09

Complex Impedance & AC Circuit Analysis

Slides 5.3 & 5.4.1–5.4.2 · the payoff chapter for phasors

Everything you learned in the DC half of this course — Ohm's law, KVL, KCL, series/parallel reduction, voltage dividers, node-voltage, mesh-current, Thévenin — is about to come back unchanged. The only thing that changes is what a number is allowed to be. Swap real resistances RR for complex impedances Z\mathbf{Z}, swap real voltages and currents for phasors, and every technique from Chapter 2 works verbatim on a circuit driven by a sinusoid. This chapter builds that bridge, and then walks it. Budget roughly 60–75 minutes: 20 on the derivations, 20 on the widget and worked examples, the rest on practice. The single most common exam error in this material is a dropped or flipped jj, so every sign below is derived, not asserted.

1. The one fact that makes all of this work

A capacitor and an inductor are defined by derivatives: iC=C dv/dti_C = C\,dv/dt and vL=L di/dtv_L = L\,di/dt. That's why a circuit containing them is a differential equation, and why Chapter 4 was all exponentials. But in sinusoidal steady state — one frequency ω\omega, everything settled — the phasor transform turns differentiation into multiplication:

ddt  ⟷  jω∫dt  ⟷  1jω\frac{d}{dt}\;\longleftrightarrow\;j\omega \qquad\qquad \int dt\;\longleftrightarrow\;\frac{1}{j\omega}
Differentiating a sinusoid scales it by ω\omega and advances its phase by 90°90° — which is exactly what multiplying its phasor by jj does.

Why 90°90°? Because j=1∠90°j = 1\angle 90°. Multiplying any complex number by jj leaves its magnitude alone and rotates it a quarter turn counter-clockwise. Multiplying by −j=1∠−90°-j = 1\angle -90° rotates it a quarter turn clockwise. Hold onto those two sentences — nearly every sign question in this chapter reduces to them.

Re Im I jI +90° multiply by j = rotate a quarter turn CCW multiply by −j = rotate a quarter turn CW
The whole chapter in one picture: jj is a rotation operator, not just "the square root of minus one."

2. Impedance of R, L and C — derived

Impedance Z\mathbf{Z} is defined as the ratio of the voltage phasor across an element to the current phasor through it, with the passive sign convention (current entering the ++ terminal):

Z  =  VI⟺V=Z I\mathbf{Z} \;=\; \frac{\mathbf{V}}{\mathbf{I}} \qquad\Longleftrightarrow\qquad \mathbf{V} = \mathbf{Z}\,\mathbf{I}
Ohm's law, phasor form. Units of ohms. Note Z\mathbf{Z} is a complex number, not a phasor — it does not correspond to any sinusoid in time and never gets converted back.
R

Resistor. v=Riv = Ri has no derivative in it, so the phasor relation is just V=R I\mathbf{V} = R\,\mathbf{I}.

ZR=R=R∠0°Z_R = R = R\angle 0°

Purely real. Voltage and current are exactly in phase — no rotation, because there's no jj.

L

Inductor. Start from the defining law and transform term by term:

vL(t)=LdiLdt    ⟶    VL=L (jω IL)=jωL ILv_L(t) = L\frac{di_L}{dt} \;\;\longrightarrow\;\; \mathbf{V}_L = L\,(j\omega\,\mathbf{I}_L) = j\omega L\,\mathbf{I}_L   ZL=jωL=ωL ∠ 90°  \boxed{\;Z_L = j\omega L = \omega L\,\angle\,90°\;}

Purely imaginary and positive. Since VL=jωL IL\mathbf{V}_L = j\omega L\,\mathbf{I}_L, the voltage phasor is the current phasor rotated +90°+90°: voltage leads current by 90°, equivalently current lags voltage by 90°. Note ∣ZL∣=ωL|Z_L| = \omega L grows with frequency — an inductor is a short at DC (ω=0⇒ZL=0\omega = 0 \Rightarrow Z_L = 0) and an open circuit as ω→∞\omega \to \infty.

