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CIR:010117The 3-2-6 ohm delta has wye legs 6/11, 18/11, 12/11 ohm and open-C voltage 9 V

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Circuits 1 Delta Wye IdentitiesCircuits 1 Unloaded Star Node Voltage

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Voltage, current, power, and resistive networksCircuits I · Mixed review

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CIR:010118The 2-4 vs 3-6 bridge is balanced at 8 V; the 9 ohm load gives Ig=-12/115 A

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Circuits 1 Wheatstone Balance RatioCircuits 1 Loaded Bridge Two Node Kcl

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Voltage, current, power, and resistive networksCircuits I · Mixed review

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CIR:010132The 6 ohm resistor cube between opposite corners has Req=5 ohm; 12 V draws 12/5 A with outer 4/5 A, middle 2/5 A, and power 144/5 W

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Circuits 1 Cube Corner EquipotentialsCircuits 1 Three Layer Resistor Reduction

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Voltage, current, power, and resistive networksCircuits I · Mixed review

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CIR:010133The infinite 1 ohm series-then-shunt ladder locks to phi=(1+sqrt5)/2; 10 V draws 5(sqrt5-1) A, first node 5(3-sqrt5) V, power 50(sqrt5-1)

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Circuits 1 Infinite Ladder Fixed PointCircuits 1 Positive Resistance Root Selection

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Voltage, current, power, and resistive networksCircuits I · Mixed review

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CIR:010134The 12 V, 6k/3k divider with a 6k meter reads 3 V; unloaded 4 V, currents 1.5/0.5/1 mA, error -1/4, meter 1.5 mW

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Circuits 1 Unloaded Resistive DividerCircuits 1 Voltmeter Shunt Loading

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Voltage, current, power, and resistive networksCircuits I · Mixed review

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CIR:010219The floating 10 V source and 3 A injection force va=56/5 V and vb=6/5 V

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Circuits 1 Supernode KclCircuits 1 Floating Source Constraint

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Node, mesh, superposition, and equivalent circuitsCircuits I · Mixed review

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CIR:010220Clockwise supermesh currents 2/5 A and 7/5 A put 42/5 V on the 1 A source

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Circuits 1 Supermesh KvlCircuits 1 Shared Source Voltage Audit

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Node, mesh, superposition, and equivalent circuitsCircuits I · Mixed review

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CIR:010221The 3 ohm absorbs 64/3 W; the separate-source powers miss the 32/3 W cross term

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Circuits 1 Source SuperpositionCircuits 1 Quadratic Power Cross Term

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Node, mesh, superposition, and equivalent circuitsCircuits I · Mixed review

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CIR:010235Millman node via 2 ohm to +10, 5 ohm to -5, 10 ohm to ground is 5 V; leaving (-5/2,2,1/2) A; resistors 25/2+20+5/2=35 W; sources deliver 25 W and 10 W

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Circuits 1 Millman Conductance Weighted NodeCircuits 1 Signed Source Resistor Power Audit

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Node, mesh, superposition, and equivalent circuitsCircuits I · Mixed review

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CIR:010236Clockwise CCVS meshes give i1=30/11 A and i2=5/11 A; shared 25/11 A, vd=30/11 V; CCVS absorbs 150/121 W, resistors 3150/121 W, source delivers 3300/121 W

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Circuits 1 Ccvs Two Mesh KvlCircuits 1 Controlled Source Power Split

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Node, mesh, superposition, and equivalent circuitsCircuits I · Mixed review

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CIR:010322The inverting summer yields vo=-1 V and the output pin sinks 7/12 mA

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Circuits 1 Inverting Weighted SumCircuits 1 Opamp Output Pin Current

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Ideal operational-amplifier circuitsCircuits I · Mixed review

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CIR:010323Matched ratios give vo=4(v2-v1); R4=30 kΩ yields |CMRR|=31/2

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Circuits 1 Difference Amp Matched GainCircuits 1 Diffamp Cmrr From Mismatch

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Ideal operational-amplifier circuitsCircuits I · Mixed review

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CIR:010337The 40 μA photodiode TIA with Rf=100 kΩ and 10 kΩ load has v-=0, vo=-4 V; the output sinks 0.44 mA and the powers are 1.6 mW and 160 μW

