Final answer:
In a series RC circuit, it takes about 4 time constants for a capacitor to charge to 80% of its maximum value, and approximately 0.693 time constants to charge the capacitor to one-half of its full equilibrium charge.
Step-by-step explanation:
The question involves calculating the time required for a capacitor to achieve a certain charge in an RC circuit. The time constant (τ) is a measure of how quickly the capacitor charges and is calculated as τ = RC, where R is the resistance and C is the capacitance. For a capacitor to charge to 80% of its maximum value, it takes about 4 time constants (τ). To charge to half of its maximum value, about 0.693 time constants are needed. This is because the voltage or charge (q) across a capacitor at any time t in an RC circuit can be expressed as q(t) = Qmax(1 - e^{-t/τ}), where Qmax is the maximum charge and e is the base of natural logarithms. At t = τ, q(t) = 0.632 · Qmax. Therefore, for half the maximum charge (0.5 · Qmax), we need to solve for t in the following equation: 0.5 · Qmax = Qmax(1 - e^{-t/τ}). This equation simplifies to 0.5 = 1 - e^{-t/τ}, and hence e^{-t/τ} = 0.5. Taking the natural logarithm of both sides gives -t/τ = ln(0.5), so t = τ · ln(2) which is approximately 0.693 · τ.