C

Capacitor. Same move, but the derivative sits on the voltage:

iC(t)=CdvCdt    ⟶    IC=C (jω VC)=jωC VCi_C(t) = C\frac{dv_C}{dt} \;\;\longrightarrow\;\; \mathbf{I}_C = C\,(j\omega\,\mathbf{V}_C) = j\omega C\,\mathbf{V}_C

Now solve for the ratio VC/IC\mathbf{V}_C/\mathbf{I}_C, and clear the jj out of the denominator using 1j=1j⋅−j−j=−j−j2=−j1=−j\dfrac{1}{j} = \dfrac{1}{j}\cdot\dfrac{-j}{-j} = \dfrac{-j}{-j^2} = \dfrac{-j}{1} = -j:

  ZC=1jωC=− jωC=1ωC ∠ −90°  \boxed{\;Z_C = \frac{1}{j\omega C} = -\,\frac{j}{\omega C} = \frac{1}{\omega C}\,\angle\,{-90°}\;}

Purely imaginary and negative. The voltage phasor is the current phasor rotated −90°-90°: current leads voltage by 90°. ∣ZC∣=1/(ωC)|Z_C| = 1/(\omega C) shrinks with frequency — a capacitor is an open circuit at DC and a short as ω→∞\omega \to \infty. That is the same "capacitors block DC, inductors pass it" fact you used for DC steady state in Chapter 4, now stated quantitatively.

The single most-failed step

1j=−j\dfrac{1}{j} = -j, not +j+j. Every capacitor sign error in this course traces back to this one identity. Sanity check it whenever you're unsure: j⋅(−j)=−j2=+1j \cdot (-j) = -j^2 = +1. ✓

I V 90° INDUCTOR — V leads I "ELI": E before I in an L
VL=jωL IL\mathbf{V}_L = j\omega L\,\mathbf{I}_L — V\mathbf{V} sits 90°90° CCW from I\mathbf{I}.
V I 90° CAPACITOR — I leads V "ICE": I before E in a C
VC=−jωC IC\mathbf{V}_C = -\dfrac{j}{\omega C}\,\mathbf{I}_C — V\mathbf{V} sits 90°90° CW from I\mathbf{I}.
Mnemonic — ELI the ICE man

In an L: E (voltage) comes before I → ELI. In a C: I comes before E → ICE. "Comes before" means leads in phase. If you can recall this, you can reconstruct both signs of jj from scratch mid-exam.

Self-check 1

A 25 μF25\ \mu\text{F} capacitor is driven at ω=2000\omega = 2000 rad/s. Its complex impedance is:

3. Rectangular form: Z=R+jX\mathbf{Z} = R + jX

Any impedance, however tangled the network behind it, collapses to two real numbers:

Z=R+jX,∣Z∣=R2+X2,∠Z=arctan⁡ ⁣(XR)\mathbf{Z} = R + jX, \qquad |Z| = \sqrt{R^2 + X^2}, \qquad \angle Z = \arctan\!\left(\frac{X}{R}\right)
RR = resistance (always ≥0\ge 0 for passive networks); XX = reactance, positive for net inductive, negative for net capacitive.

Because RR and XX are the legs of a right triangle whose hypotenuse is ∣Z∣|Z|, this picture is universally called the impedance triangle. It is the fastest way to read off what a circuit will do: the triangle's angle θ=∠Z\theta = \angle Z is the phase by which current lags the voltage, since I=V/Z\mathbf{I} = \mathbf{V}/\mathbf{Z} subtracts ∠Z\angle Z from ∠V\angle \mathbf{V}.

Re Im R jX Z θ > 0 X > 0 · INDUCTIVE · I lags V
Net inductive: X>0X>0, so ∠Z\angle Z is in the first quadrant and current lags.
Re Im R jX Z θ < 0 X < 0 · CAPACITIVE · I leads V
Net capacitive: X<0X<0, so ∠Z\angle Z is in the fourth quadrant and current leads.
ElementImpedancePolarAt low ω\omegaAt high ω\omega
RRRRR∠0°R\angle 0°RRRR
LLjωLj\omega LωL∠90°\omega L\angle 90°→0\to 0 (short)→∞\to \infty (open)
CC1jωC=−jωC\dfrac{1}{j\omega C} = -\dfrac{j}{\omega C}1ωC∠−90°\dfrac{1}{\omega C}\angle{-90°}→∞\to \infty (open)→0\to 0 (short)
Self-check 2

A 22 mH inductor is placed in series with a capacitor CC and driven at ω=5000\omega = 5000 rad/s. What value of CC makes the pair behave as a short circuit (Zeq=0\mathbf{Z}_{eq} = 0)?

