LC circuit — oscillations (electrical analog of SHM): Worked examples

Intuition What this page is for

The parent note LC circuit note gave you the machinery: $\overset{q}{¨} = - \frac{1}{L C} q$ , the solution $q = q_{0} cos (ω t)$ , and the energy split. This page is a workout. We list every type of question an LC circuit can throw at you, then solve one of each — including the weird edge cases (zero current start, energy-sharing instants, the limits $L \to 0$ and $C \to \infty$ , and an exam twist that mixes energy with timing).

Before we start, five symbols we reuse constantly, all built in the parent note:

$q$ = charge on the capacitor (unit: coulomb, C). Think "how full the tank is." Its maximum value over a cycle is written $q_{0}$ — the amplitude of the charge oscillation.
$i = \frac{d q}{d t}$ = current, the rate charge flows (unit: ampere, A). Think "flow speed." Its maximum is $i_{0} = q_{0} ω$ (peak charge times angular frequency).
$ω = \frac{1}{L C}$ = angular frequency (unit: rad/s). How fast the sloshing cycles.
$T = \frac{2 π}{ω} = 2 π L C$ = period (unit: s) — the time for one complete slosh, so that $ω T = 2 π$ . We use $T$ constantly below (e.g. " $T /8$ "), so it is defined here up front; Ex 1 will re-derive it from the base numbers.
$U_{E} = \frac{q ^{2}}{2 C}$ = electric energy stored in the capacitor, and $U_{B} = \frac{1}{2} L i^{2}$ = magnetic energy stored in the inductor (both in joules, J). These two "buckets" trade energy back and forth; their sum stays fixed.

Figure — LC circuit — oscillations (electrical analog of SHM)

The scenario matrix

Every LC problem falls into one of these cells. Each worked example below is tagged with the cell(s) it covers. (Here $T$ is the period defined above.)

Cell	What it tests	Degenerate / edge part	Example
A Frequency & period	plug $L, C$ into $ω, T, f$	large/small $L$ and large/small $C$	Ex 1
B Peak current	energy swap $U_{E} \to U_{B}$	—	Ex 2
C Value at a given time	evaluate $q (t), i (t)$	signs in each quarter-cycle	Ex 3
D Energy sharing instant	when $U_{E} = U_{B}$ , or a given ratio	$T /8$ , general fraction	Ex 4
E Different initial condition	switch closed while current already flows	$q (0) = 0$ start (pure sine)	Ex 5
F Limiting behaviour	$L \to 0$ , $C \to \infty$ , $R = 0$ vs $R \neq = 0$	degenerate + damped + critical/over-damped	Ex 6
G Real-world word problem	radio tuning / design a frequency	pick component to hit target	Ex 7
H Exam twist	combine timing + energy + phase	multi-step	Ex 8

We reuse one base circuit for A–D and H so numbers stay familiar: $L = 2 mH = 2 \times 1 0^{- 3} H, C = 5 μ F = 5 \times 1 0^{- 6} F .$

From these once and for all: $ω = \frac{1}{L C} = \frac{1}{( 2 \times 1 0 ^{- 3} ) ( 5 \times 1 0 ^{- 6} )} = \frac{1}{1 0 ^{- 8}} = 1 0^{4} rad/s .$

Ex 1 — Cell A: frequency & period (and what large $L$ and large $C$ do)

Worked example Statement

For the base circuit ( $L = 2 mH$ , $C = 5 μ F$ ), find $ω$ , $f$ , and $T$ . Then predict what happens to $T$ if you (a) swap in an inductor four times larger, and (b) swap in a capacitor four times larger (keeping the other component at its base value each time).

Forecast: guess first — does a bigger inductor make the oscillation faster or slower? What about a bigger capacitor? (Hint: $L$ is the "mass" $m$ , and $1/ C$ is the "spring constant" $k$ in the analogy.)