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Circuits 1 Transimpedance Virtual GroundCircuits 1 Opamp Output Pin Current Split

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Ideal operational-amplifier circuitsCircuits I · Mixed review

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CIR:010338The three-opamp in-amp with R=10 kΩ and Rg=5 kΩ maps (1.2 V, 1.8 V) to vo1=0, vo2=3 V, vo3=3 V with differential gain 5 and common-mode gain 0

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Circuits 1 Instrumentation First Stage BridgeCircuits 1 Unity Subtractor Common Mode Null

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Ideal operational-amplifier circuitsCircuits I · Mixed review

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CIR:010339The ideal integrator with RC=1 s and vo(0)=1 V maps the 2 then -1 V input to 1-2t then t-2, zeros at 1/2 and 2, and vo(3)=1 V

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Circuits 1 Ideal Inverting IntegratorCircuits 1 Piecewise State Continuity

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Ideal operational-amplifier circuitsCircuits I · Mixed review

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CIR:010424The RC pulse peaks at t=2 with height 4(1-e^{-2}) and area 8 V s

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Circuits 1 RC Pulse Piecewise ResponseCircuits 1 First Order Area Identity

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Capacitors, inductors, and first-order transientsCircuits I · Mixed review

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CIR:010425The RL equal on/off ripple locks to 1 A low and 2 A high, mean 3/2 A

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Circuits 1 RL Periodic Exponential MapCircuits 1 Inductor Ripple Mean Energy

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Capacitors, inductors, and first-order transientsCircuits I · Mixed review

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CIR:010426Charge sharing ends at 2 V with 12 μJ lost; R→0 is singular, not lossless

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Circuits 1 Capacitor Charge SharingCircuits 1 Resistive Energy Deficit

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Capacitors, inductors, and first-order transientsCircuits I · Mixed review

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CIR:010440The relaxed RC ramp vs=t yields vC=t-1+e^{-t}; at t=1, vC=e^{-1}, capacitor energy e^{-2}/2, and resistor energy -1/2+2/e-1/(2e^2)

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Circuits 1 RC Ramp Forced ResponseCircuits 1 Resistor Capacitor Energy Split

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Capacitors, inductors, and first-order transientsCircuits I · Mixed review

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CIR:010441The RL turn-on of 10 cos t with R=3 Ω and L=1 H gives i=3cos t+sin t-3e^{-3t}, i'(0)=10, steady peak √10, and resistor average 15 W

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Circuits 1 RL Cosine Turn ONCircuits 1 RL Steady Phasor Average Power

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Capacitors, inductors, and first-order transientsCircuits I · Mixed review

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CIR:010527Source-free parallel LC exchanges a constant 5/2 J between C=1 F and L=4 H

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Parallel LC Undamped OscillationLC Energy Exchange Conservation

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Second-order natural and step responsesCircuits I · Mixed review

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CIR:010528Critical parallel RLC: v=4(1+t)e^{-2t} and iC+iR+iL is identically zero

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Parallel Rlc Critical DampingBranch Current Kcl Audit

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Second-order natural and step responsesCircuits I · Mixed review

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CIR:010542Series RLC poles -3±4j force R=6 Ω and C=1/25 F; the 10 V step is 10[1-e^{-3t}(cos4t+(3/4)sin4t)] with first peak 10(1+e^{-3π/4})

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Circuits 1 Series Rlc Pole TO Element MapCircuits 1 Underdamped Step Peak Overshoot

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Second-order natural and step responsesCircuits I · Mixed review

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CIR:010543Ringdown period π/2 and same-sign ratio e^π identify ωd=4, α=2, R=4 Ω, C=1/20 F, poles -2±4j, ζ=1/√5, Q=√5/2

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Circuits 1 Ringdown Period Decrement IdentificationCircuits 1 Series Rlc Damping Q From Poles

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Second-order natural and step responsesCircuits I · Mixed review

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CIR:010544A 6 V s series impulse on relaxed R=4 Ω, L=2 H, C=1/10 F forces i(0+)=3 A, i=e^{-t}(3cos2t-(3/2)sin2t), vC=15e^{-t}sin2t, energy 9 J