4. Combining impedances — the DC rules, unchanged

This is the part that should feel like a gift. KVL and KCL are statements about sums, and phasors add linearly, so both laws survive the transform intact. Everything built on top of them therefore survives too:

Zseries=Z1+Z2+⋯1Zparallel=1Z1+1Z2+⋯\mathbf{Z}_{\text{series}} = \mathbf{Z}_1 + \mathbf{Z}_2 + \cdots \qquad\qquad \frac{1}{\mathbf{Z}_{\text{parallel}}} = \frac{1}{\mathbf{Z}_1} + \frac{1}{\mathbf{Z}_2} + \cdots two in parallel:Zeq=Z1Z2Z1+Z2divider:V1=Vs Z1Z1+Z2\text{two in parallel:}\quad \mathbf{Z}_{eq} = \frac{\mathbf{Z}_1\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \qquad\qquad \text{divider:}\quad \mathbf{V}_1 = \mathbf{V}_s\,\frac{\mathbf{Z}_1}{\mathbf{Z}_1 + \mathbf{Z}_2}
Identical in form to the resistor rules — only the arithmetic is complex.
Practical tip — pick your form

Complex arithmetic is easy in the right form and miserable in the wrong one. Add and subtract in rectangular (a+jba + jb): just add the parts. Multiply and divide in polar (M∠θM\angle\theta): multiply magnitudes, add angles; divide magnitudes, subtract angles. Series combination is addition → rectangular. Product-over-sum needs both: do the sum in rectangular, convert, then do the product and quotient in polar.

R Z_R = R L Z_L = jωL C Z_C = −j/(ωC) Vs I Z_total = R + jωL − j/(ωC) · one loop, so the same I flows through all three
The phasor-domain redraw. Once every element carries an impedance label, this is a three-resistor series loop as far as the algebra is concerned.
Worked example — building Z\mathbf{Z} at a given ω\omegaslides 5.3

A 100 Ω100\ \Omega resistor, a 0.10.1 H inductor and a 12.5 μF12.5\ \mu\text{F} capacitor are in series, driven at ω=1000\omega = 1000 rad/s. Find Z\mathbf{Z} in rectangular and polar form, and say whether the circuit is inductive or capacitive.

1

Convert each element at this one frequency.

ZR=100 ΩZ_R = 100\ \Omega ZL=jωL=j(1000)(0.1)=j100 ΩZ_L = j\omega L = j(1000)(0.1) = j100\ \Omega ZC=−jωC=−j(1000)(12.5×10−6)=−j0.0125=−j80 ΩZ_C = -\frac{j}{\omega C} = -\frac{j}{(1000)(12.5\times10^{-6})} = -\frac{j}{0.0125} = -j80\ \Omega

Do the 1/(ωC)1/(\omega C) division as a magnitude first, then attach the minus sign — it's far harder to lose the sign that way than by carrying 1/(jωC)1/(j\omega C) through the arithmetic.

2

Series → add, in rectangular form. Real parts with real, imaginary with imaginary:

Z=100+j100−j80=100+j20 Ω\mathbf{Z} = 100 + j100 - j80 = 100 + j20\ \Omega
3

Convert to polar for interpretation:

∣Z∣=1002+202=10400=102.0 Ω,∠Z=arctan⁡ ⁣20100=11.31°|Z| = \sqrt{100^2 + 20^2} = \sqrt{10400} = 102.0\ \Omega, \qquad \angle Z = \arctan\!\frac{20}{100} = 11.31°
Z=100+j20 Ω=102.0 ∠ 11.31° Ω\mathbf{Z} = 100 + j20\ \Omega = 102.0\,\angle\,11.31°\ \Omega. Since X=+20 Ω>0X = +20\ \Omega > 0 the network is net inductive — the inductor's +j100+j100 more than cancels the capacitor's −j80-j80 — so the current lags the source voltage by 11.31°11.31°.

