Consider a circuit with an input voltage, a resistor, and a capacitor, as shown in
• Imagine that the input voltage initially has size 12 volts at t = 0. At t = 10 it
drops down to a value of 5 volts, which it then maintains indefinitely. Setting
R = 3 and C = 2 formulate an ODE that describes the voltage across the
capacitor over time.
• Solve the ODE using Laplace Transforms.
To formulate the ODE that describes the voltage across the capacitor over time, we can apply Kirchhoff's voltage law (KVL) to the circuit.
Let's denote the voltage across the capacitor as Vc(t).
At t = 0, the initial input voltage is 12 volts. According to KVL, the voltage across the resistor is equal to the voltage across the capacitor at any given time. Therefore, at t = 0, Vc(0) = 12 volts.
From t = 0 to t = 10, the voltage drops from 12 volts to 5 volts, which implies that there is a change in voltage Vc(t) of 12 - 5 = 7 volts. We can now set up an equation to describe this change in voltage over time.
Let's denote the time constant of the RC circuit (the product of resistance and capacitance) as τ = R * C.
During the initial decay phase (t < 10), the rate of change of the voltage across the capacitor is governed by the equation:
dVc(t) / dt = - (Vc(t) - 5) / τ
After t = 10, the voltage across the capacitor remains constant at 5 volts. Therefore, the rate of change becomes zero, and the equation simplifies to:
dVc(t) / dt = 0
Now, let's solve the ODE using Laplace Transforms.
1. Taking the Laplace Transform of both sides of the equation, we get:
sVc(s) - Vc(0) = - (1 / τ) * (Vc(s) - 5) + 0
2. Substituting the values Vc(0) = 12, τ = R * C = 3 * 2 = 6, and simplifying the equation, we obtain:
sVc(s) - 12 = - (1 / 6) * Vc(s) + 5 / 6
3. Rearranging the equation to solve for Vc(s):
(s + 1 / 6) * Vc(s) = 17 / 6 + 12
4. Simplifying further, we have:
(s + 1 / 6) * Vc(s) = 49 / 6
5. Dividing both sides by s + 1 / 6:
Vc(s) = (49 / 6) / (s + 1 / 6)
Now, we have obtained the Laplace Transform of the voltage across the capacitor. To solve for Vc(t), we need to take the inverse Laplace Transform.
1. To perform partial fraction decomposition on the Laplace Transform, we write:
(49 / 6) / (s + 1 / 6) = A / (s + 1 / 6)
2. Solving for A, we get:
A = 49 / 6
3. Now, we can rewrite the Laplace Transform as:
Vc(s) = (49 / 6) / (s + 1 / 6) = (49 / 6) * (1 / (s + 1 / 6))
4. Taking the inverse Laplace Transform, we obtain:
Vc(t) = (49 / 6) * e^(-1/6t)
So, the solution for the ODE using Laplace Transforms is Vc(t) = (49 / 6) * e^(-1/6t).