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Circuits 1 Series Inductor Voltage Impulse JumpCircuits 1 Source Free Series Rlc Impulse Response

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Second-order natural and step responsesCircuits I · Mixed review

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CIR:010629Balanced Y: S3=3456+j2592 VA at 0.8 lag, VAB=120sqrt3∠30°, ZΔ=24+j18

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Balanced Y Positive SequenceThree Phase Complex Power

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Phasors, sinusoidal steady state, and AC powerCircuits I · Mixed review

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CIR:010630Ideal 2:1 transformer reflects 12+j16 Ω and Sin=SL=12+j16 VA

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Ideal Transformer Turns RatioImpedance Reflection Power

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Phasors, sinusoidal steady state, and AC powerCircuits I · Mixed review

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CIR:010631Conjugate match delivers 25 W; the best resistor is 5 Ω at 200/9 W

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Conjugate Match Maximum PowerResistive Load Optimum

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Phasors, sinusoidal steady state, and AC powerCircuits I · Mixed review

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CIR:010645Four-wire unbalanced Y: IA=10 A, IB=-5/2-j5√3/2 A, IC=-15/2+j15√3/2 A, IN=j5√3 A, powers 1200+600+1800=3600 W

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Circuits 1 Four Wire Unbalanced Y Phase CurrentsCircuits 1 Neutral Return And Phase Power Sum

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Phasors, sinusoidal steady state, and AC powerCircuits I · Mixed review

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CIR:010646Unbalanced delta lines IA=16-j4√3, IB=-15-j3√3, IC=-1+j7√3 A with magnitudes 4√19, 6√7, 2√37 and powers 1440+720+960=3120 W

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Circuits 1 Unbalanced Delta Phase And Line CurrentsCircuits 1 Delta Line Current Kcl And Resistor Power

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Phasors, sinusoidal steady state, and AC powerCircuits I · Mixed review

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908011Three resistors are connected in series between node a and node b: R_1=10\,\Omega from a to node c, R_2=15\,\Omega…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908012Four resistors are connected in parallel between node a and the reference node 0: R_1=8\,\Omega, R_2=24\,\Omega, R_…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908013A series string runs from node a to the reference node 0: R_1=20\,\mathrm{k}\Omega from a to node b, R_2=30\,\mathr…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908014An ideal independent current source i_s=36\,\mathrm{mA} is connected from the reference node 0 to node a, with the…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908015Between node a and node b, resistor R_1=9\,\Omega is connected from a to an intermediate node c, and resistors R_2=…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908016An ideal independent voltage source of 15\,\mathrm{V} is connected from the reference node 0 to node a, with its po…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908017A single loop is formed by four elements in series.

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908018An ideal independent current source of 6\,\mathrm{mA} is connected from the reference node 0 to node a, with the so…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908019A single loop contains three elements in series with an independent source.

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908020An ideal independent voltage source of 12\,\mathrm{V} is connected from the reference node 0 to node a, with its po…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Problem library

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Topic choices for Circuits I are shortened. Search still covers every problem.

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CIR:010117The 3-2-6 ohm delta has wye legs 6/11, 18/11, 12/11 ohm and open-C voltage 9 V

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Circuits 1 Delta Wye IdentitiesCircuits 1 Unloaded Star Node Voltage

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Voltage, current, power, and resistive networksCircuits I · Mixed review

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CIR:010118The 2-4 vs 3-6 bridge is balanced at 8 V; the 9 ohm load gives Ig=-12/115 A

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Circuits 1 Wheatstone Balance RatioCircuits 1 Loaded Bridge Two Node Kcl

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Voltage, current, power, and resistive networksCircuits I · Mixed review

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CIR:010132The 6 ohm resistor cube between opposite corners has Req=5 ohm; 12 V draws 12/5 A with outer 4/5 A, middle 2/5 A, and power 144/5 W

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Circuits 1 Cube Corner EquipotentialsCircuits 1 Three Layer Resistor Reduction

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Voltage, current, power, and resistive networksCircuits I · Mixed review

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CIR:010133The infinite 1 ohm series-then-shunt ladder locks to phi=(1+sqrt5)/2; 10 V draws 5(sqrt5-1) A, first node 5(3-sqrt5) V, power 50(sqrt5-1)