5. Build the triangle yourself

Reactance is a function of frequency, so "is this circuit inductive?" has no answer until you name ω\omega. The widget below makes that concrete: pick which elements are in the series string, set their values, then sweep the frequency and watch jXLjX_L and −jXC-jX_C fight each other on the complex plane. Look for the frequency where the vertical arrows exactly cancel — that's resonance, and Z\mathbf{Z} collapses onto the real axis.

Impedance triangle builder
ω = —
X_L = —
X_C = —
Z = —
|Z| = —
∠Z = —
char: —
Re (Ω) Im (Ω) R jX_L −jX_C Z

6. Full phasor-domain circuit analysis (5.4)

With impedance in hand, AC analysis is a five-step recipe, and steps 3–4 are the DC techniques you already own. There is no new circuit theory in this section — only bookkeeping.

StepWhat you do
1Read off ω\omega from the source. Convert every source to a phasor (put all sources in the same trig form first — the course converts sin⁡\sin to cos⁡\cos via sin⁡(ωt+θ)=cos⁡(ωt+θ−90°)\sin(\omega t + \theta) = \cos(\omega t + \theta - 90°)).
2Replace every RR, LL, CC with its impedance evaluated at that ω\omega. The circuit is now a resistive-looking network of complex numbers.
3Reduce with series/parallel and dividers wherever the topology allows.
4Where it doesn't, apply node-voltage, mesh-current, superposition or Thévenin exactly as in Chapter 2 — all still valid, now over C\mathbb{C}.
5Convert the phasor answers back to the time domain: V=Vm∠θ  →  v(t)=Vmcos⁡(ωt+θ)\mathbf{V} = V_m\angle\theta \;\to\; v(t) = V_m\cos(\omega t + \theta). Skip this step if the question asks only for phasors or impedance.
One frequency at a time

Impedance is defined at a single ω\omega. If a circuit has two sources at different frequencies, you cannot build one impedance network for both — you must use superposition, analyze each frequency in its own separate phasor circuit, convert each result to the time domain, and only then add the time functions. Adding phasors from different frequencies is meaningless.

Worked example — series RLC at resonance, fully solvedslides 5.4

A source vs(t)=100cos⁡(1000t)v_s(t) = 100\cos(1000t) V drives a series combination of R=50 ΩR = 50\ \Omega, L=0.1L = 0.1 H and C=10 μFC = 10\ \mu\text{F}. Find the current and all three element voltages, as phasors and in the time domain.

1

The source is already in cosine form, so Vs=100∠0°\mathbf{V}_s = 100\angle 0° V and ω=1000\omega = 1000 rad/s.

2

Impedances at ω=1000\omega = 1000:

ZR=50 Ω,ZL=j(1000)(0.1)=j100 Ω,ZC=−j(1000)(10×10−6)=−j0.01=−j100 ΩZ_R = 50\ \Omega,\qquad Z_L = j(1000)(0.1) = j100\ \Omega,\qquad Z_C = -\frac{j}{(1000)(10\times10^{-6})} = -\frac{j}{0.01} = -j100\ \Omega
3

Single loop → series → add:

Z=50+j100−j100=50+j0=50 ∠ 0° Ω\mathbf{Z} = 50 + j100 - j100 = 50 + j0 = 50\,\angle\,0°\ \Omega

The reactances cancel exactly. This circuit is at series resonance: at ω0=1/LC=1/(0.1)(10−5)=1000\omega_0 = 1/\sqrt{LC} = 1/\sqrt{(0.1)(10^{-5})} = 1000 rad/s the network looks purely resistive to the source, even though it is full of energy-storage elements.

4

Phasor Ohm's law for the loop current:

I=VsZ=100∠0°50∠0°=2 ∠ 0° A\mathbf{I} = \frac{\mathbf{V}_s}{\mathbf{Z}} = \frac{100\angle 0°}{50\angle 0°} = 2\,\angle\,0°\ \text{A}

Zero phase difference between Vs\mathbf{V}_s and I\mathbf{I} — unity power factor. That's the defining signature of resonance, and it's what you'd look for on an oscilloscope.

