LC circuit — oscillations (electrical analog of SHM): Question bank

Reminders of the machinery you'll need (all earned in the parent note):

$q$ = charge on the capacitor (the "position"), $i = \overset{q}{˙}$ = current (the "velocity").
$q (t) = q_{0} cos (ω t)$ , $i (t) = - q_{0} ω sin (ω t)$ , with $ω = \frac{1}{L C}$ .
$ω$ = angular frequency (radians of the cycle swept per second). The period $T$ is the time for one full cycle: since one full cycle is $2 π$ radians, $T = \frac{2 π}{ω} = 2 π L C$ . So " $t = T /4$ " just means "a quarter of the way through one cycle."
Energies: $U_{E} = \frac{q ^{2}}{2 C}$ (capacitor, electric field), $U_{B} = \frac{1}{2} L i^{2}$ (inductor, magnetic field), total fixed at $\frac{q _{0}^{2}}{2 C}$ .

Before the traps, fix the pictures in your mind.

Figure 1 — the $9 0^{\circ}$ dance. The cyan curve is the charge $q (t)$ and the amber curve is the current $i (t)$ , drawn on the same time axis. Look for one thing: where the cyan curve is flat (a peak or trough), the amber curve crosses zero, and vice-versa. The double-headed white arrow marks the quarter-cycle gap between the two peaks — that is the $9 0^{\circ}$ phase difference made visible.

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

Figure 2 — the two buckets. Same time axis, now in energy. Cyan is the capacitor's energy $U_{E}$ , amber is the inductor's energy $U_{B}$ , and the dashed white line on top is their sum. Look for two things: the crest of one bucket always sits over the trough of the other (energy is being handed back and forth), and the dashed total never moves — it is a flat line, which is energy conservation with your own eyes. The white dots mark the equal-sharing instants where the two curves cross at half-height.

Figure 3 — a whisper of resistance. The faint cyan wave is the ideal $R = 0$ oscillation; the bold amber wave is the same circuit with a small resistor added. Look for two changes: the amber wave shrinks inside the dashed white envelope (energy is now leaking away), and its peaks drift slightly to the right of the cyan peaks — the frequency has dropped from $ω$ to $ω^{'}$ . Both effects are the subject of the edge-case questions below.

True or false — justify

Charge and current in an LC circuit both peak at the same instant.

False. Current is the slope of

q (t)

; at the top of a cosine the slope is zero, so when

q

is maximum the current is zero. They are

9 0^{\circ}

apart — see the offset peaks in Figure 1.

The total energy in an ideal LC circuit stays constant only on average, not at every instant.

False. It is constant at every instant:

\frac{q _{0}^{2}}{2 C} (cos^{2} ω t + sin^{2} ω t) = \frac{q _{0}^{2}}{2 C}

always. In Figure 2 the two energy humps always sum to the same flat line. There is nothing to dissipate it since

R = 0

Doubling the capacitance makes the oscillation faster.

False.

ω = 1/ L C

, so more

C

means smaller

ω

— slower. A bigger

C

is a "softer spring" (

1/ C

shrinks), so it takes longer to swing back.

If you start the circuit with the capacitor uncharged AND no current, it will oscillate anyway.

False. With

q_{0} = 0

and

i_{0} = 0

there is zero energy in the system, so

q (t) = 0

for all time. You need to inject energy first (charge the cap or run a current).

The inductor's magnetic energy is largest exactly when the capacitor is fully charged.

False. It's the opposite:

U_{B}

peaks when

q = 0

(current maximum), and

U_{E}

peaks when

i = 0

(charge maximum). In Figure 2 the crest of one bucket sits over the trough of the other.

At the moment

q = 0

, the circuit has no energy.

False. At

q = 0

the electric energy is zero, but all the energy sits in the inductor's magnetic field (

U_{B}

is at its maximum). Energy just moved stores.

The current reverses direction twice per period.

True.

i = - q_{0} ω sin (ω t)

passes through zero and flips sign at

t = T /4

and

t = 3 T /4

(where the sine crosses zero). Those are the two turnarounds in each cycle: the charge sloshes one way, momentarily stops, then sloshes back the other way.

In the analogy, the inductance

L

plays the role of the spring constant.

False.

L

plays the role of the mass

m

(inertia — it resists changes in current, just as mass resists changes in velocity). The spring constant maps to

1/ C

Removing the resistor from a real circuit is what allows perpetual oscillation.

True (idealised). With

R = 0

there is no dissipation term, so the damping vanishes and the amplitude never decays. Real wires always have some

R

, so true perpetual motion is a limit, not a lab fact.

