Circuit Analysis (DC and AC Steady State)

Given a 12 V source and series resistors 2 kΩ and 4 kΩ, find the voltage across the 4 kΩ.

Series divider.
V_{4k} = 12\cdot\frac{4}{2+4}=8\text{ V}

Three resistors 3 kΩ, 6 kΩ, and 2 kΩ in parallel. Find the equivalent resistance.

Use

R{eq}^{-1}=\sum 1/Ri
R_{eq}=\left(\frac{1}{3000}+\frac{1}{6000}+\frac{1}{2000}\right)^{-1}=1000\ \Omega

In the circuit: source 10 V feeds series 1 kΩ then a parallel branch of 2 kΩ // 3 kΩ to ground. Find total current from the source.

First

R{//}=(2k//3k)=1.2\text{ k}\Omega. Total RT=1k+1.2k=2.2\text{ k}\Omega
I=\frac{10}{2.2k}=4.545\text{ mA}

Use current division: Node fed by 5 mA splits into parallel 2 kΩ and 8 kΩ. Find current in 2 kΩ.

I{2k}=IT\frac{R{_{o\operatorname{th}er}}}{R{_{all}}}=5\text{ mA}\cdot\frac{8k}{2k+8k}=4\text{ mA} - current divider formula

Find Thevenin equivalent seen by a 1 kΩ load: a 12 V source in series with 3 kΩ.

Open-circuit V:

V{th}=12\text{ V}. R{th}=3\text{ k}\Omega (deactivate source).
With 1 kΩ load: I=\frac{12}{3k+1k}=3\text{ mA}

Norton equivalent of the same (12 V in series with 3 kΩ).

IN=\frac{12}{3k}=4\text{ mA},\quad RN=3\text{ k}\Omega - This represents the current and resistance in a Norton equivalent circuit derived from a Thevenin equivalent. The Norton equivalent describes how a circuit can be simplified into a current source and parallel resistance.

Superposition: A node has two sources: 10 V through 2 kΩ and 5 V through 1 kΩ to the same node, node to ground via 3 kΩ. Find node voltage.

Deactivate one source at a time (voltage→short).
With 10 V active: divider to node via 2 kΩ||3 kΩ →

V1=10\cdot\frac{3||2}{(3||2)+2} = 10\cdot\frac{1.2}{3.2}=3.75\text{ V} With 5 V active through 1 kΩ and 3 kΩ to ground: V2=5\cdot\frac{3}{1+3}=3.75\text{ V}

V=V1+V2=7.5\text{ V}

Source transformation: 2 mA current source in parallel with 5 kΩ → Thevenin form.

V_{th}=I\cdot R=2\text{ mA}\cdot5k=10\text{ V} in series with 5 kΩ. - This process converts a current source and parallel resistor into an equivalent voltage source and series resistor, maintaining the same load current behavior.

Max power transfer: If

R{th}=1.5\text{ k}\Omega and V{th}=9\text{ V}, find R_L and maximum power.

RL=R{th}=1.5\text{ k}\Omega. The maximum power transfer theorem states that maximum power is delivered to a load when the load resistance equals the Thevenin resistance of the source circuit, leading to a calculated maximum power output.

P{max}=\frac{V{th}^2}{4R_{th}}=\frac{81}{6000}=13.5\text{ mW}

Two meshes share a 4 kΩ. Mesh1 has 12 V source and 2 kΩ; Mesh2 has 8 V source (opposite polarity) and 3 kΩ. Write mesh equations (clockwise currents).

\text{M1: }(2k+4k)i1-4k\,i2=12

\text{M2: }-4k\,i1+(3k+4k)i2=8

Solve quick nodal: Node V has resistors to 10 V via 5 kΩ, to 0 V via 10 kΩ, and to 5 V via 10 kΩ. Find V.

KCL: \frac{V-10}{5k}+\frac{V-0}{10k}+\frac{V-5}{10k}=0
Multiply 10k: 2(V-10)+V+(V-5)=0\Rightarrow 4V=25\Rightarrow V=6.25\text{ V}

Find power in a 4 kΩ resistor that has 8 V across it.

P=V^2/R=64/4000=16\text{ mW}

A 10 μF capacitor initially uncharged is connected to 12 V through 6 kΩ. Find v_C(t).

RC step [HB]. \tau=RC=6k\cdot10\mu=0.06\text{ s}
v_C(t)=12(1-e^{-t/0.06})\text{ V}

For the same circuit at t=0.12\text{ s}, find capacitor voltage.

v_C=12(1-e^{-0.12/0.06})=12(1-e^{-2})=12(1-0.1353)=10.38\text{ V} At t=0.12 s, the voltage across the capacitor can be calculated using the result from the RC charging equation, yielding approximately 10.38 V.