5

Element voltages, each Vk=ZkI\mathbf{V}_k = Z_k\mathbf{I} with the same I\mathbf{I}:

VR=(50)(2∠0°)=100 ∠ 0° V\mathbf{V}_R = (50)(2\angle 0°) = 100\,\angle\,0°\ \text{V} VL=(j100)(2∠0°)=j200=200 ∠ 90° V\mathbf{V}_L = (j100)(2\angle 0°) = j200 = 200\,\angle\,90°\ \text{V} VC=(−j100)(2∠0°)=−j200=200 ∠ −90° V\mathbf{V}_C = (-j100)(2\angle 0°) = -j200 = 200\,\angle\,{-90°}\ \text{V}
6

Independent check — KVL around the loop (this is how you catch a flipped jj):

VR+VL+VC=100+j200−j200=100+j0=100∠0°=Vs    ✓\mathbf{V}_R + \mathbf{V}_L + \mathbf{V}_C = 100 + j200 - j200 = 100 + j0 = 100\angle 0° = \mathbf{V}_s \;\;\checkmark

Had you written ZC=+j100Z_C = +j100, the sum would have come out 100+j400≠Vs100 + j400 \ne \mathbf{V}_s and the error would be caught instantly. Always close the KVL loop with your phasors.

7

Back to time domain:

i(t)=2cos⁡(1000t) A,vR(t)=100cos⁡(1000t) Vi(t) = 2\cos(1000t)\ \text{A},\quad v_R(t) = 100\cos(1000t)\ \text{V} vL(t)=200cos⁡(1000t+90°) V,vC(t)=200cos⁡(1000t−90°) Vv_L(t) = 200\cos(1000t + 90°)\ \text{V},\quad v_C(t) = 200\cos(1000t - 90°)\ \text{V}
I=2∠0°\mathbf{I} = 2\angle 0° A, VR=100∠0°\mathbf{V}_R = 100\angle 0° V, VL=200∠90°\mathbf{V}_L = 200\angle 90° V, VC=200∠−90°\mathbf{V}_C = 200\angle{-90°} V. KVL verified.
Not a mistake — a real phenomenon

Notice that ∣VL∣=∣VC∣=200|\mathbf{V}_L| = |\mathbf{V}_C| = 200 V while the source is only 100100 V. Individual reactive element voltages can and do exceed the source voltage near resonance; KVL is not violated because they are 180°180° out of phase with each other and cancel in the sum. If a DC instinct tells you "that must be wrong," override it — but do run the KVL check, because a genuine sign error looks superficially similar.

Self-check 3

A series circuit has Z=30+j40 Ω\mathbf{Z} = 30 + j40\ \Omega and is driven by Vs=100∠0°\mathbf{V}_s = 100\angle 0° V. The current phasor is:

7. Practice

Every solution below was worked two independent ways before being written down — the second route is shown explicitly in each one, because that habit is what will save you on an exam where nobody can tell you the answer.

Problem 1 — Series RLC, full phasor solvecore

A source vs(t)=120cos⁡(2000t)v_s(t) = 120\cos(2000t) V drives a series loop containing R=60 ΩR = 60\ \Omega, L=50L = 50 mH and C=12.5 μFC = 12.5\ \mu\text{F}.

  1. Find Z\mathbf{Z} in rectangular and polar form.
  2. Find the phasor current I\mathbf{I} and write i(t)i(t).
  3. Find VR\mathbf{V}_R, VL\mathbf{V}_L, VC\mathbf{V}_C and verify KVL.
  4. Does the current lead or lag the source?

(1) Impedances at ω=2000\omega = 2000 rad/s.

ZL=j(2000)(0.05)=j100 Ω,ZC=−j(2000)(12.5×10−6)=−j0.025=−j40 ΩZ_L = j(2000)(0.05) = j100\ \Omega, \qquad Z_C = -\frac{j}{(2000)(12.5\times10^{-6})} = -\frac{j}{0.025} = -j40\ \Omega Z=60+j100−j40=60+j60 Ω\mathbf{Z} = 60 + j100 - j40 = 60 + j60\ \Omega ∣Z∣=602+602=602=84.85 Ω,∠Z=arctan⁡ ⁣6060=45°|Z| = \sqrt{60^2+60^2} = 60\sqrt2 = 84.85\ \Omega, \qquad \angle Z = \arctan\!\frac{60}{60} = 45°

So Z=84.85∠45° Ω\mathbf{Z} = 84.85\angle 45°\ \Omega.