Spot the error

"Since

ω = 1/ L C

, the period is

T = 2 π / L C

Wrong.

T = 2 π / ω = 2 π L C

. Inverting

ω

moves

L C

into the numerator, so

T

grows with

L

and

C

, not shrinks.

"Energy is shared equally between

L

and

C

t = T /4

Wrong. At

T /4

(

ω t = π /2

) the charge is zero and all energy is magnetic. Equal sharing needs

cos^{2} = sin^{2}

, i.e.

ω t = π /4

, which is

t = T /8

"The capacitor's voltage and the inductor's voltage add up to the battery voltage."

Wrong. There is no battery in an LC circuit. Kirchhoff's loop gives

V_{C} + V_{L} = 0

; the two voltages are equal and opposite at every instant.

U_{B} = \frac{1}{2} L i^{2}

, so the magnetic energy amplitude is

\frac{1}{2} L q_{0}^{2}

Wrong. The current amplitude is

i_{0} = q_{0} ω

, so

U_{B, m a x} = \frac{1}{2} L (q_{0} ω)^{2} = \frac{1}{2} L q_{0}^{2} ω^{2}

. Using

ω^{2} = 1/ L C

this equals

q_{0}^{2} /2 C

— the same as

U_{E, m a x}

, as energy conservation demands.

"The back-EMF is

V_{L} = L i

Wrong. An inductor opposes the change in current, so

V_{L} = L \frac{d i}{d t}

— proportional to the rate of change, not the current itself. See Inductance and Self-Induction.

"Because current lags charge by

9 0^{\circ}

, the current is

q_{0} ω cos (ω t)

Wrong sign/function. Differentiating

q_{0} cos (ω t)

gives

- q_{0} ω sin (ω t)

, a negative sine, not a cosine. To see the lag concretely, compare the peaks: the charge

q_{0} cos (ω t)

peaks at

t = 0

, while the current

- q_{0} ω sin (ω t)

reaches its positive peak later, at

ω t = 3 π /2

(i.e.

t = 3 T /4

). Because the current's matching peak arrives a quarter-cycle after the charge's, we say current lags charge by

9 0^{\circ}

. (Equivalently

- sin θ = cos (θ + 9 0^{\circ})

, but reading it off the peaks is the safest way to keep lead/lag straight.)

Why questions

Why does the charge overshoot zero instead of just stopping there?

Because at

q = 0

the current (energy) is maximum, and the inductor resists any sudden change in that current. Picture the pendulum at the bottom of its swing: it's moving fastest, and its inertia won't let it stop. Here the inductor is that inertia — it keeps current flowing past

q = 0

, and that continuing current is charge piling onto the plates the other way. The overshoot is inertia refusing to brake instantly.

Why do we say

1/ C

(not

C

) is the analog of the spring constant

k

A spring's restoring force is

F = - k x

: displacement

x

produces a push proportional to

k

. Here the "displacement" is the charge

q

, and the restoring "push" is the capacitor voltage

V_{C} = q / C

. So the proportionality constant between displacement and restoring push is

1/ C

, which is what

k

means. A small

C

therefore behaves like a stiff spring: even a little charge builds a large voltage that slams the current back, giving a fast, high-

ω

oscillation.

Why is Kirchhoff's Voltage Law the right starting tool here, not energy conservation?

Kirchhoff's Voltage Law is a statement about voltages at every instant, so it produces a differential equation in

q (t)

— a rule that ties

q

, its slope, and its curvature together at all times. That's what pins down the full waveform

q_{0} cos (ω t)

. Energy conservation is a single number that's true across the whole motion; it tells you peak current from peak charge, but it can't distinguish a cosine from any other curve with the same energy, so it can never generate the time-dependence on its own.

Why must charge and current be exactly

9 0^{\circ}

apart, not some other phase?

Current is defined as the derivative (slope) of charge, and differentiation shifts a cosine into

- sin

, which is the same shape moved a quarter-cycle over. So the phase gap isn't a physical coincidence you could tune — it's baked into the meaning of "rate of change." Wherever

q

is flat (a peak or trough), its slope is zero, so

i = 0

; wherever

q

races through zero, its slope is steepest, so

∣ i ∣

peaks. Figure 1 makes this slope-to-height correspondence visible.

Why does a bigger inductor slow the oscillation?

