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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

Circuits 1 Delta Wye IdentitiesCircuits 1 Unloaded Star Node Voltage

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

Circuits 1 Wheatstone Balance RatioCircuits 1 Loaded Bridge Two Node Kcl

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

Circuits 1 Cube Corner EquipotentialsCircuits 1 Three Layer Resistor Reduction

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)

Circuits 1 Infinite Ladder Fixed PointCircuits 1 Positive Resistance Root Selection

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

Circuits 1 Unloaded Resistive DividerCircuits 1 Voltmeter Shunt Loading

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

Circuits 1 Supernode KclCircuits 1 Floating Source Constraint

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

Circuits 1 Supermesh KvlCircuits 1 Shared Source Voltage Audit

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

Circuits 1 Source SuperpositionCircuits 1 Quadratic Power Cross Term

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

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

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

Circuits 1 Ccvs Two Mesh KvlCircuits 1 Controlled Source Power Split

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

Circuits 1 Inverting Weighted SumCircuits 1 Opamp Output Pin Current

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

Circuits 1 Difference Amp Matched GainCircuits 1 Diffamp Cmrr From Mismatch

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

Circuits 1 Transimpedance Virtual GroundCircuits 1 Opamp Output Pin Current Split

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

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

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

Circuits 1 Ideal Inverting IntegratorCircuits 1 Piecewise State Continuity

Ideal operational-amplifier circuitsCircuits I · Mixed review

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CIR:010424Mathematical problem title

Circuits 1 RC Pulse Piecewise ResponseCircuits 1 First Order Area Identity

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

Circuits 1 RL Periodic Exponential MapCircuits 1 Inductor Ripple Mean Energy

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

Circuits 1 Capacitor Charge SharingCircuits 1 Resistive Energy Deficit

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

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CIR:010440Mathematical problem title

Circuits 1 RC Ramp Forced ResponseCircuits 1 Resistor Capacitor Energy Split

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

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CIR:010441Mathematical problem title

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

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

Parallel LC Undamped OscillationLC Energy Exchange Conservation

Second-order natural and step responsesCircuits I · Mixed review

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CIR:010528Mathematical problem title

Parallel Rlc Critical DampingBranch Current Kcl Audit

Second-order natural and step responsesCircuits I · Mixed review

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CIR:010542Mathematical problem title

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

Second-order natural and step responsesCircuits I · Mixed review

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CIR:010543Mathematical problem title

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

Second-order natural and step responsesCircuits I · Mixed review

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Showing 24 of 330 matching problems.

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The RC pulse peaks at t=2 with height

4(1-e^{-2})

and area 8 V s

The 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)

The 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

Critical parallel RLC:

v=4(1+t)e^{-2t}

and iC+iR+iL is identically zero

Series 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})

Ringdown 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