (kΩ×mA=V; kΩ×μF=ms). Check degrees-vs-radians before every phasor problem.
Continuity & RMSvCv_CvC​, iLi_LiL​ continuous at t=0t=0t=0, everything else can jump. Power formulas want RMS, not peak. Superposition never for power directly.
VerifyKCL at one node + KVL around one loop + ∑p=0\sum p = 0∑p=0 catches most arithmetic slips.

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