Derive the expression for V(t) of a parallel R-L circuit when excited by a AC source with Laplace Transforms approach
To derive the expression for V(t) of a parallel R-L circuit using the Laplace Transform approach, we start by writing the governing differential equation for the circuit.
The voltage across the resistor (VR) is given by Ohm's Law: VR = I(t) * R
The voltage across the inductor (VL) is given by Faraday's Law: VL = L * dI(t)/dt
The total voltage across the circuit (VT) is the input voltage source V(t).
Since the circuit is in parallel, the total current (I(t)) through the circuit is the same as the current through the resistor and the current through the inductor.
Therefore, we can write the equation: VT = VR + VL
Now, let's apply the Laplace Transform to both sides of the equation. Using the standard Laplace Transform notations, we get:
V(s) = R * I(s) + L * s * I(s)
To solve for I(s), we need to represent the input voltage source V(t) in the Laplace domain. Let's assume that the input voltage source is of the form V(t) = Vm * sin(ωt).
The Laplace Transform of sin(ωt) is given by: F(s) = ω / (s^2 + ω^2)
Therefore, the Laplace Transform of V(t) is:
V(s) = Vm * ω / (s^2 + ω^2)
Now, we can substitute this expression for V(s) into the equation V(s) = R * I(s) + L * s * I(s), and solve for I(s):
V(s) = R * I(s) + L * s * I(s)
Vm * ω / (s^2 + ω^2) = (R + L * s) * I(s)
I(s) = (Vm * ω) / ((s^2 + ω^2) * (R + L * s))
Finally, to find the inverse Laplace Transform of I(s), we use partial fraction decomposition. This involves expressing the right-hand side (RHS) of the equation in terms of simpler fractions. However, the inverse Laplace Transform of the resulting terms will depend on the specific values of R, L, and ω.
In conclusion, the expression for V(t) of a parallel R-L circuit when excited by an AC source using the Laplace Transform approach involves finding the inverse Laplace Transform of the current I(s) obtained above. The exact form of the expression will depend on the specific values of R, L, and ω.