An inductor 50 mH in series with 100 Ω, step of 5 V at t=0. Find current i(t).

RL step [HB]. \tau=\frac{L}{R}=0.05/100=0.0005\text{ s}
i(t)=\frac{V}{R}\left(1-e^{-t/\tau}\right)=0.05(1-e^{-t/0.0005})\ \text{A}

Energy stored in a 2 mH inductor at 3 A.

W=\frac{1}{2}Li^2=\frac{1}{2}(0.002)(9)=0.009\text{ J} - The energy stored in an inductor is calculated using the formula for electromagnetic energy, which is dependent on the inductance and the square of the current.

Energy in a 22 μF capacitor at 5 V.

W=\frac{1}{2}CV^2=0.5\cdot22\mu\cdot25=0.000275\text{ J}

AC: Find impedance of series R=100 Ω and C=10 μF at f=159.15\text{ Hz}.

X_C=\frac{1}{2\pi f C}=100\ \Omega → Z=100-j100=\;141.4\angle-45^\circ\ \Omega - The impedance in an AC circuit is the total opposition to current flow, which includes both resistance (R) and reactance (X). In this case, the impedance is expressed in rectangular form and can also be represented in polar form.

AC: For Z above with 120∠0° Vrms, find Irms magnitude.

|I|=|V|/|Z|=120/141.4=0.849\text{ A} - The rms current magnitude is calculated using Ohm's law for AC circuits, where the voltage is divided by the impedance magnitude, resulting in approximately 0.849 A.

AC: Compute complex power for V=120\angle0^\circ\text{ V} and I=0.849\angle45^\circ\text{ A}.

S=VI^*=120\angle0^\circ\cdot0.849\angle-45^\circ=101.9\angle-45^\circ\ \text{VA}
P=|S|\cos45^\circ=72.1\text{ W},\ Q=|S|\sin(-45^\circ)=-72.1\text{ var}

Power factor correction: Load 2 kW at pf=0.8 lag on 120 V, 60 Hz. Find required shunt C to raise pf to 1.

Load current angle \phi=\cos^{-1}0.8=36.87^\circ.

QL=P\tan\phi=2\cdot0.75=1.5\text{ kvar}. Need QC=1.5\text{ kvar}

XC=\frac{V^2}{QC}=\frac{(120)^2}{1500}=9.6\ \Omega →
C=\frac{1}{2\pi f X_C}=\frac{1}{2\pi\cdot60\cdot9.6}=276\ \mu\text{F}

Series RLC: R=20 Ω, L=50 mH, C=100 μF. Find resonant frequency.

\omega0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.05\cdot100\mu}}=447.2\ \text{rad/s} f0=\omega_0/2\pi=71.2\text{ Hz} - The resonant frequency of a series RLC circuit is the frequency at which the inductive and capacitive reactances are equal, resulting in maximum current flow. For the given values, it is calculated as f_0=71.2 ext{ Hz}.

Delta–Wye: A delta of 6 Ω per side. Convert to wye.

Each wye leg R=\frac{R_\Delta}{3}=\;2\ \Omega

Wye–Delta: A balanced wye of 12 Ω per leg → delta side value?

Each delta side

R=RY+RY=\;24\ \Omega

Ideal op-amp inverting: R_{in}=10\text{ k}\Omega,\ R_f=47\text{ k}\Omega. Gain and output for V_{in}=0.2\text{ V}.

Gain A_v=-R_f/R_{in}=-4.7 → V_o=-0.94\text{ V}

Ideal op-amp non-inverting: R_1=10\text{ k}\Omega,\ R_2=30\text{ k}\Omega (R2 from output to – input, R1 from – input to ground). Vin=0.3 V. Find Vo.

A_v=1+\frac{R_2}{R_1}=1+3=4 → V_o=1.2\text{ V}

Op-amp summing inverter: R_f=20\text{ k}\Omega, inputs: V_1=0.5\text{ V} via 10 kΩ, V_2=0.2\text{ V} via 20 kΩ. Find output.

V_o=-R_f\left(\frac{V_1}{10k}+\frac{V_2}{20k}\right)=-20k(0.00005+0.00001)=-1.2\text{ V}

Diode piecewise-linear: 5 V source, series 1 kΩ, silicon diode (0.7 V drop). Find current.

Assume on: I=\frac{5-0.7}{1k}=4.3\text{ mA} → valid (I>0).