$ω = \frac{1}{L C} = 1 0^{4}$ rad/s. Why this step? $ω$ is what every other timing quantity is built from — get it first.
$f = \frac{ω}{2 π} = \frac{1 0 ^{4}}{2 π} \approx 1591.5$ Hz. Why this step? $ω = 2 π f$ ; $f$ counts full cycles per second, which is what "frequency" means in everyday speech.
$T = \frac{1}{f} = \frac{2 π}{ω} = 2 π L C \approx 6.283 \times 1 0^{- 4}$ s $\approx 0.628$ ms. Why this step? Period is the reciprocal of frequency — time for one full slosh.
(a) Four times larger $L$ : $T \propto L$ , so $T$ grows by $4 = 2$ . New $T \approx 1.257$ ms. Why this step? Only the $L$ factor changed. Bigger "mass" ⟹ slower ⟹ longer period. Your forecast should have said slower.
(b) Four times larger $C$ : $T \propto C$ , so again $T$ grows by $4 = 2$ . New $T \approx 1.257$ ms. Why this step? $1/ C$ is the "spring constant." Bigger $C$ ⟹ softer spring ⟹ weaker pull-back ⟹ slower oscillation. So large $L$ and large $C$ act the same way on the timing — both slow it down — while small $L$ or small $C$ speed it up. This covers all four corners of the "large/small $L$ or $C$ " edge.

Verify: Units of $L C = H \cdot F$ . Since $H = V\cdots/A$ and $F = A\cdots/V$ , the product is $s^{2}$ , so $L C$ is in seconds. ✓ And $f \cdot T = 1591.5 \times 6.283 \times 1 0^{- 4} \approx 1.000$ . ✓ Both (a) and (b) give the same $4 = 2$ factor ✓.

Ex 2 — Cell B: peak current from energy

Worked example Statement

Same $L, C$ . The capacitor is charged to $q_{0} = 2 \times 1 0^{- 4}$ C, then the switch closes at rest. Find the peak current $i_{0}$ .

Forecast: all the energy starts electric. Where is current biggest — where charge is full, or where it is empty?

At the start, energy is entirely electric: $U_{E}^{m a x} = \frac{q _{0}^{2}}{2 C}$ . Why this step? At $t = 0$ the current is zero, so $U_{B} = 0$ ; all energy sits in the capacitor.
At the swap point, all of it becomes magnetic: $\frac{q _{0}^{2}}{2 C} = \frac{1}{2} L i_{0}^{2}$ . Why this step? Energy is conserved (no resistor), so peak $U_{E}$ = peak $U_{B}$ .
Solve: $i_{0} = q_{0} \frac{1}{L C} = q_{0} ω$ . Why this step? Cancel the $\frac{1}{2}$ 's, isolate $i_{0}$ , and recognise $1/ L C = ω$ .
$i_{0} = (2 \times 1 0^{- 4}) (1 0^{4}) = 2$ A. Why this step? Substitute the known $q_{0}$ and $ω$ to turn the symbolic result into an actual number — the answer the question asked for.

Verify: cross-check via the current solution $i (t) = - q_{0} ω sin (ω t)$ whose amplitude is $q_{0} ω = 2 \times 1 0^{- 4} \times 1 0^{4} = 2$ A. Same number, no energy argument needed. ✓ Units: C·(rad/s) = C/s = A. ✓

Here we must be careful with signs, because a full cycle passes through four quarter-turns and $q, i$ change sign differently in each. The figure below tracks them.

Worked example Statement

With $q_{0} = 2 \times 1 0^{- 4}$ C, $ω = 1 0^{4}$ rad/s, find $q$ and $i$ at $t = \frac{T}{8}$ and at $t = \frac{3 T}{8}$ .

Forecast: at $T /8$ we are one-eighth into the cosine. Is $q$ still positive? Is $i$ negative (draining) or positive?

$ω t$ at $t = T /8$ : since $ω T = 2 π$ , we get $ω t = \frac{2 π}{8} = \frac{π}{4}$ . Why this step? Everything depends on the angle $ω t$ , not $t$ directly. Convert once.
$q = q_{0} cos (π /4) = 2 \times 1 0^{- 4} \cdot \frac{1}{2} \approx 1.414 \times 1 0^{- 4}$ C (positive — capacitor still partly charged, same polarity). Why this step? $π /4$ is in the first quarter where cosine is positive.
$i = - q_{0} ω sin (π /4) = - 2 \cdot \frac{1}{2} \approx - 1.414$ A (negative — current is discharging the capacitor). Why this step? The minus sign in $i = - q_{0} ω sin (ω t)$ says current flows to empty the capacitor in the first quarter. See the orange curve dipping negative in the figure.
At $t = 3 T /8$ : $ω t = \frac{3 π}{4}$ (second quarter). $q = q_{0} cos (3 π /4) \approx - 1.414 \times 1 0^{- 4}$ C (now negative — capacitor is charging up the other way). $i = - q_{0} ω sin (3 π /4) \approx - 1.414$ A (still negative, still same flow direction). Why this step? This is the case-coverage point: past $T /4$ the charge flips sign but the current has not yet flipped. This is exactly the 90° phase offset.

