Consider a circuit with an input voltage, a resistor, and a capacitor,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 for the given circuit, we can use the Kirchhoff's voltage law (KVL).
Considering the voltage drop across the resistor (VR) and the voltage drop across the capacitor (VC), the KVL equation can be written as:
VR + VC = Vin(t)
Where Vin(t) represents the input voltage at time t.
To express this equation in terms of derivatives with respect to time, we need to differentiate VC with respect to time. The voltage across a capacitor is given by:
VC = 1/C * ∫(i(t)) dt
Where i(t) represents the current flowing through the capacitor.
Now, using Ohm's Law, we can relate the current i(t) with the voltage across the resistor:
VR = R * i(t)
Substituting these equations into the KVL equation, we get:
R * i(t) + 1/C * ∫(i(t)) dt = Vin(t)
Next, we differentiate the entire equation with respect to time:
R * di(t)/dt + 1/C * i(t) = dVin(t)/dt
Since the input voltage Vin(t) initially has a value of 12 volts at t = 0, then drops to 5 volts at t = 10, we can express Vin(t) as follows:
Vin(t) = 12 u(t) - 7 u(t-10)
Where u(t) is the Heaviside step function.
Substituting this expression into the differential equation, we get the final form of the ODE:
R * di(t)/dt + 1/C * i(t) = dVin(t)/dt
R * di(t)/dt + 1/C * i(t) = 12 δ(t) - 7 δ(t-10)
To solve this ODE using Laplace Transforms, we need to take the Laplace Transform of both sides of the equation. Let's denote the Laplace Transform of i(t) as I(s):
L{R * di(t)/dt} + L{1/C * i(t)} = L{12 δ(t) - 7 δ(t-10)}
Using the properties of the Laplace Transform and the standard Laplace Transform table, we can evaluate each term on the left-hand side:
R * (s * I(s) - i(0)) + 1/C * I(s) = 12 - 7 * e^(-10s)
Rearranging the equation to solve for I(s), we get:
I(s) = (12 - 7 * e^(-10s)) / (R * s + 1/C)
Finally, to find the inverse Laplace Transform of I(s) and obtain the solution i(t), we can use the inverse Laplace Transform table or software tools.