Differential equations

Question

A series RLC circuit with R = 6 ohm, C = 0.02 Farad and L = 0.1 has no applied voltage. Find the subsequent current in the circuit if the initial charge, on the capacitor is q° and the initial current is zero

Answer 1

Kirchoff's Loop Law states that the Algebraic sum of the Voltage drops in a simple closed electric circuit is zero. Derive from this law a Differential Equation for the Current I in a simple series circuit consisting of a Resistor (R), Inductor (L), and Capacitor (C) as well as an Electromotive Force or emf of E(t) created by a Battery or Generator. R, C, and L are in Ohms, Farads, and Henrys respectively; E(t) is in Volts and I is in Ampères.

The Voltage drops across a Resistor, Capacitor, and Inductor are given respectively by RI, (1/C)q, and L(dI/dt) with q the charge on the Capacitor. The Voltage drop across an emf is given by -E(t). Kirchoff's Loop Law then gives RI + L(dI/dt) + (1/C)q − E(t) = 0. The relationship between I and q is I = dq/dt.

Differentiate RI + L(dI/dt) + (1/C)q − E(t) = 0 with respect to t: R(dI/dt) + L(d²I/dt²) + (1/C)(dq/dt) − dE(t)/dt = 0 or (d²I/dt²) + (R/L)(dI/dt) + (1/LC)I = (1/L)dE(t)/dt.

[Also note that a Differential Equation for the Charge on the Capacitor is obtained by use of I = dq/dt and dI/dt = d²q/dt² in the equation RI + L(dI/dt) + (1/C)q − E(t) = 0: (R/L)(dq/dt) + (d²q/dt²) + (1/LC)q = (1/L)E(t) or

d²q/dt² + (R/L)(dq/dt) + (1/LC)q = (1/L)E(t).]

For R, C, and L at 6 Ohms, 0.02 Farad, and 0.1 Henry, the Differential Equation for the subsequent Current in the circuit is d²I/dt² + (6/0.1)(dI/dt) + (1/(0.1×0.02))I = (1/0.1)(dE(0)/dt) or d²I/dt² + 60(dI/dt) + 500I = 0.

To solve d²I/dt² + 60(dI/dt) + 500I = 0, write the characteristic equation as λ² + 60λ + 500 = 0 or (λ+50)(λ+10) = 0. Since the roots λ = -50 & λ = -10 are real and distinct, the solution is I(t) = c₁e^-50t + c₂e^-10t.

Given that I(0) = 0, obtain c₁ + c₂ = 0. Differentiation of I(t) = c₁e^-50t + c₂e^-10t yields

dI/dt = -50c₁e^-50t − 10c₂e^-10t. It then follows that dI/dt at t = 0 equals -50c₁ − 10c₂.

Next, obtain 2 initial conditions for I in a simple series RCL circuit with known emf when initial conditions for the Current and Charge on the Capacitor are given at t = 0. This is done by writing the circuit Current and Capacitor Charge at Time t by I(t) and q(t). I(0) is given as one initial condition for the Current. Solve RI + L(dI/dt) + (1/C)q − E(t) = 0 for dI/dt and then set t to 0: dI/dt at t=0 equals (-R/L)I(0) + (-1/LC)q(0) +(1/L)E(0). Rewrite as dI/dt at t=0 equals (1/L)E(0) − (R/L)I(0) − (1/LC)q(0) to obtain the second initial condition for I.

[Note also that, given q(0) and I(0), q(0) gives one initial condition for q(t). With dq/dt = I, dq/dt_(t=0) = I(0) gives the second initial condition for the charge on the Capacitor.]

One can then write dI/dt_(t=0) = (1/0.1)(0) − (6/0.1)(0) − (1/(0.1×0.02))q° or -500q°. Then -500q° =-50c₁ − 10c₂. With c₁ + c₂ = 0, obtain c_{2 =} -c₁ and -500q° = -50c₁ + 10c₁ which gives c₁ = 12.5q° and c₂ = -12.5q°. I is then 12.5q°(e^-50t − e^-10t).

[Note that this expression for I is also found by first finding the charge on the capacitor. With R=6, C=0.02, L=0.1, and E=0, d²q/dt² + (R/L)(dq/dt) + (1/LC)q = (1/L)E(t) (which is d²I/dt² + 60(dI/dt) + 500I = 0 with I replaced by q), the solution of this equation is q = c₁e^-50t + c₂e^-10t. Given initial condition q(0) = q°, write

q° = c₁ + c₂. Also I(0) = (dq/dt)_(t=0) = 0, so that -50c₁ − 10c₂ = 0. Solve for c₁ & c₂ simultaneously to obtain

c₁ = -q°/4 and c₂ = 5q°/4. Then q = (-q°/4)e^-50t + (5q°/4)e^-10t. I is then dq/dt or 12.5q°e^-50t −12.5q°e^-10t or

12.5q°(e^-50t − e^-10t) as before.]