Verify: energy must be conserved at both times. At $T /8$ : $U_{E} + U_{B} = \frac{( 1.414 \times 1 0 ^{- 4} ) ^{2}}{2 \cdot 5 \times 1 0 ^{- 6}} + \frac{1}{2} (2 \times 1 0^{- 3}) (1.414)^{2} \approx 2 \times 1 0^{- 3} + 2 \times 1 0^{- 3} = 4 \times 1 0^{- 3}$ J. Compare total $\frac{q _{0}^{2}}{2 C} = \frac{( 2 \times 1 0 ^{- 4} ) ^{2}}{1 0 ^{- 5}} = 4 \times 1 0^{- 3}$ J. ✓

Recall from the top of the page: $U_{E} = \frac{q ^{2}}{2 C}$ is the capacitor's electric energy and $U_{B} = \frac{1}{2} L i^{2}$ is the inductor's magnetic energy. Here we ask when the two buckets hold specified fractions. On the ellipse figure above, " $U_{E} = U_{B}$ " is the point where the state has swung 45° around.

Worked example Statement

(a) First time $U_{E} = U_{B}$ . (b) First time the capacitor holds one-quarter of the total energy, $U_{E} = \frac{1}{4} U_{total}$ .

Forecast: (a) should land at some clean fraction of $T$ . For (b), less capacitor energy means the wave has swung further — later or earlier than (a)?

Total energy $U_{total} = \frac{q _{0}^{2}}{2 C}$ , and $U_{E} = U_{total} cos^{2} (ω t)$ ; likewise $U_{B} = U_{total} sin^{2} (ω t)$ . Why this step? $U_{E} = \frac{q ^{2}}{2 C} = \frac{q _{0}^{2}}{2 C} cos^{2} (ω t)$ — the cosine-squared carries all the time dependence; the leftover fraction is the sine-squared, which is exactly $U_{B}$ .
(a) $U_{E} = U_{B} \Rightarrow cos^{2} ω t = sin^{2} ω t \Rightarrow tan^{2} ω t = 1 \Rightarrow ω t = π /4$ . Why this step? Equal shares means $cos^{2} = sin^{2}$ ; dividing gives $tan^{2} = 1$ . We use $tan$ because it packages the cosine/sine ratio into one condition.
(a) $t = \frac{π /4}{ω} = \frac{π}{4 ω} = \frac{T}{8}$ . Why this step? $ω t = π /4$ and $ω T = 2 π$ give $t / T = 1/8$ .
(b) $U_{E} = \frac{1}{4} U_{total} \Rightarrow cos^{2} ω t = \frac{1}{4} \Rightarrow cos ω t = \frac{1}{2} \Rightarrow ω t = π /3$ . Why this step? We want the first positive time, so take the smaller angle whose cosine is $+ \frac{1}{2}$ .
(b) $t = \frac{π /3}{ω} = \frac{T}{6}$ . Why this step? $ω t = π /3 = \frac{2 π}{6}$ , so $t = T /6$ . Since $T /6 > T /8$ , the capacitor drops to a quarter later than it hits half — matches the forecast (wave swung further).

Verify: at $t = T /6$ , $ω t = π /3$ : $cos^{2} (π /3) = (0.5)^{2} = 0.25$ ✓ and $sin^{2} (π /3) + cos^{2} (π /3) = 1$ ✓. At $t = T /8$ , $cos^{2} (π /4) = 0.5 = sin^{2} (π /4)$ ✓.

Ex 5 — Cell E: different initial condition (current flowing at $t = 0$ )

Worked example Statement

Now the switch is closed at the instant the capacitor is fully discharged ( $q (0) = 0$ ) but current is already flowing at its peak value $i_{0}$ . (Recall $q_{0}$ is the amplitude of the charge oscillation, and $i_{0} = q_{0} ω$ is the corresponding peak current.) Write $q (t)$ and $i (t)$ , and find when the capacitor first reaches full charge.