Bridge rectifier: Approximate average DC output for 12 V_rms secondary (ideal diodes, capacitorless).

V_{avg}\approx\frac{2V_m}{\pi}=\frac{2(12\sqrt{2})}{\pi}=10.8\text{ V}

Time constant recognition: In RC ladder with Thevenin seen by C as 3 kΩ, C=22 μF. Find τ.

\tau=R_{th}C=3000\cdot22\mu=0.066\text{ s}

RL decay: Inductor 100 mH, series R=50 Ω. Initial current 0.5 A, switch opens (free decay). Find current at 20 ms.

\tau=L/R=0.1/50=2\text{ ms}
i(t)=0.5\,e^{-t/\tau}=0.5 e^{-20/2}=0.5e^{-10}\approx0.0000227\text{ A}

Nodal with dependent source: Node V to ground via 4 kΩ; to 10 V via 8 kΩ; dependent current source leaving node =0.001V A. Find V.

KCL: \frac{V-10}{8k}+\frac{V-0}{4k}+0.001V=0
Multiply 8k: (V-10)+2V+8k(0.001)V=0\Rightarrow 3V-10+8V=0
11V=10\Rightarrow V=0.909\text{ V}

Find Norton current by shorting load: 10 V source → 1 kΩ → node, from node 2 kΩ to ground. Short node to ground.

Short kills 2 kΩ branch; current is 10/1k=10\text{ mA}

Find Thevenin resistance by looking-in after deactivating sources: prior circuit.

With source shorted, 1 kΩ || 2 kΩ → R_{th}=667\ \Omega

Current division with impedances: Two branches in parallel: Z_1=50-j50\ \Omega and Z_2=100+j0\ \Omega, total current 2∠0° A. Find branch currents.

I_1=I_T\frac{Z_2}{Z_1+Z_2}=2\frac{100}{(50-j50)+100}=2\frac{100}{150-j50}

Compute: I_1=2\cdot\frac{100(150+j50)}{(150)^2+(50)^2}=2\cdot\frac{15000+j5000}{25000}=1.2+ j0.4\text{ A}

I_2=2-(1.2+j0.4)=0.8-j0.4\text{ A}

RMS vs peak: A sine has

V{rms}=120\text{ V}. Find V{peak}

Vp=\sqrt{2}V{rms}=169.7\text{ V}

Real/Reactive/Complex power for V=208\angle0^\circ\ \text{V}, I=10\angle-36.87^\circ\ \text{A}.

S=VI^*=208\cdot 10\angle 36.87^\circ=2080\angle36.87^\circ\ \text{VA}
P=2080\cos36.87^\circ=1664\text{ W},\ Q=2080\sin36.87^\circ=1248\text{ var}

Resonant RLC current at f0: R=10 Ω, L=25 mH, C chosen for resonance at 100 Hz. With 20 Vrms, find I_rms.

At resonance, XL=XC → purely resistive.
I=V/R=20/10=2\text{ A}

Find C for resonance with R=10 Ω, L=25 mH at 100 Hz.

\omega0=2\pi 100, C=\frac{1}{\omega0^2 L}=\frac{1}{(2\pi 100)^2\cdot0.025}=101.3\ \mu\text{F}

Thévenin via nodal: Source 5 V through 1 kΩ to node; node to ground through 2 kΩ; output is node to ground. Find

V{th}, R{th}.

Open-circuit V_{th}=5\cdot\frac{2}{1+2}=3.333\text{ V}.

Deactivate source → R_{th}=1k||2k=667\ \Omega

Given a dependent voltage source

Vx=3Ix in series with 2 kΩ, and I_x is current in a 1 kΩ to ground at node, outline Thevenin find.

(1) Excite with test source at port.

(2) Write nodal equations including V_x=3I_x and I_x=(V/1k).

(3) Solve V/I to get R_{th}.

(4) Open-circuit voltage gives V_{th}.

Capacitor AC impedance: C=4.7 μF at 60 Hz. Find magnitude of Xc.

X_C=\frac{1}{2\pi f C}=\frac{1}{2\pi\cdot60\cdot4.7\mu}=563\ \Omega

Inductor AC impedance: L=120 mH at 400 Hz. Find XL.

X_L=2\pi f L=2\pi\cdot400\cdot0.12=301.6\ \Omega

Two-port quick concept: For a buffer (voltage follower) what is input and output resistance ideally?

R{in}\to\infty,\quad R{out}\to0 (prevents loading) - The input resistance is infinitely high and the output resistance is zero, ensuring no loading effect on the preceding stage.

Loading: A source 10 V with internal 1 kΩ feeds a 2 kΩ voltmeter (modeled as resistor). Measured voltage?