(2) Current — route A, polar division.

I=120∠0°84.85∠45°=1.414 ∠ −45° A\mathbf{I} = \frac{120\angle 0°}{84.85\angle 45°} = 1.414\,\angle\,{-45°}\ \text{A}

Route B, rectangular division (multiply by the conjugate — an independent path through the arithmetic):

I=12060+j60⋅60−j6060−j60=120(60−j60)602+602=7200−j72007200=1−j1\mathbf{I} = \frac{120}{60+j60}\cdot\frac{60-j60}{60-j60} = \frac{120(60-j60)}{60^2+60^2} = \frac{7200 - j7200}{7200} = 1 - j1

And ∣1−j1∣=2=1.414|1-j1| = \sqrt2 = 1.414 with arctan⁡(−1/1)=−45°\arctan(-1/1) = -45° — the two routes agree.

i(t)=1.414cos⁡(2000t−45°) Ai(t) = 1.414\cos(2000t - 45°)\ \text{A}

(3) Element voltages (using the convenient rectangular form I=1−j1\mathbf{I} = 1 - j1):

VR=60(1−j1)=60−j60=84.85∠−45° V\mathbf{V}_R = 60(1-j1) = 60 - j60 = 84.85\angle{-45°}\ \text{V} VL=j100(1−j1)=j100−j2100=100+j100=141.4∠45° V\mathbf{V}_L = j100(1-j1) = j100 - j^2 100 = 100 + j100 = 141.4\angle 45°\ \text{V} VC=−j40(1−j1)=−j40+j240=−40−j40=56.57∠−135° V\mathbf{V}_C = -j40(1-j1) = -j40 + j^2 40 = -40 - j40 = 56.57\angle{-135°}\ \text{V}

KVL check (the second independent verification):

(60−j60)+(100+j100)+(−40−j40)=(60+100−40)+j(−60+100−40)=120+j0  =  Vs  ✓(60 - j60) + (100 + j100) + (-40 - j40) = (60+100-40) + j(-60+100-40) = 120 + j0 \;=\; \mathbf{V}_s \;\checkmark

(4) X=+60 Ω>0X = +60\ \Omega > 0, so the network is net inductive and ∠I=−45°\angle\mathbf{I} = -45° relative to the source.

Z=60+j60=84.85∠45° Ω\mathbf{Z} = 60+j60 = 84.85\angle 45°\ \Omega; I=1.414∠−45°\mathbf{I} = 1.414\angle{-45°} A, i.e. i(t)=1.414cos⁡(2000t−45°)i(t) = 1.414\cos(2000t - 45°) A; VR=84.85∠−45°\mathbf{V}_R = 84.85\angle{-45°} V, VL=141.4∠45°\mathbf{V}_L = 141.4\angle 45° V, VC=56.57∠−135°\mathbf{V}_C = 56.57\angle{-135°} V. The current lags the source by 45°45°.
Problem 2 — Parallel R‖L, two ways to the source currentcombination

A source Vs=48∠0°\mathbf{V}_s = 48\angle 0° V at ω=1000\omega = 1000 rad/s drives a 40 Ω40\ \Omega resistor in parallel with a 3030 mH inductor.

  1. Find the equivalent impedance Zeq\mathbf{Z}_{eq} in polar and rectangular form.
  2. Find the branch currents IR\mathbf{I}_R and IL\mathbf{I}_L, and the source current Is\mathbf{I}_s.
  3. Confirm your Zeq\mathbf{Z}_{eq} using the currents from part 2.

(1) Route A — product over sum. ZL=j(1000)(0.03)=j30 ΩZ_L = j(1000)(0.03) = j30\ \Omega.