Forecast: the parent note used $cos$ because charge started at maximum. If charge starts at zero, which function should describe $q (t)$ ?

General solution: $q (t) = q_{0} cos (ω t + ϕ)$ for some phase $ϕ$ , where $q_{0}$ is the (still unknown) amplitude of the charge. Why this step? Every solution of $\overset{q}{¨} = - ω^{2} q$ is a shifted cosine; the initial conditions pick $ϕ$ and $q_{0}$ .
Apply $q (0) = 0$ : $q_{0} cos ϕ = 0 \Rightarrow ϕ = - π /2$ (so charge rises, not falls, right after $t = 0$ ). Why this step? We choose $ϕ = - π /2$ rather than $+ π /2$ so that current is initially positive (charging up).
Then $q (t) = q_{0} cos (ω t - π /2) = q_{0} sin (ω t)$ . Why this step? The identity $cos (θ - π /2) = sin θ$ . So a pure sine describes charge that starts empty. This is the mirror image of the parent's cosine case. On the phase ellipse, we simply start at a different point on the same ellipse.
Current: $i (t) = \frac{d q}{d t} = q_{0} ω cos (ω t) = i_{0} cos (ω t)$ with $i_{0} = q_{0} ω$ . At $t = 0$ : $i = i_{0}$ , and since the problem gives the peak current $i_{0}$ , this fixes the amplitude via $q_{0} = i_{0} / ω$ . Why this step? Differentiating $sin$ gives $cos$ , so now current leads with the cosine. The link $i_{0} = q_{0} ω$ (from the top of the page) is what converts the given peak current back into the charge amplitude — the two are not independent. Charge and current are still exactly 90° apart.
Capacitor first full when $sin (ω t) = 1 \Rightarrow ω t = π /2 \Rightarrow t = \frac{π}{2 ω} = \frac{T}{4}$ . Why this step? Starting empty, it takes a quarter cycle to fill — the same quarter-cycle timing as SHM going from equilibrium to maximum displacement.

Verify: check $\overset{q}{¨} = - ω^{2} q$ for $q = q_{0} sin ω t$ : $\overset{q}{¨} = - q_{0} ω^{2} sin ω t = - ω^{2} q$ ✓. And $q (0) = q_{0} sin 0 = 0$ ✓, $i (0) = i_{0} cos 0 = i_{0}$ ✓, and $i_{0} = q_{0} ω$ ✓.

Ex 6 — Cell F: limiting behaviour ( $L \to 0$ , $C \to \infty$ , and the full damped picture)

Worked example Statement

Explain what happens to the oscillation as (a) $L \to 0$ , (b) $C \to \infty$ , (c) a resistance $R$ is added. In (c), derive the damped frequency $ω_{d}$ from the equation of motion, give the critical-damping threshold, and describe the over-/critically-damped regimes where the circuit no longer oscillates.

Forecast: which limit makes the oscillation infinitely fast, and which makes it stop entirely? And when $R$ grows, at what value does the "oscillation" die completely rather than merely fade?

