Final answer:
The time required for the maximum energy in the capacitor to fall to half in the given RLC circuit is approximately 1.021 seconds, calculated using the time constant τ = L/R and the natural logarithm of 2.
Step-by-step explanation:
In an oscillating series RLC circuit with an 8.08 Ω resistor and an 11.9 H inductor, the time required for the maximum energy present in the capacitor to fall to half of its initial value can be found using the formula for the decay of energy in an RLC circuit. The decay of energy is characterized by the time constant of the circuit, which is τ = L/R. The energy in the circuit after a time t is given by E(t) = E_0 e^{-t/τ}, where E_0 is the initial energy. To find the time it takes for the energy to reduce to half its initial value, we set E(t) to E_0/2 and solve for t, which yields t = τ ln(2).
For an RLC circuit, the time constant τ is calculated as τ = L/R. In this case, with R = 8.08 Ω and L = 11.9 H, τ = 11.9 H / 8.08 Ω = 1.473 Ω·s. The time it takes for the energy to fall to half its initial value is then t = 1.473 · ln(2) = approximately 1.021 seconds.
Therefore, the time required for the energy to fall to half its value in this RLC circuit is approximately 1.021 seconds.