---
title: "LC circuit analogy, equal maximum energies"
subject: "Engineering Physics"
chapter: 1
key: "semester_1/engineering_physics:lc-max-energies-equal"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/lc-max-energies-equal
site: BCT Engineering
---

# LC circuit analogy, equal maximum energies

Engineering Physics · Chapter 1: Oscillation

- Asked in: 2073 Chaitra, 5 marks; 2072 Chaitra, 5 marks; 2071 Chaitra, 5 marks; 2069 Chaitra, 5 marks; 2067 Asoj, 5 marks (asked 5 times)
- Video: [LC Oscillations - Comparison with SHM](https://www.youtube.com/watch?v=Fa97J8eqT5E) — ibis physics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/lc-max-energies-equal
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1

## Question

Prove that an LC circuit is an analogue of the spring-mass system (simple harmonic motion) and hence prove that the maximum energy stored in the capacitor (electric field) is equal to the maximum energy stored in the inductor (magnetic field).

## Short answer

The LC circuit differential equation d²q/dt² + (1/LC)q = 0 is identical to mass-spring SHM. With q = q0 cosω0t, maximum capacitor energy q0²/(2C) equals maximum inductor energy ½L(q0ω0)² = q0²/(2C). Thus energy oscillates between electric and magnetic fields.

## Answer

An LC circuit (an ideal inductor L and capacitor C, with no resistance) oscillates electrically in exactly the same mathematical way as a mass on a spring oscillates mechanically.

#### Differential equation
```text
   +q |----|-q        Capacitor C discharges through L.
      |    |          At any instant: q = charge, i = dq/dt.
      C    L
      |____|
```
By Kirchhoff's voltage law, (voltage across C) + (voltage across L) = 0:

```text
q/C + L di/dt = 0
L d²q/dt² + (1/C) q = 0
d²q/dt² + ω0² q = 0 ,   ω0 = 1/√(LC)
```
For a mass on a spring: m d²x/dt² + kx = 0, i.e. d²x/dt² + ω0² x = 0 with ω0 = √(k/m). The two equations have the same form, so q varies simple harmonically: q = q0 cos(ω0 t).

| Spring-mass | LC circuit |
|---|---|
| displacement x | charge q |
| velocity v = dx/dt | current i = dq/dt |
| mass m (inertia) | inductance L |
| spring constant k | 1/C |
| PE = ½kx² | electric energy q²/2C |
| KE = ½mv² | magnetic energy ½Li² |

So the LC circuit is the electrical analogue of SHM.

#### Maximum energies
1. Charge: q = q0 cos ω0t, so the current is i = dq/dt = −q0 ω0 sin ω0t.
2. Energy in the capacitor: U_E = q²/2C = (q0²/2C) cos² ω0t. Its maximum value is q0²/2C (when i = 0).
3. Energy in the inductor: U_B = ½Li² = ½L q0² ω0² sin² ω0t. Its maximum value is ½L q0² ω0² (when q = 0).
4. Put ω0² = 1/LC: (U_B)max = ½L q0² /(LC) = q0²/2C.

```text
(U_E)max = q0²/2C = (U_B)max
```
Hence the maximum energy stored in the electric field of the capacitor equals the maximum energy stored in the magnetic field of the inductor.

Also U_E + U_B = (q0²/2C)(cos² ω0t + sin² ω0t) = q0²/2C = constant. Energy is exchanged between the two fields while the total stays the same, just as KE and PE are exchanged in SHM.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Applying Kirchhoff's voltage law to an ideal LC circuit gives which equation?
   A. L d2q/dt2 + R dq/dt + q/C = 0
   B. L d2q/dt2 - q/C = 0
   C. C d2q/dt2 + q/L = 0
   D. L d2q/dt2 + q/C = 0
   Answer: D. With no resistance, q/C + L di/dt = 0 and i = dq/dt, which gives L q'' + q/C = 0. (Key point: Kirchhoff equation)

2. In the LC / spring-mass analogy, which pairing is correct?
   A. L corresponds to mass m and 1/C to spring constant k
   B. L corresponds to k and C to m
   C. C corresponds to mass m and L to k
   D. R corresponds to mass m and 1/C to k
   Answer: A. Inductance acts as inertia (m), and 1/C acts as stiffness (k), so w0 = 1/sqrt(LC) matches sqrt(k/m). (Key point: Analogy)

3. Why is (U_B)max = q0^2/2C?
   A. Because the capacitor and inductor have equal values
   B. (U_B)max = (1/2)L q0^2 w0^2 and w0^2 = 1/LC
   C. Because the current is zero at the maximum
   D. Because the total energy is zero
   Answer: B. Substituting w0^2 = 1/LC into (1/2)L q0^2 w0^2 gives q0^2/2C, equal to (U_E)max. (Key point: Maximum energies)

4. What happens to U_E + U_B as time passes in the ideal LC circuit?
   A. It decays exponentially
   B. It oscillates at frequency w0
   C. It stays constant at q0^2/2C
   D. It is zero at all times
   Answer: C. cos^2 + sin^2 = 1 so the sum is constant; energy only moves between the fields. (Key point: Energy exchange)

## Related in Engineering Physics

- [Energy stored in an inductor](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aenergy-density-inductor-magnetic-field) — Engineering Physics, chapter 8, asked 4×
- [Electric and magnetic energy densities](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aenergy-density-electric-magnetic-proportional) — Engineering Physics, chapter 8
- [Maximum current in LC circuit](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Alc-maximum-current-numerical) — Engineering Physics, chapter 1
- [Energy stored in coil, time constants](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acoil-energy-stored-time-constants) — Engineering Physics, chapter 8
- [Induced magnetic field, displacement current](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adisplacement-current-induced-field-circular-plates) — Engineering Physics, chapter 9
- [Energy stored in parallel capacitors](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aparallel-capacitors-energy-stored) — Engineering Physics, chapter 7

## Related in other subjects

- [Magnetic circuit, analogy with electric circuit](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Amagnetic-circuit-ohms-law) — Electrical Machines, chapter 1
- [F-V and F-I analogy of mass-spring-damper system](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Afv-fi-m-k1-k2-b2) — Control Systems, chapter 2
- [Two-mass spring system X1/F and F-V analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fv-m1-m2-k1-k2) — Control Systems, chapter 2
- [Three-mass system X2/F with F-V and F-I analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fv-fi-three-mass) — Control Systems, chapter 2, asked 3×