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Circuits 1 Infinite Ladder Fixed PointCircuits 1 Positive Resistance Root Selection

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Voltage, current, power, and resistive networksCircuits I · Mixed review

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CIR:010134The 12 V, 6k/3k divider with a 6k meter reads 3 V; unloaded 4 V, currents 1.5/0.5/1 mA, error -1/4, meter 1.5 mW

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Circuits 1 Unloaded Resistive DividerCircuits 1 Voltmeter Shunt Loading

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Voltage, current, power, and resistive networksCircuits I · Mixed review

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CIR:010219The floating 10 V source and 3 A injection force va=56/5 V and vb=6/5 V

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Circuits 1 Supernode KclCircuits 1 Floating Source Constraint

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Node, mesh, superposition, and equivalent circuitsCircuits I · Mixed review

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CIR:010220Clockwise supermesh currents 2/5 A and 7/5 A put 42/5 V on the 1 A source

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Circuits 1 Supermesh KvlCircuits 1 Shared Source Voltage Audit

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Node, mesh, superposition, and equivalent circuitsCircuits I · Mixed review

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CIR:010221The 3 ohm absorbs 64/3 W; the separate-source powers miss the 32/3 W cross term

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Circuits 1 Source SuperpositionCircuits 1 Quadratic Power Cross Term

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Node, mesh, superposition, and equivalent circuitsCircuits I · Mixed review

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CIR:010235Millman node via 2 ohm to +10, 5 ohm to -5, 10 ohm to ground is 5 V; leaving (-5/2,2,1/2) A; resistors 25/2+20+5/2=35 W; sources deliver 25 W and 10 W

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Circuits 1 Millman Conductance Weighted NodeCircuits 1 Signed Source Resistor Power Audit

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Node, mesh, superposition, and equivalent circuitsCircuits I · Mixed review

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CIR:010236Clockwise CCVS meshes give i1=30/11 A and i2=5/11 A; shared 25/11 A, vd=30/11 V; CCVS absorbs 150/121 W, resistors 3150/121 W, source delivers 3300/121 W

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Circuits 1 Ccvs Two Mesh KvlCircuits 1 Controlled Source Power Split

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Node, mesh, superposition, and equivalent circuitsCircuits I · Mixed review

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CIR:010322The inverting summer yields vo=-1 V and the output pin sinks 7/12 mA

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Circuits 1 Inverting Weighted SumCircuits 1 Opamp Output Pin Current

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Ideal operational-amplifier circuitsCircuits I · Mixed review

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CIR:010323Matched ratios give vo=4(v2-v1); R4=30 kΩ yields |CMRR|=31/2

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Circuits 1 Difference Amp Matched GainCircuits 1 Diffamp Cmrr From Mismatch

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Ideal operational-amplifier circuitsCircuits I · Mixed review

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CIR:010337The 40 μA photodiode TIA with Rf=100 kΩ and 10 kΩ load has v-=0, vo=-4 V; the output sinks 0.44 mA and the powers are 1.6 mW and 160 μW

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Circuits 1 Transimpedance Virtual GroundCircuits 1 Opamp Output Pin Current Split

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Ideal operational-amplifier circuitsCircuits I · Mixed review

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CIR:010338The three-opamp in-amp with R=10 kΩ and Rg=5 kΩ maps (1.2 V, 1.8 V) to vo1=0, vo2=3 V, vo3=3 V with differential gain 5 and common-mode gain 0

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Circuits 1 Instrumentation First Stage BridgeCircuits 1 Unity Subtractor Common Mode Null

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Ideal operational-amplifier circuitsCircuits I · Mixed review

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CIR:010339The ideal integrator with RC=1 s and vo(0)=1 V maps the 2 then -1 V input to 1-2t then t-2, zeros at 1/2 and 2, and vo(3)=1 V

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Circuits 1 Ideal Inverting IntegratorCircuits 1 Piecewise State Continuity

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Ideal operational-amplifier circuitsCircuits I · Mixed review

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CIR:010424The RC pulse peaks at t=2 with height 4(1-e^{-2}) and area 8 V s

