Question

In an RLC circuit, L=0.1 H, R=0.6 Ω, C=0.4 F, and E(t)=60 V. Find the charge at time t. Assume the initial charge on the capacitor is 0C and the initial current is 7.5 A.

A) q(t) = −16.125e⁻³ˣ sin(4x) − 24e⁻³ˣ cos(4x)
B) q(t) = −16.125e⁻³ˣ sin(4x) − 24e⁻³ˣ cos(4x) + 24
C) q(t) = e⁻³ˣ sin(4x) − 24e⁻³ˣ cos(4x) + 24
D) q(t) = e³ˣ sin(4x) − 24e³ˣ cos(4x) + 24

Answer 1

In an RLC circuit, the charge at time t is given by the equation q(t) = Qe^-t/RC, where Q is the initial charge on the capacitor and RC is the time constant of the circuit. The time constant is calculated as RC = R * C, where R is the resistance and C is the capacitance.

Step-by-step explanation:

In an RLC circuit, the charge at time t is given by the equation q(t) = Qe-t/RC, where Q is the initial charge on the capacitor and RC is the time constant of the circuit. The time constant is calculated as RC = R * C, where R is the resistance and C is the capacitance. In this case, R = 0.6 Ω and C = 0.4 F, so the time constant is RC = 0.6 * 0.4 = 0.24 s.

Given E(t) = 60 V, the charge equation becomes q(t) = 60e-t/0.24.

The charge equation can also be written as q(t) = Q * e-t/t0, where t0 = RC = 0.24 s.