Question

Demonstrate the process of solving a first-order RC circuit using differential equations.

Answer 1

Sure! Let's demonstrate the process of solving a first-order RC (Resistor-Capacitor) circuit using differential equations.

A first-order RC circuit consists of a resistor (R) and a capacitor (C) connected in series to a voltage source. The circuit can be represented by the following schematic:

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    ---- R ---- C ----
   |                   |
   V (Voltage Source)  |
   |                   |
    -------------------


Let's denote the voltage across the capacitor as Vc(t), and the current flowing through the circuit as I(t). According to Kirchhoff's voltage law (KVL), the voltage across the resistor is Vr(t) = V - Vc(t), where V is the voltage supplied by the source.

The governing equation for the circuit is based on the relationship between the current, the voltage across the resistor, and the voltage across the capacitor:

Ohm's law: Vr(t) = R * I(t)
Capacitor voltage: Vc(t) = (1/C) * ∫[0 to t] I(t) dt

Now, we need to find the differential equation that relates Vc(t) and its derivative. The derivative of Vc(t) with respect to time (t) gives us the rate of change of voltage across the capacitor:

d(Vc(t))/dt = (1/C) * I(t)

To relate I(t) to Vc(t), we can use Ohm's law (Vr(t) = R * I(t)) and substitute Vc(t) = V - Vr(t):

d(Vc(t))/dt = (1/C) * (V - Vc(t))/R

Now, we have the differential equation that governs the behavior of the first-order RC circuit:

d(Vc(t))/dt = (1/RC) * (V - Vc(t))

This is a first-order linear ordinary differential equation (ODE) with a constant coefficient, and its solution can be found using various methods. One common approach is to use separation of variables and integrate:

1/(V - Vc(t)) * d(Vc(t)) = (1/RC) * dt

Integrate both sides:

∫(1/(V - Vc(t)) * d(Vc(t))) = ∫(1/RC) * dt

Now, integrate:

ln|V - Vc(t)| = (1/RC) * t + K

where K is the constant of integration.

Next, we can isolate Vc(t):

|V - Vc(t)| = e^((1/RC) * t + K)

Since the voltage across the capacitor cannot be negative, we can remove the absolute value:

V - Vc(t) = e^((1/RC) * t + K)

Now, solve for Vc(t):

Vc(t) = V - e^((1/RC) * t + K)

To determine the constant K, we need the initial condition, which is the initial voltage across the capacitor at t = 0, denoted as Vc(0). Substituting t = 0 into the equation:

Vc(0) = V - e^((1/RC) * 0 + K)
Vc(0) = V - e^K

Now, solve for K:

e^K = V - Vc(0)
K = ln(V - Vc(0))

Finally, the complete solution for the voltage across the capacitor is:

Vc(t) = V - e^((1/RC) * t + ln(V - Vc(0)))
Vc(t) = V - (V - Vc(0)) * e^(-t/RC)

And there you have it! The equation Vc(t) = V - (V - Vc(0)) * e^(-t/RC) represents the voltage across the capacitor in a first-order RC circuit as a function of time. This solution describes how the capacitor charges or discharges over time when connected to a voltage source through a resistor.