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Circuits 1 RC Pulse Piecewise ResponseCircuits 1 First Order Area Identity

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Capacitors, inductors, and first-order transientsCircuits I · Mixed review

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CIR:010425The RL equal on/off ripple locks to 1 A low and 2 A high, mean 3/2 A

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Circuits 1 RL Periodic Exponential MapCircuits 1 Inductor Ripple Mean Energy

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Capacitors, inductors, and first-order transientsCircuits I · Mixed review

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CIR:010426Charge sharing ends at 2 V with 12 μJ lost; R→0 is singular, not lossless

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Circuits 1 Capacitor Charge SharingCircuits 1 Resistive Energy Deficit

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Capacitors, inductors, and first-order transientsCircuits I · Mixed review

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CIR:010440The relaxed RC ramp vs=t yields vC=t-1+e^{-t}; at t=1, vC=e^{-1}, capacitor energy e^{-2}/2, and resistor energy -1/2+2/e-1/(2e^2)

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Circuits 1 RC Ramp Forced ResponseCircuits 1 Resistor Capacitor Energy Split

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Capacitors, inductors, and first-order transientsCircuits I · Mixed review

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CIR:010441The RL turn-on of 10 cos t with R=3 Ω and L=1 H gives i=3cos t+sin t-3e^{-3t}, i'(0)=10, steady peak √10, and resistor average 15 W

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Circuits 1 RL Cosine Turn ONCircuits 1 RL Steady Phasor Average Power

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Capacitors, inductors, and first-order transientsCircuits I · Mixed review

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CIR:010527Source-free parallel LC exchanges a constant 5/2 J between C=1 F and L=4 H

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Parallel LC Undamped OscillationLC Energy Exchange Conservation

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Second-order natural and step responsesCircuits I · Mixed review

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CIR:010528Critical parallel RLC: v=4(1+t)e^{-2t} and iC+iR+iL is identically zero

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Parallel Rlc Critical DampingBranch Current Kcl Audit

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Second-order natural and step responsesCircuits I · Mixed review

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CIR:010542Series RLC poles -3±4j force R=6 Ω and C=1/25 F; the 10 V step is 10[1-e^{-3t}(cos4t+(3/4)sin4t)] with first peak 10(1+e^{-3π/4})

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Circuits 1 Series Rlc Pole TO Element MapCircuits 1 Underdamped Step Peak Overshoot

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Second-order natural and step responsesCircuits I · Mixed review

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CIR:010543Ringdown period π/2 and same-sign ratio e^π identify ωd=4, α=2, R=4 Ω, C=1/20 F, poles -2±4j, ζ=1/√5, Q=√5/2

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Circuits 1 Ringdown Period Decrement IdentificationCircuits 1 Series Rlc Damping Q From Poles

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Second-order natural and step responsesCircuits I · Mixed review

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CIR:010544A 6 V s series impulse on relaxed R=4 Ω, L=2 H, C=1/10 F forces i(0+)=3 A, i=e^{-t}(3cos2t-(3/2)sin2t), vC=15e^{-t}sin2t, energy 9 J

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Circuits 1 Series Inductor Voltage Impulse JumpCircuits 1 Source Free Series Rlc Impulse Response

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Second-order natural and step responsesCircuits I · Mixed review

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CIR:010629Balanced Y: S3=3456+j2592 VA at 0.8 lag, VAB=120sqrt3∠30°, ZΔ=24+j18

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Balanced Y Positive SequenceThree Phase Complex Power

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Phasors, sinusoidal steady state, and AC powerCircuits I · Mixed review

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CIR:010630Ideal 2:1 transformer reflects 12+j16 Ω and Sin=SL=12+j16 VA

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Ideal Transformer Turns RatioImpedance Reflection Power

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Phasors, sinusoidal steady state, and AC powerCircuits I · Mixed review

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CIR:010631Conjugate match delivers 25 W; the best resistor is 5 Ω at 200/9 W

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Conjugate Match Maximum PowerResistive Load Optimum

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Phasors, sinusoidal steady state, and AC powerCircuits I · Mixed review

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CIR:010645Four-wire unbalanced Y: IA=10 A, IB=-5/2-j5√3/2 A, IC=-15/2+j15√3/2 A, IN=j5√3 A, powers 1200+600+1800=3600 W