(a) $L \to 0$ : $T = 2 π L C \to 0$ , so $ω = 1/ L C \to \infty$ . Why this step? $L$ is the "mass." Zero mass ⟹ no inertia to slow the swap ⟹ infinitely fast oscillation. Physically the current can change instantly. Degenerate: not a real oscillator, it's a limit.
(b) $C \to \infty$ : $T = 2 π L C \to \infty$ , $ω \to 0$ . Why this step? $1/ C$ is the "spring constant." Infinite $C$ ⟹ zero springiness ⟹ nothing pulls the charge back ⟹ the oscillation period stretches to infinity (never returns). Degenerate: effectively no restoring force.
(c) Add $R$ : Kirchhoff's Voltage Law round the loop now gives $L \overset{q}{¨} + R \overset{q}{˙} + \frac{q}{C} = 0$ — the damped case (see LCR Circuit). Why this step? The resistor adds a voltage drop $R \overset{q}{˙}$ (its $V = i R$ with $i = \overset{q}{˙}$ ), which was absent in the ideal circuit.
Derive $ω_{d}$ : try $q = e^{s t}$ . Substituting, $L s^{2} + R s + \frac{1}{C} = 0$ , the characteristic equation. Its roots are $s = - \frac{R}{2 L} \pm (\frac{R}{2 L})^{2} - \frac{1}{L C} .$ When the term under the root is negative, write it as $\pm j ω_{d}$ with $ω_{d} = \frac{1}{L C} - (\frac{R}{2 L})^{2}, T_{d} = \frac{2 π}{ω _{d}} .$ Why this step? Guessing $q = e^{s t}$ turns the differential equation into an ordinary quadratic (that is why we use the exponential — it is the one function whose derivatives are copies of itself). The real part $- R /2 L$ gives amplitude decay $e^{- R t /2 L}$ ; the imaginary part gives the oscillation at $ω_{d}$ . Because $ω_{d} < ω$ , damping makes the swing slightly slower ( $T_{d} > T$ ).
Critical threshold: the root becomes real (no oscillation) exactly when $(\frac{R}{2 L})^{2} = \frac{1}{L C}$ , i.e. $R_{crit} = 2 \frac{L}{C} .$ Why this step? This is the boundary case where $ω_{d} = 0$ . For $R < R_{crit}$ the circuit oscillates while decaying (underdamped). At $R = R_{crit}$ it is critically damped — returns to zero as fast as possible without overshooting. For $R > R_{crit}$ it is overdamped — charge oozes back to zero with no oscillation at all. These are the important non-oscillatory edge cases.
Sanity ordering: as $R \to 0$ , $ω_{d} \to 1/ L C = ω$ and $T_{d} \to 2 π L C$ , recovering the ideal LC result. Why this step? A limit check — any correct damped formula must collapse back to the undamped one when the new ingredient ( $R$ ) is switched off; this confirms we did not break the $R = 0$ case.

Verify: for our base circuit, $R_{crit} = 2 L / C = 2 (2 \times 1 0^{- 3}) / (5 \times 1 0^{- 6}) = 2400 = 40 Ω$ . Take $R = 8 Ω$ (well below, so underdamped): $\frac{1}{L C} = 1 0^{8}$ , $(\frac{R}{2 L})^{2} = (2000)^{2} = 4 \times 1 0^{6}$ , so $ω_{d} = 1 0^{8} - 4 \times 1 0^{6} = 9.6 \times 1 0^{7} \approx 9798$ rad/s $< 1 0^{4}$ ✓ (slower). $T_{d} = 2 π / ω_{d} \approx 6.413 \times 1 0^{- 4}$ s $> 6.283 \times 1 0^{- 4}$ s ✓. At $R = 40 Ω$ : $(\frac{R}{2 L})^{2} = (1 0^{4})^{2} = 1 0^{8} = \frac{1}{L C}$ , so $ω_{d} = 0$ — exactly critical ✓.

Ex 7 — Cell G: real-world word problem (radio tuning)

Worked example Statement

A radio's tuning circuit uses a fixed inductor $L = 0.25 μ H$ . What capacitance $C$ makes it resonate at $f = 100$ MHz (an FM station)? (See Resonance and AC Circuits — resonance is exactly $ω = 1/ L C$ .)

Forecast: to hit a high frequency, should $C$ be large or tiny?

Start from $f = \frac{1}{2 π L C}$ and solve for $C$ . Why this step? We know $f$ and $L$ , want $C$ — rearrange the tuning formula.
Square: $f^{2} = \frac{1}{4 π ^{2} L C} \Rightarrow C = \frac{1}{4 π ^{2} f ^{2} L}$ . Why this step? $L C$ appears under a reciprocal; squaring removes the root cleanly.
Plug in $f = 1 0^{8}$ Hz, $L = 0.25 \times 1 0^{- 6}$ H: $C = \frac{1}{4 π ^{2} ( 1 0 ^{8} ) ^{2} ( 0.25 \times 1 0 ^{- 6} )} = \frac{1}{4 π ^{2} \cdot 1 0 ^{16} \cdot 0.25 \times 1 0 ^{- 6}} .$ Denominator $= 4 π^{2} \cdot 2.5 \times 1 0^{9} \approx 9.870 \times 1 0^{10}$ , so $C \approx 1.013 \times 1 0^{- 11}$ F $\approx 10.1$ pF. Why this step? Direct substitution; keep track of powers of ten carefully.
Interpretation: a tiny ~10 pF capacitor — high frequency needs small $C$ , matching the forecast. Why this step? Reading the number back as physics confirms the design intuition and catches gross power-of-ten slips.