Zeq=(40)(j30)40+j30=j120040+j30=1200∠90°50∠36.87°=24 ∠ 53.13° Ω\mathbf{Z}_{eq} = \frac{(40)(j30)}{40 + j30} = \frac{j1200}{40+j30} = \frac{1200\angle 90°}{50\angle 36.87°} = 24\,\angle\,53.13°\ \Omega

(Here ∣40+j30∣=1600+900=50|40+j30| = \sqrt{1600+900} = 50 and arctan⁡(30/40)=36.87°\arctan(30/40) = 36.87°.) In rectangular form,

Zeq=24(cos⁡53.13°+jsin⁡53.13°)=24(0.6+j0.8)=14.4+j19.2 Ω\mathbf{Z}_{eq} = 24(\cos 53.13° + j\sin 53.13°) = 24(0.6 + j0.8) = 14.4 + j19.2\ \Omega

Route B — admittances. Parallel elements add in Y=1/Z\mathbf{Y} = 1/\mathbf{Z}:

Y=140+1j30=0.025−j0.03333 S\mathbf{Y} = \frac{1}{40} + \frac{1}{j30} = 0.025 - j0.03333\ \text{S}

(using 1/j=−j1/j = -j). Then

Zeq=1Y=0.025+j0.033330.0252+0.033332=0.025+j0.033331.7361×10−3=14.4+j19.2 Ω\mathbf{Z}_{eq} = \frac{1}{\mathbf{Y}} = \frac{0.025 + j0.03333}{0.025^2 + 0.03333^2} = \frac{0.025 + j0.03333}{1.7361\times10^{-3}} = 14.4 + j19.2\ \Omega

The two routes agree. Note ∣Zeq∣=24 Ω|Z_{eq}| = 24\ \Omega is smaller than either branch — parallel combination always reduces magnitude, exactly as with resistors.

(2) Branch currents. Both branches see the full Vs\mathbf{V}_s:

IR=48∠0°40=1.2 ∠ 0° A=1.2+j0\mathbf{I}_R = \frac{48\angle 0°}{40} = 1.2\,\angle\,0°\ \text{A} = 1.2 + j0 IL=48∠0°j30=4830⋅1j=−j1.6=1.6 ∠ −90° A\mathbf{I}_L = \frac{48\angle 0°}{j30} = \frac{48}{30}\cdot\frac{1}{j} = -j1.6 = 1.6\,\angle\,{-90°}\ \text{A}

KCL at the top node:

Is=IR+IL=1.2−j1.6=1.44+2.56  ∠arctan⁡ ⁣−1.61.2=2.0 ∠ −53.13° A\mathbf{I}_s = \mathbf{I}_R + \mathbf{I}_L = 1.2 - j1.6 = \sqrt{1.44+2.56}\;\angle\arctan\!\frac{-1.6}{1.2} = 2.0\,\angle\,{-53.13°}\ \text{A}

(3) Independent confirmation. If Zeq\mathbf{Z}_{eq} is right, it must also reproduce Is\mathbf{I}_s directly:

Is=VsZeq=48∠0°24∠53.13°=2 ∠ −53.13° A  ✓\mathbf{I}_s = \frac{\mathbf{V}_s}{\mathbf{Z}_{eq}} = \frac{48\angle 0°}{24\angle 53.13°} = 2\,\angle\,{-53.13°}\ \text{A}\;\checkmark

Two entirely different routes — combine-then-divide vs. divide-then-add — landed on the same answer.

Zeq=14.4+j19.2 Ω=24∠53.13° Ω\mathbf{Z}_{eq} = 14.4 + j19.2\ \Omega = 24\angle 53.13°\ \Omega; IR=1.2∠0°\mathbf{I}_R = 1.2\angle 0° A, IL=1.6∠−90°\mathbf{I}_L = 1.6\angle{-90°} A, Is=2.0∠−53.13°\mathbf{I}_s = 2.0\angle{-53.13°} A (lagging — the combination is inductive).
Problem 3 — Identify the black box, then correct itinverse · stretch

A sealed two-terminal box is driven at ω=400\omega = 400 rad/s. With V=200∠0°\mathbf{V} = 200\angle 0° V applied, the measured current is I=4∠−53.13°\mathbf{I} = 4\angle{-53.13°} A.

  1. Find Z\mathbf{Z} and model the box as a resistor in series with a single reactive element. Which element is it, and what is its value?
  2. What capacitance placed in series with the box would make the whole thing look purely resistive at this frequency, and what would the current become?

(1) Route A — polar division.