L

is the electrical inertia: the voltage it demands to change the current is

L d i / d t

, so a large

L

means even a modest change in current costs a big voltage. Current therefore builds and drains sluggishly, and the whole slosh takes longer. In the mechanical analogy this is a heavier mass — harder to get moving, harder to stop — and

ω = 1/ L C

duly shrinks as

L

grows, stretching the period

T = 2 π L C

Why is

ω = 1/ L C

also called the resonant frequency?

Left alone, this LC pair oscillates at exactly

ω = 1/ L C

— its "natural note," set only by how much inertia (

L

) and springiness (

1/ C

) it has. If you now drive it with an AC source, matching the drive to this natural note lets each push arrive in step with the motion it reinforces, so the amplitude climbs to a peak. That resonant peak is precisely why

1/ L C

is the frequency tuned to in radios — the subject of Resonance and AC Circuits.

Edge cases

What happens to

ω

and

T

C \to \infty

(a huge capacitor)?

ω = 1/ L C \to 0

and

T = 2 π L C \to \infty

. An infinitely "soft spring" never builds enough voltage to pull the charge back, so the period stretches to infinity — effectively no oscillation.

What happens to

ω

and

T

C \to 0

(vanishing capacitance)?

The opposite singular limit:

ω = 1/ L C \to \infty

and

T = 2 π L C \to 0

. An infinitely "stiff spring" (

1/ C \to \infty

) slams the charge back instantly, so the oscillation becomes infinitely fast. Physically the tiny capacitor holds almost no charge, so there is barely any energy to trade and the idealised LC model breaks down at this extreme.

What happens as

L \to 0

(an ideal wire, no inductor)?

ω \to \infty

and

T \to 0

: infinitely fast, mirroring the

C \to 0

case. Physically there is no inertia to store magnetic energy, so the circuit degenerates — the capacitor would discharge instantly through the wire, and the LC model breaks down.

If you introduce a small resistance

R

, does the oscillation stop being harmonic immediately?

No, but it is no longer strictly harmonic. The amplitude decays exponentially, and the frequency shifts to

ω^{'} = \frac{1}{L C} - (\frac{R}{2 L})^{2}

, which is slightly below the ideal

ω = 1/ L C

. So the waveform oscillates at

ω^{'}

under a shrinking envelope (see Figure 3). This is the underdamped regime of the LCR Circuit and Damped Oscillations.

Keep increasing

R

— is there a value where oscillation stops entirely?

Yes. When

R = 2 L / C

the term inside the square root,

\frac{1}{L C} - (\frac{R}{2 L})^{2}

, hits exactly zero, so

ω^{'} = 0

. This is the critically damped case: the charge returns to zero as fast as possible without overshooting — no back-and-forth at all. It is the sharp boundary between "still wiggles" and "never wiggles."

What about even larger resistance,

R > 2 L / C

Now

\frac{1}{L C} < (\frac{R}{2 L})^{2}

, so the quantity under the root is negative and

ω^{'}

is no longer a real oscillation frequency at all. This is the overdamped regime: the charge oozes slowly back to zero as a sum of two decaying exponentials, with no oscillation. In the limit

R \to \infty

the current is choked off completely and the charge is essentially frozen — the swing is buried in molasses.

The capacitor starts fully discharged but a current

i_{0}

is already flowing. Does it still oscillate?

Yes. The energy now starts entirely in the inductor (

U_{B}

maximum), and

q (t) = \frac{i _{0}}{ω} sin (ω t)

— a sine solution. Charge and current still oscillate, just with a shifted starting phase.

Is there any instant when both

U_{E} = 0

and

U_{B} = 0

simultaneously?

No. Since

U_{E} + U_{B} = q_{0}^{2} /2 C

is a fixed positive constant, they can never both vanish at once. When one is zero the other holds the entire energy.

t = T /2

, how do

q

and

i

compare with

t = 0

q (T /2) = q_{0} cos (π) = - q_{0}

(fully charged the opposite way) and

i (T /2) = - q_{0} ω sin (π) = 0

(momentarily at rest). It's the mirror image of the start — the swing at the far top of the other side.

Recall One-line self-test

Cover every answer above and speak your reasoning aloud before revealing. If any answer is just "true" or "false" in your head with no because, you haven't learned it yet.

Connections

Parent topic
Simple Harmonic Motion — the analogy that resolves half these traps.
Capacitance and Energy in Capacitors · Inductance and Self-Induction — where $U_{E}$ and $V_{L}$ come from.
Kirchhoff's Voltage Law — the loop rule behind the equation of motion.
Damped Oscillations / LCR Circuit — the $R \neq = 0$ edge cases (underdamped, critical, overdamped).
Resonance and AC Circuits — why $ω = 1/ L C$ is the resonant frequency.