Divider: V=10\cdot\frac{2k}{1k+2k}=6.667\text{ V}

Current measurement loading: Ammeter modeled as 0.1 Ω in series with 10 Ω load and 5 V source. True current vs measured?

True I=5/10=0.5\text{ A}; measured I=5/(10+0.1)=0.495\text{ A} (slight low).

Transient: RC discharge from 10 V with R_th=5 kΩ, C=1 μF. Voltage after 2 ms?

\tau=5k\cdot1\mu=5\text{ ms}
v(t)=10e^{-t/\tau}=10e^{-2/5}=10e^{-0.4}=6.70\text{ V}

Compute average power in resistor: i(t)=3\sqrt{2}\sin(\omega t)\ \text{A} through 10 Ω.

I_{rms}=3\text{ A} \Rightarrow P=I^2R=9\cdot10=90\text{ W}

Phasor from time function: v(t)=50\cos(1000t-30^\circ). Give V phasor (rms).

Peak 50 → rms 35.36.
\tilde{V}=35.36\angle-30^\circ\ \text{V}

Given V=120\angle20^\circ\ \text{V},\ I=4\angle-10^\circ\ \text{A}. Compute S,P,Q,pf.

S=480\angle30^\circ\ \text{VA};\ P=480\cos30^\circ=416\text{ W}
Q=480\sin30^\circ=240\text{ var};\ \text{pf}=\cos30^\circ=0.866\ \text{lag}

Find Norton of a black box measured: open-circuit V=6 V, short-circuit I=30 mA.

IN=30\text{ mA},\ RN=V/I=6/0.03=200\ \Omega

Measure tau experimentally: Step input → output reaches 63.2% in 4 ms. What is \tau?

By definition, \tau=4\text{ ms}

First-order step response form (concept).

y(t)=y(\infty)+\big(y(0^+)-y(\infty)\big)e^{-t/\tau} [HB]

Dependent source power check: If a dependent current source value is

2Vx A with Vx=3\text{ V} and terminal voltage across source is -4\text{ V} (reference + at top). Power absorbed?

i=6\text{ A},\ p=v i=-4\cdot6=-24\text{ W} → delivers 24 W.

Find equivalent of series: Z1= 40∠30° Ω and Z2= 20∠-60° Ω.

Convert to rectangular, add:
Z1≈(34.64+j20), Z2≈(10 - j17.32).
Sum ≈(44.64 + j2.68)=44.72∠3.44° Ω.

Parallel of Z1= 50∠-45° Ω and Z2= 100∠0° Ω.

Use product over sum or admittances:

Y1=1/50\angle45^\circ=0.02\angle45^\circ Y2=0.01\angle0^\circ
Add: convert to rectangular and invert → Z_{eq}\approx 66.9\angle-15.6^\circ\ \Omega

Given a phasor current I=5\angle-53.13^\circ\ \text{A} through Z=10+j10\ \Omega. Find voltage phasor.

V=IZ=5\angle-53.13^\circ\cdot14.14\angle45^\circ=70.7\angle-8.13^\circ\ \text{V}

Design a divider to get 2 V from 10 V with total resistance ≈ 10 kΩ.

Let 

R{_{top}}+R{_{bot}}=10k,\ \frac{Vo}{V{_{I}}}=\frac{R_{bot}}{10k}=0.2 → R{bot}=2k,\ R_{top}=8k

Find average value of half-wave rectified sine with peak 10 V.

V{avg}=\frac{Vp}{\pi}=3.183\text{ V}

Find RMS of square wave ±5 V.

V_{rms}=5\text{ V}

Capacitor current given voltage:

vC(t)=10(1-e^{-t/0.02})u(t), C=100 μF. Find iC(t) for t>0

i=C\,dv/dt=100\mu\cdot\frac{10}{0.02}e^{-t/0.02}=0.05e^{-t/0.02}\ \text{A}

Inductor voltage given current: iL(t)=2e^{-t/0.01}u(t), L=5 mH. Find vL(t).

v=L\,di/dt=0.005\cdot\left(-\frac{2}{0.01}\right)e^{-t/0.01}=-1e^{-t/0.01}\ \text{V}

KCL/KVL sign convention (concept).

Passive sign convention: power absorbed when current enters the positive-labeled terminal. KCL: currents leaving/entering sum to 0; KVL: oriented voltage drops sum to source rises around a loop.

When to use source transformation vs superposition (concept).

Use source transformation to simplify series/parallel combinations with a single source; use superposition when multiple independent sources act on linear networks to find contributions cleanly.