---
title: "Maximum current in LC circuit"
subject: "Engineering Physics"
chapter: 1
key: "semester_1/engineering_physics:lc-maximum-current-numerical"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/lc-maximum-current-numerical
site: BCT Engineering
---

# Maximum current in LC circuit

Engineering Physics · Chapter 1: Oscillation

- Asked in: 2070 Chaitra, 5 marks
- Video: [An oscillating LC circuit consisting of a 1.4 nF capacitor and... | Physics Electricity & Magnetism](https://www.youtube.com/watch?v=k8hcmPOA348) — Computer Engineering life
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/lc-maximum-current-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1

## Question

A 2μF capacitor is charged up to 50V. The battery is disconnected and a 50mH coil is connected across the capacitor so that LC oscillation occurs. Calculate the maximum value of the current in the circuit.

## Short answer

Using energy conservation ½CV₀² = ½LI_max² gives I_max = V₀√(C/L). With C=2μF, L=50mH, V₀=50V, I_max = 50√(2×10⁻⁶/50×10⁻³) ≈ 0.316 A.

## Answer

Given: C = 2 μF = 2 × 10⁻⁶ F, V0 = 50 V, L = 50 mH = 50 × 10⁻³ H.

**Method (energy conservation).** When the coil is connected, all the energy is in the capacitor: U = ½CV0². At the instant the current is maximum, the capacitor is empty and all energy is in the inductor: ½LI_max². There is no resistance, so the two are equal.

```text
½ C V0² = ½ L I_max²
I_max = V0 √(C/L)
```

#### Steps
1. Energy initially stored:
```text
U = ½ × 2×10⁻⁶ × (50)² = 2.5 × 10⁻³ J
```
2. C/L = 2×10⁻⁶ / 50×10⁻³ = 4 × 10⁻⁵ F/H.
3. √(C/L) = 6.325 × 10⁻³ A/V.
4. I_max = 50 × 6.325 × 10⁻³ = 0.316 A.

Check: ½ × 0.05 × (0.316)² = 2.5 × 10⁻³ J.

**Answer: I_max ≈ 0.316 A (about 0.32 A).**

## Check yourself

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

1. Which principle gives I_max?
   A. Ohm's law
   B. Charge conservation only
   C. Energy conservation: (1/2)CV0^2 = (1/2)L I_max^2
   D. Faraday's law with R
   Answer: C. No resistance, so energy is conserved. (Key point: Method)

2. I_max =?
   A. V0 sqrt(L/C)
   B. V0 / sqrt(LC)
   C. V0 C/L
   D. V0 sqrt(C/L)
   Answer: D. Solved from energy equality. (Key point: Formula)

3. Initial energy is?
   A. 2.5 x 10^-3 J
   B. 5 x 10^-3 J
   C. 1.25 x 10^-3 J
   D. 2.5 J
   Answer: A. 0.5 x 2e-6 x 2500. (Key point: Initial energy)

4. I_max =?
   A. 0.5 A
   B. 0.316 A
   C. 3.16 A
   D. 0.0316 A
   Answer: B. 50 x sqrt(4e-5) = 0.316 A. (Key point: Result)

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