Worked examples · Sheet 1

Free oscillations in LC circuits

Theory used: Oscillations of the LC circuit Conservation of energy in the LC circuit Analogy with the mass-spring oscillator

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

An oscillating circuit consists of an inductor of inductance $L = 25 mH$ and a capacitor of capacitance $C = 4.0 μ F$ , connected by wires of negligible resistance. Compute the natural angular frequency, the frequency and the period of the oscillations.

ω_{0} = \frac{1}{L C}, f_{0} = \frac{ω _{0}}{2 π}, T = \frac{1}{f _{0}} = 2 π L C

L C ω_{0} f_{0} T = 25 \cdot 1 0^{- 3} \cdot 4.0 \cdot 1 0^{- 6} s^{2} = 1.0 \cdot 1 0^{- 7} s^{2} = \frac{1}{1.0 \cdot 1 0 ^{- 7}} rad/s \approx 3162 rad/s \approx 3.2 \cdot 1 0^{3} rad/s = \frac{3162}{2 π} Hz \approx 503 Hz \approx 5.0 \cdot 1 0^{2} Hz = \frac{1}{503} s \approx 2.0 \cdot 1 0^{- 3} s = 2.0 ms

The resistance of the circuit is negligible, so the theorem on the oscillations of the LC circuit applies, by which the charge of the capacitor oscillates in harmonic motion with angular frequency $ω_{0} = \frac{1}{L C}$ . Frequency and period follow from the relations $f_{0} = \frac{ω _{0}}{2 π}$ and $T = \frac{1}{f _{0}}$ . The product $L C$ is measured in $s^{2}$ , because $1 H = 1 V \cdot s/A$ and $1 F = 1 A \cdot s/V$ ; therefore $\frac{1}{L C}$ is the reciprocal of a time, as an angular frequency must be. The data have two significant figures, and so do the results.

Exercise 2

A capacitor of capacitance $C = 2.0 μ F$ is charged with the potential difference $Δ V = 12 V$ and at time $t = 0$ it is connected to an inductor of inductance $L = 50 mH$ by closing a switch. The resistance of the circuit is negligible. Compute the initial charge, the energy of the circuit, the maximum current, and the first instant at which the current reaches its maximum value.

The capacitor charged with

Q_{0} = C Δ V

is connected to the inductor by closing the switch at time

t = 0

Q_{0} = C Δ V, U = \frac{Q _{0}^{2}}{2 C} = \frac{1}{2} C Δ V^{2}, \frac{1}{2} L I_{0}^{2} = U ⟹ I_{0} = \frac{Q _{0}}{L C}, t_{1} = \frac{T}{4} = \frac{π}{2} L C

Q_{0} U L C I_{0} t_{1} = 2.0 \cdot 1 0^{- 6} \cdot 12 C = 2.4 \cdot 1 0^{- 5} C = 24 μ C = \frac{1}{2} \cdot 2.0 \cdot 1 0^{- 6} \cdot 1 2^{2} J = 1.44 \cdot 1 0^{- 4} J \approx 0.14 mJ = 50 \cdot 1 0^{- 3} \cdot 2.0 \cdot 1 0^{- 6} s = 1.0 \cdot 1 0^{- 7} s \approx 3.16 \cdot 1 0^{- 4} s = \frac{2.4 \cdot 1 0 ^{- 5}}{3.16 \cdot 1 0 ^{- 4}} A \approx 0.076 A = 76 mA = \frac{π}{2} \cdot 3.16 \cdot 1 0^{- 4} s \approx 5.0 \cdot 1 0^{- 4} s = 0.50 ms

The initial charge follows from the definition of capacitance, and the initial energy is entirely in the electric field of the capacitor, because at the instant of closing the current is zero. The resistance is negligible, so by the conservation of energy in the LC circuit the energy remains $U$ at every instant. When the capacitor is discharged all the energy is in the inductor and the current is maximum; setting $\frac{1}{2} L I_{0}^{2}$ equal to $U$ we obtain $I_{0}$ , which coincides with $ω_{0} Q_{0}$ . As a check, $\frac{1}{2} \cdot 50 \cdot 1 0^{- 3} \cdot 0.075 9^{2} J \approx 1.44 \cdot 1 0^{- 4} J$ .

The charge is $q (t) = Q_{0} cos (ω_{0} t)$ and the current $i (t) = - ω_{0} Q_{0} sin (ω_{0} t)$ is maximum in magnitude when $sin (ω_{0} t) = \pm 1$ . The first positive instant is $ω_{0} t_{1} = \frac{π}{2}$ , that is, a quarter of a period after the switch is closed, and it is about $0.50 ms$ .

Exercise 3

In an ideal $L C$ circuit with $L = 90 mH$ and $C = 10 μ F$ the capacitor has at time $t = 0$ the charge $Q_{0} = 50 μ C$ and the current is zero. Determine the first instant at which the energy is shared equally between the capacitor and the inductor, and compute at that instant the charge of the capacitor and the magnitude of the current.

\frac{q ^{2}}{2 C} = \frac{1}{2} \cdot \frac{Q _{0}^{2}}{2 C} ⟹ ∣ q ∣ = \frac{Q _{0}}{2}, \frac{1}{2} L i^{2} = \frac{1}{2} \cdot \frac{1}{2} L I_{0}^{2} ⟹ ∣ i ∣ = \frac{I _{0}}{2} = \frac{ω _{0} Q _{0}}{2}

Q_{0} cos (ω_{0} t) = \frac{Q _{0}}{2} ⟹ ω_{0} t = \frac{π}{4} ⟹ t = \frac{π}{4 ω _{0}} = \frac{π}{4} L C

ω_{0} t q ∣ i ∣ = \frac{1}{90 \cdot 1 0 ^{- 3} \cdot 10 \cdot 1 0 ^{- 6}} rad/s = \frac{1}{9.0 \cdot 1 0 ^{- 7}} rad/s \approx 1054 rad/s = \frac{π}{4 \cdot 1054} s \approx 7.5 \cdot 1 0^{- 4} s = 0.75 ms = \frac{50}{2} μ C \approx 35 μ C = \frac{1054 \cdot 50 \cdot 1 0 ^{- 6}}{2} A \approx 0.037 A = 37 mA

By the conservation of energy in the LC circuit the total energy is always $\frac{Q _{0}^{2}}{2 C}$ . The capacitor holds half of it when $q^{2} = \frac{Q _{0}^{2}}{2}$ ; at that instant the other half is in the inductor, so $i^{2} = \frac{I _{0}^{2}}{2}$ , with $I_{0} = ω_{0} Q_{0}$ the amplitude of the current.

The initial conditions are those of the theorem on the oscillations of the $L C$ circuit, therefore $q (t) = Q_{0} cos (ω_{0} t)$ . The charge drops to $\frac{Q _{0}}{2}$ for the first time when $cos (ω_{0} t) = \frac{2}{2}$ with $ω_{0} t$ in the first quadrant, that is, for $ω_{0} t = \frac{π}{4}$ , after one eighth of a period. The result does not depend on $Q_{0}$ , because the sharing of the energy depends only on the phase of the oscillation. In the analogy with the mass-spring oscillator it is the instant at which kinetic energy and elastic potential energy are equal.