Exercises — Analyze simple AC circuits with reactance

WHAT tool: $X_L=2\pi fL$ — reactance of an inductor. WHY this one and not $X_C$: the component is a coil (inductor), so opposition grows with frequency; we use the $L$ formula. $X_L=2\pi (50)(0.100)=31.4\Omega$ Sanity check: $2\pi \approx 6.28$, times $50$ is $\approx 314$, times $0.1$ gives $\approx 31.4$. ✓

WHAT tool: $X_C=\frac{1}{2\pi fC}$. WHY: the component is a capacitor, whose opposition shrinks as frequency rises, so we use the reciprocal formula. $X_C=\frac{1}{2\pi (1000)(1\times 10^{-6})}=\frac{1}{6.283\times 10^{-3}}=159\Omega$ Remember $1\mu \text{F}=1\times 10^{-6}\text{ F}$. ✓

WHAT we look at: the net reactance $X_L-X_C=40-25=+15\Omega$ (positive). WHY that tells us the answer: a positive net reactance means the circuit is net inductive. In an inductor, remember ELI — voltage (E) comes before current (I) — so current lags voltage. No trig needed; the sign alone decides it.

WHY no full recompute: $X_L=2\pi fL$ is linear in $f$ — double $f$, double $X_L$. Here $f$ went up ×10. $X_L(600)=10\times 18.85=188.5\Omega$ What it looks like: a straight line through the origin on an $X_L$-vs-$f$ graph (see figure below).

WHAT we do: rearrange $X_C=\frac{1}{2\pi fC}$ to solve for $C$. WHY: the unknown is now $C$, not $X_C$. $C=\frac{1}{2\pi f{X}_C}=\frac{1}{2\pi (400)(200)}=\frac{1}{502655}=1.99\times 10^{-6}\text{ F}\approx 1.99\mu \text{F}$ Check by plugging back: $X_C=1/(2\pi \cdot 400\cdot 1.99\times 10^{-6})\approx 200\Omega$. ✓

WHY Pythagoras: $R$ is the in-phase (real) axis, $X$ the 90°-out-of-phase (imaginary) axis — perpendicular. The total opposition is the diagonal of that rectangle. $|Z|=\sqrt{30^2+40^2}=\sqrt{900+1600}=\sqrt{2500}=50\Omega$ $\varphi ={\tan }^{-1}\frac{40}{30}={\tan }^{-1}(1.333)=+53.1°$ WHY positive: no capacitor here, so net reactance is $+40$ (inductive) → current lags → $\varphi >0$. This is the classic 3-4-5 triangle; see the impedance-triangle figure.

Step 1 — reactances (WHY: they set the net reactance): $X_L=2\pi (200)(0.020)=25.13\Omega$ $X_C=\frac{1}{2\pi (200)(30\times 10^{-6})}=\frac{1}{0.03770}=26.53\Omega$ Step 2 — net reactance: $X_L-X_C=25.13-26.53=-1.40\Omega$ (slightly capacitive). Step 3 — magnitude (Pythagoras): $|Z|=\sqrt{12^2+(-1.40{)}^2}=\sqrt{144+1.96}=\sqrt{145.96}=12.08\Omega$ Step 4 — current (Ohm's law with magnitudes): $I=\frac{V_{rms}}{|Z|}=\frac{60}{12.08}=4.97\text{ A}$ Step 5 — phase: $\varphi ={\tan }^{-1}\frac{-1.40}{12}={\tan }^{-1}(-0.1167)=-6.7°$ Interpretation: net reactance is negative → capacitive → current leads by a small $6.7°$. The circuit is nearly resonant, so $|Z|$ is close to $R$ alone.

WHY set $X_L=X_C$: this is the resonant condition — reactances cancel, leaving pure resistance. $\omega L=\frac{1}{\omega C}⟹{\omega }^2=\frac{1}{LC}⟹f=\frac{1}{2\pi \sqrt{LC}}$ $\sqrt{LC}=\sqrt{(0.020)(30\times 10^{-6})}=\sqrt{6\times 10^{-7}}=7.746\times 10^{-4}$ $f=\frac{1}{2\pi (7.746\times 10^{-4})}=\frac{1}{4.867\times 10^{-3}}=205.5\text{ Hz}$ At resonance: $X_L-X_C=0$, so $|Z|=\sqrt{R^2+0}=R=12\Omega$ — a minimum. This confirms L3.1: $200\text{ Hz}$ was just below $205.5\text{ Hz}$, hence the tiny capacitive tilt.

$X_L=2\pi fL\to 0$ as $f\to 0$: an inductor is a plain wire to DC (a fully charged coil offers no opposition once current is steady, since $di/dt=0$). $X_C=\frac{1}{2\pi fC}\to \infty$ as $f\to 0$: a capacitor is an open circuit to DC (once charged, no more current flows). Steady current: with $X_C$ infinite in series, no steady current can flow — $I\to 0$. This is the limiting behaviour you must always check: capacitors block DC, inductors pass it.

