The charge on the capacitor (q(t)) is represented by 30t, and the current (i(t)) is given by -(6/5) * 30e^(-5t/6) + 30, calculated through LRC circuit analysis.
Certainly! Let's go through the detailed calculations to find the charge q(t) on the capacitor and the current i(t) in the given LRC-series circuit.
The differential equation governing the circuit is:
L * (di/dt) + Ri + (1/C) * q = E(t)
Given values:
L = 5/6 H, R = 10 Ω, C = 1/60 F, E(t) = 300 V, q(0) = 0 C, i(0) = 0 A
First, we can express the differential equation in terms of q(t):
(5/6) * (di/dt) + 10i + (60/1) * q = 300
Now, let's find the homogeneous solution to this differential equation:
L * (di_h/dt) + Ri_h + (1/C) * q_h = 0
(5/6) * (di_h/dt) + 10i_h + 60 * q_h = 0
The homogeneous solution is found by assuming i_h(t) = Ae^(rt) and q_h(t) = Be^(rt), where A and B are constants.
Substituting these into the homogeneous equation and solving, we get:
i_h(t) = -(6/5) * Ae^(-5t/6)
q_h(t) = (1/5) * Be^(-5t/6)
Now, let's find the particular solution to the complete differential equation. We assume i_p(t) = A (a constant) to simplify calculations.
Substituting this into the differential equation, we get:
(5/6) * (di_p/dt) + 10i_p + 60 * q_p = 300
(5/6) * (0) + 10A + (60/1) * q_p = 300
Solving for A, we find A = 30.
Now, we have the complete solution:
i(t) = i_h(t) + i_p(t) = -(6/5) * 30e^(-5t/6) + 30
q(t) = q_h(t) + q_p(t) = (1/5) * 0e^(-5t/6) + 30t
Now, apply the initial conditions to find B:
q(0) = 0 = (1/5) * B + 0
B = 0
The final expressions are:
i(t) = -(6/5) * 30e^(-5t/6) + 30
q(t) = 30t