Verify: feed the exact $C = \frac{1}{4 π ^{2} f ^{2} L}$ back into $f = \frac{1}{2 π L C}$ : it returns exactly $1 0^{8}$ Hz $= 100$ MHz ✓. Units: $1/ H \cdot F$ is $1/ s$ , i.e. hertz. ✓

Ex 8 — Cell H: exam twist (timing + energy + phase together)

Worked example Statement

Base circuit, $q_{0} = 2 \times 1 0^{- 4}$ C. Starting from full charge at rest, find the first time $t^{*}$ at which the inductor's magnetic energy equals three times the capacitor's electric energy, and give the current there.

Forecast: $U_{B}$ dominating over $U_{E}$ means we are past the halfway sharing point ( $T /8$ ) but before the swap ( $T /4$ ). So $t^{*}$ lies between $T /8$ and $T /4$ — agree?

Condition: $U_{B} = 3 U_{E} \Rightarrow sin^{2} ω t = 3 cos^{2} ω t \Rightarrow tan^{2} ω t = 3$ . Why this step? $U_{B} \propto sin^{2}$ , $U_{E} \propto cos^{2}$ ; their ratio is $tan^{2}$ . Using $tan$ collapses two conditions into one.
$tan ω t = 3 \Rightarrow ω t = \frac{π}{3}$ (first positive solution). Why this step? $tan (π /3) = 3$ ; we take the smallest positive angle for the first time.
$t^{*} = \frac{π /3}{ω} = \frac{π}{3 \cdot 1 0 ^{4}} \approx 1.047 \times 1 0^{- 4}$ s. Check bracket: $T /8 = 0.785 \times 1 0^{- 4}$ s, $T /4 = 1.571 \times 1 0^{- 4}$ s — yes $t^{*}$ sits between them ✓ (forecast confirmed). Why this step? Convert angle to time and sanity-check against the two landmark instants.
Current: $i = - q_{0} ω sin (π /3) = - (2) \cdot \frac{3}{2} = - 3 \approx - 1.732$ A. Why this step? Use the current solution at $ω t = π /3$ ; the sign is negative (discharging first quarter).

Verify: at $ω t = π /3$ : $U_{B} / U_{E} = sin^{2} (π /3) / cos^{2} (π /3) = 0.75/0.25 = 3$ ✓. Energy check: $U_{B} = \frac{1}{2} L i^{2} = \frac{1}{2} (2 \times 1 0^{- 3}) (1.732)^{2} = 3 \times 1 0^{- 3}$ J, and total is $4 \times 1 0^{- 3}$ J, so $U_{B}$ should be $\frac{3}{4} \cdot 4 \times 1 0^{- 3} = 3 \times 1 0^{- 3}$ J ✓.

Recall Quick self-test on the matrix

Which cell does "find $C$ for a 100 MHz radio" belong to? ::: Cell G (real-world design). In Ex 3, why is $q < 0$ at $3 T /8$ but the current still negative? ::: Charge has flipped polarity (2nd quarter) but current still 90° behind, still draining that direction. Why does $L \to 0$ give infinite frequency? ::: $L$ is the inertia; zero inertia means the current can swap instantly, $T = 2 π L C \to 0$ . First time $U_{E} = \frac{1}{4} U_{total}$ ? ::: $t = T /6$ (from $cos ω t = \frac{1}{2}$ ). At what $R$ does the circuit stop oscillating entirely? ::: At $R_{crit} = 2 L / C$ (critical damping); above it, overdamped. With small $R$ added, is the period longer or shorter? ::: Longer, because $ω_{d} = 1/ L C - (R /2 L)^{2} < ω$ , so $T_{d} = 2 π / ω_{d} > T$ .

Connections

Parent topic (Hinglish) — the theory these examples drill.
Simple Harmonic Motion — the quarter-cycle timings (Ex 5) come straight from SHM.
Capacitance and Energy in Capacitors — $U_{E} = q^{2} /2 C$ used in Ex 2, 4, 8.
Inductance and Self-Induction — $U_{B} = \frac{1}{2} L i^{2}$ used in Ex 2, 8.
Kirchhoff's Voltage Law — the loop equation behind every solution.
Damped Oscillations / LCR Circuit — the $R \neq = 0$ limit, damped period, and critical damping in Ex 6.
Resonance and AC Circuits — the tuning formula in Ex 7.