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Circuits 1 Four Wire Unbalanced Y Phase CurrentsCircuits 1 Neutral Return And Phase Power Sum

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Phasors, sinusoidal steady state, and AC powerCircuits I · Mixed review

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CIR:010646Unbalanced delta lines IA=16-j4√3, IB=-15-j3√3, IC=-1+j7√3 A with magnitudes 4√19, 6√7, 2√37 and powers 1440+720+960=3120 W

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Circuits 1 Unbalanced Delta Phase And Line CurrentsCircuits 1 Delta Line Current Kcl And Resistor Power

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Phasors, sinusoidal steady state, and AC powerCircuits I · Mixed review

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908011Three resistors are connected in series between node a and node b: R_1=10\,\Omega from a to node c, R_2=15\,\Omega…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908012Four resistors are connected in parallel between node a and the reference node 0: R_1=8\,\Omega, R_2=24\,\Omega, R_…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908013A series string runs from node a to the reference node 0: R_1=20\,\mathrm{k}\Omega from a to node b, R_2=30\,\mathr…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908014An ideal independent current source i_s=36\,\mathrm{mA} is connected from the reference node 0 to node a, with the…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908015Between node a and node b, resistor R_1=9\,\Omega is connected from a to an intermediate node c, and resistors R_2=…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908016An ideal independent voltage source of 15\,\mathrm{V} is connected from the reference node 0 to node a, with its po…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908017A single loop is formed by four elements in series.

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908018An ideal independent current source of 6\,\mathrm{mA} is connected from the reference node 0 to node a, with the so…

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908019A single loop contains three elements in series with an independent source.

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Voltage, current, power, and resistive networksCircuits I · Calculation

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CIR:9908020An ideal independent voltage source of 12\,\mathrm{V} is connected from the reference node 0 to node a, with its po…

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Laws ResistiveCircuits 1 Problem Solving

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Voltage, current, power, and resistive networksCircuits I · Calculation

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Problem results

CIR:9908001The current entering a conductor at terminal a is i(t)=6e^{-3t} A for t\ge 0, with t in seconds.

CIR:9908002The charge that has entered a device through terminal p is q(t)=(2t^3-9t^2+12t) mC, with t in seconds.

CIR:9908003The current entering a two-terminal element at the terminal marked plus is The charge that has entered the plus ter…

CIR:9908004A charge q=50\,\mu\mathrm{C} is moved slowly from node n to node p.

CIR:9908005A two-terminal element has voltage v_{ab}(t)=8\cos(50t)\,\mathrm{V}, with the plus reference at terminal a and the…

CIR:9908006Element X has terminals a and b.

CIR:9908007An ideal independent voltage source v_s=20\,\mathrm{V} is connected between node a and the reference node 0, with i…

CIR:9908008A resistor R=1.5\,\mathrm{k}\Omega is connected between node a and node b.

CIR:9908009Three nodes a, b, and the reference node 0 are joined by five branches.

CIR:9908010A single closed loop contains four elements A, B, C, and D in that clockwise order.

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Problem results

CIR:9908001The current entering a conductor at terminal a is i(t)=6e^{-3t} A for t\ge 0, with t in seconds.

CIR:9908002The charge that has entered a device through terminal p is q(t)=(2t^3-9t^2+12t) mC, with t in seconds.

CIR:9908003The current entering a two-terminal element at the terminal marked plus is The charge that has entered the plus ter…

CIR:9908004A charge q=50\,\mu\mathrm{C} is moved slowly from node n to node p.

CIR:9908005A two-terminal element has voltage v_{ab}(t)=8\cos(50t)\,\mathrm{V}, with the plus reference at terminal a and the…

CIR:9908006Element X has terminals a and b.

CIR:9908007An ideal independent voltage source v_s=20\,\mathrm{V} is connected between node a and the reference node 0, with i…

CIR:9908008A resistor R=1.5\,\mathrm{k}\Omega is connected between node a and node b.

CIR:9908009Three nodes a, b, and the reference node 0 are joined by five branches.

CIR:9908010A single closed loop contains four elements A, B, C, and D in that clockwise order.