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Lesson 3: RLC Electrical Circuits

Estimated time: 35-45 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Kirchhoff's Voltage Law and the RLC Equation

A series RLC circuit contains a resistor R, inductor L, and capacitor C connected in series with a voltage source E(t). Kirchhoff's voltage law states the sum of voltage drops equals the applied voltage:

RLC Circuit Equation: Lq'' + Rq' + (1/C)q = E(t), where q(t) is the charge on the capacitor, i(t) = q'(t) is the current, and the voltage drops are: L di/dt across the inductor, Ri across the resistor, and q/C across the capacitor.

This has exactly the same form as the spring-mass equation mx'' + cx' + kx = F(t), with L playing the role of mass, R playing the role of damping, and 1/C playing the role of the spring constant.

Free Response (E(t) = 0)

The characteristic equation is Lr² + Rr + 1/C = 0 with discriminant D = R² - 4L/C.

Circuit Damping Classification:

Overdamped: R² > 4L/C (large resistance, no oscillation)

Critically damped: R² = 4L/C (fastest decay without oscillation)

Underdamped: R² < 4L/C (oscillating current and charge)

Example 1: Classifying an RLC Circuit

Classify: L = 1 H, R = 4 Ω, C = 1/5 F.

Step 1: Compute 4L/C = 4(1)(5) = 20.

Step 2: R² = 16 < 20.

Answer: Underdamped. The circuit oscillates with decaying amplitude.

Example 2: Underdamped Circuit

Solve q'' + 4q' + 5q = 0, q(0) = 2 (coulombs), q'(0) = i(0) = 0.

Step 1: r² + 4r + 5 = 0, r = (-4 ± √(16-20))/2 = -2 ± i.

Step 2: q(t) = e-2t(c1cos t + c2sin t).

Step 3: q(0) = c1 = 2.

Step 4: q'(t) = e-2t[(-2c1+c2)cos t + (-c1-2c2)sin t]. q'(0) = -2c1 + c2 = -4 + c2 = 0, c2 = 4.

Answer: q(t) = e-2t(2 cos t + 4 sin t). Current: i(t) = q'(t) = e-2t(0 · cos t - 10 sin t) = -10e-2t sin t.

Forced Response with DC Source

If E(t) = E0 (constant), the particular solution for charge is qp = CE0.

Example 3: DC Source

Solve q'' + 6q' + 8q = 24, q(0) = 0, q'(0) = 0 (L=1, R=6, 1/C=8).

Step 1: qh: r² + 6r + 8 = (r+2)(r+4) = 0. qh = c1e-2t + c2e-4t. (Overdamped.)

Step 2: Particular: guess qp = A. Then 8A = 24, A = 3.

Step 3: q = c1e-2t + c2e-4t + 3. q(0) = c1 + c2 + 3 = 0. q'(0) = -2c1 - 4c2 = 0.

Step 4: c1 = 2c2, so 2c2 + c2 = -3, c2 = -1. Then c1 = -2 -- wait, let me recheck. From c1 = 2c2: 2c2+c2=-3 gives 3c2=-3, so c2=-1, c1=-2.

Answer: q(t) = -2e-2t - e-4t + 3. As t → ∞, q → 3 = CE0.

Forced Response with AC Source

When E(t) = E0 cos(ωt), we use undetermined coefficients just as with forced vibrations.

Steady-State Current (AC): The steady-state current has amplitude I = E0/Z where Z = √[R² + (ωL - 1/(ωC))²] is the impedance. Maximum current occurs when ωL = 1/(ωC), i.e., ω = 1/√(LC).

Example 4: AC Source

Find the steady-state charge for q'' + 2q' + 5q = 10cos(t) (L=1, R=2, 1/C=5).

Step 1: Guess qp = A cos t + B sin t.

Step 2: qp'' + 2qp' + 5qp = (-A+2B+5A)cos t + (-B-2A+5B)sin t = (4A+2B)cos t + (-2A+4B)sin t.

Step 3: 4A + 2B = 10 and -2A + 4B = 0. From the second: A = 2B. Then 8B+2B = 10, B = 1, A = 2.

Answer: qp(t) = 2cos t + sin t. Steady-state current: ip = qp' = -2sin t + cos t.

Resonance in Circuits

Electrical resonance occurs when the impedance is minimized. For the undamped LC circuit (R = 0), driving at frequency ω = 1/√(LC) produces resonance with unbounded charge, analogous to pure mechanical resonance. With resistance, the amplitude is large but bounded.

Check Your Understanding

1. An RLC circuit has L = 2 H, R = 12 Ω, C = 1/18 F. Classify the damping.

Answer: 4L/C = 4(2)(18) = 144. R² = 144. Since R² = 4L/C, the circuit is critically damped.

2. What physical quantity in a circuit is analogous to velocity in a mechanical system?

Answer: Current i(t) = q'(t) is analogous to velocity x'(t), since charge is analogous to displacement.

3. For the circuit 0.5q'' + 3q' + 4q = 0, find the roots and classify.

Answer: r = (-3 ± √(9-8))/1 = -3 ± 1. r = -2, -4. Two distinct negative real roots: overdamped. q(t) = c1e-2t + c2e-4t.

Key Takeaways

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