Z=VI=200∠0°4∠−53.13°=50 ∠ +53.13° Ω\mathbf{Z} = \frac{\mathbf{V}}{\mathbf{I}} = \frac{200\angle 0°}{4\angle{-53.13°}} = 50\,\angle\,{+53.13°}\ \Omega =50(cos⁡53.13°+jsin⁡53.13°)=50(0.6+j0.8)=30+j40 Ω= 50(\cos 53.13° + j\sin 53.13°) = 50(0.6 + j0.8) = 30 + j40\ \Omega

Route B — rectangular division. Write the current in rectangular form, I=4(cos⁡(−53.13°)+jsin⁡(−53.13°))=2.4−j3.2\mathbf{I} = 4(\cos(-53.13°) + j\sin(-53.13°)) = 2.4 - j3.2, and divide by multiplying through by the conjugate:

Z=2002.4−j3.2⋅2.4+j3.22.4+j3.2=200(2.4+j3.2)2.42+3.22=480+j64016=30+j40 Ω\mathbf{Z} = \frac{200}{2.4 - j3.2}\cdot\frac{2.4+j3.2}{2.4+j3.2} = \frac{200(2.4+j3.2)}{2.4^2+3.2^2} = \frac{480 + j640}{16} = 30 + j40\ \Omega

The two routes agree. X=+40 ΩX = +40\ \Omega is positive, so the reactive element is an inductor (a capacitor would give negative XX). From X=ωLX = \omega L:

L=Xω=40400=0.1 H=100 mHL = \frac{X}{\omega} = \frac{40}{400} = 0.1\ \text{H} = 100\ \text{mH}

So the box is R=30 ΩR = 30\ \Omega in series with L=100L = 100 mH.

(2) Adding a series capacitor contributes −j/(ωC)-j/(\omega C). Purely resistive means the total reactance is zero:

40−1ωC=0  ⟹  1ωC=40  ⟹  C=1(400)(40)=6.25×10−5 F=62.5 μF40 - \frac{1}{\omega C} = 0 \;\Longrightarrow\; \frac{1}{\omega C} = 40 \;\Longrightarrow\; C = \frac{1}{(400)(40)} = 6.25\times10^{-5}\ \text{F} = 62.5\ \mu\text{F}

Then Znew=30+j40−j40=30∠0° Ω\mathbf{Z}_{\text{new}} = 30 + j40 - j40 = 30\angle 0°\ \Omega and

Inew=200∠0°30∠0°=6.67 ∠ 0° A\mathbf{I}_{\text{new}} = \frac{200\angle 0°}{30\angle 0°} = 6.67\,\angle\,0°\ \text{A}

Cross-check on the capacitor value via the resonance formula, an independent route: the series LCLC is resonant when ω=1/LC\omega = 1/\sqrt{LC}, i.e. C=1/(ω2L)=1/((400)2(0.1))=1/16000=62.5 μFC = 1/(\omega^2 L) = 1/((400)^2(0.1)) = 1/16000 = 62.5\ \mu\text{F} ✓ — same answer from a different relation.

Z=30+j40 Ω=50∠53.13° Ω\mathbf{Z} = 30 + j40\ \Omega = 50\angle 53.13°\ \Omega: a 30 Ω30\ \Omega resistor in series with a 100100 mH inductor. Adding C=62.5 μFC = 62.5\ \mu\text{F} in series makes it purely resistive (30∠0° Ω30\angle 0°\ \Omega), and the current rises to 6.67∠0°6.67\angle 0° A — in phase with the voltage.

8. Error checklist before you hand in

Run these four in order

1. Sign of every jj. Inductor ++, capacitor −-. If you carried 1/(jωC)1/(j\omega C) anywhere, re-check that you used 1/j=−j1/j = -j.
2. Close a loop or a node. Sum your element voltages around a mesh (should equal the source) or your branch currents into a node (should equal zero). This catches sign errors essentially every time.
3. Sanity-check the phase direction. Net inductive ⇒ current lags ⇒ ∠I\angle\mathbf{I} more negative than ∠V\angle\mathbf{V}. Net capacitive ⇒ current leads. If your XX and your phase disagree, one of them is wrong.
4. Magnitude plausibility. ∣Z∣|Z| for a series string is at least as big as its largest real part; ∣Z∣|Z| for a parallel combination is smaller than either branch. A parallel result bigger than both branches means an algebra slip.