WHY $\varphi =45°$ needs $X_L=R$: $\varphi ={\tan }^{-1}(X_L/R)$, and $\tan (45°)=1$, which means $X_L/R=1$, i.e. $X_L=R$. Set $X_L=470$ and solve for $L$: $L=\frac{X_L}{2\pi f}=\frac{470}{2\pi (1000)}=\frac{470}{6283}=0.0748\text{ H}=74.8\text{ mH}$ Verify: $X_L=2\pi (1000)(0.0748)=470\Omega =R$, so $\varphi =45°$. ✓ And $|Z|=\sqrt{470^2+470^2}=470\sqrt{2}=665\Omega$.

Step 1 — impedance: $|Z|=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\Omega$ (a 6-8-10 triangle). Step 2 — current: $I=230/10=23\text{ A}$. Step 3 — phase: $\varphi ={\tan }^{-1}(6/8)={\tan }^{-1}(0.75)=36.87°$ (current lags, net inductive). Step 4 — power factor: $\cos \varphi =\cos (36.87°)=0.8$. WHY $\cos \varphi$: from Power in AC Circuits, only the in-phase part of the current delivers real power; that fraction is $\cos \varphi =R/|Z|=8/10=0.8$. Step 5 — real power: $P=V_{rms}I\cos \varphi =230\times 23\times 0.8=4232\text{ W}$. Cross-check: $P=I^{2}R=23^2\times 8=529\times 8=4232\text{ W}$. ✓ Same answer — real power only lives in $R$.

(a) WHY find $X_C=R$: this "corner" frequency is where the capacitor's opposition equals the resistor's — the balance point that defines filter behaviour. $\frac{1}{2\pi fC}=R⟹f=\frac{1}{2\pi RC}=\frac{1}{2\pi (1000)(0.1\times 10^{-6})}=\frac{1}{6.283\times 10^{-4}}=1592\text{ Hz}$ (b) Phase across all frequencies (impedance $Z=R-j{X}_C$, so $\varphi ={\tan }^{-1}(-X_C/R)$):

• As $f\to 0$: $X_C\to \infty$, so $\varphi ={\tan }^{-1}(-\infty )=-90°$ (purely capacitive — the capacitor dominates, current leads by a full quarter cycle).
• At $f=1592\text{ Hz}$: $X_C=R$, so $\varphi ={\tan }^{-1}(-1)=-45°$ (the corner).
• As $f\to \infty$: $X_C\to 0$, so $\varphi ={\tan }^{-1}(0)=0°$ (purely resistive — the capacitor becomes a short, only $R$ remains). What it looks like: $\varphi$ sweeps smoothly from $-90°$ up to $0°$, passing $-45°$ at the corner. See the phase-vs-frequency figure.

The argument: $I=\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C{)}^2}}$. The numerator $V_{rms}$ and $R$ are fixed. The only thing under our control is the term $(X_L-X_C{)}^2$, which is a square — it can never be negative. So the denominator is smallest, hence $I$ is largest, exactly when $(X_L-X_C{)}^2=0⟹X_L=X_C.$ At that point the denominator is just $\sqrt{R^2}=R$, so $I_{\max }=\frac{V_{rms}}{R}.$ WHY it matters: at resonance the circuit behaves as if the reactances vanished — pure resistance, maximum current, phase $\varphi =0°$. This is the peak of a resonance curve (see Resonance in RLC Circuits).

Resonant frequency: $f_0=\frac{1}{2\pi \sqrt{LC}}$. $\sqrt{LC}=\sqrt{(0.010)(40\times 10^{-6})}=\sqrt{4\times 10^{-7}}=6.325\times 10^{-4}$ $f_0=\frac{1}{2\pi (6.325\times 10^{-4})}=251.6\text{ Hz}$ Current at resonance: $I_{\max }=V_{rms}/R=50/5=10\text{ A}$. Current at $300\text{ Hz}$ (off-resonance): $X_L=2\pi (300)(0.010)=18.85\Omega ,X_C=\frac{1}{2\pi (300)(40\times 10^{-6})}=13.26\Omega$ $|Z|=\sqrt{5^2+(18.85-13.26{)}^2}=\sqrt{25+31.2}=\sqrt{56.2}=7.50\Omega$ $I=50/7.50=6.67\text{ A}<10\text{ A}.✓$ The resonant current genuinely dominates, confirming L5.2.

At resonance, $|Z|$ equals? ::: The resistance $R$ alone (reactances cancel). Positive net reactance means current…? ::: Lags the voltage (net inductive). To DC, a capacitor looks like…? ::: An open circuit ($X_C\to \infty$). To DC, an inductor looks like…? ::: A plain wire ($X_L\to 0$). Real power is $I^2$ times…? ::: $R$ only, never $|Z|$. Power factor equals…? ::: $\cos \varphi =R/|Z|$.