Answer:
a) [tex]t = 4.6\tau[/tex]
b) [tex]L = 0.0187 \: H[/tex]
Explanation:
The current flowing in a R-L series circuit is given by
[tex]I = I_{0} (1 - e^{\frac{-t}{\tau} })[/tex]
Where τ is the time constant and is given by
[tex]\tau = \frac{L}{R}[/tex]
Where L is the inductance and R is the resistance
Assuming the current has reached steady state when it is at 99% of its maximum value,
[tex]0.99I_{0} = I_{0} (1 - e^{\frac{-t}{\tau} })\\0.99 = (1 - e^{\frac{-t}{\tau} })\\1 - 0.99 = e^{\frac{-t}{\tau}}\\ln(0.01) = ln(e^{\frac{-t}{\tau}})\\-4.6 = \frac{-t}{\tau}\\t = 4.6\tau\\[/tex]
Therefore, it would take t = 4.6τ to reach the steady state.
(b) If an emergency power circuit needs to reach steady state within 1.2 ms of turning on and the circuit has a total resistance of 72 Ω, what values of the total inductance of the circuit are needed to satisfy the requirement?
[tex]t = 4.6\frac{L}{R}\\t = 4.6\frac{L}{R}\\0.0012 = 4.6\frac{L}{72}\\0.0864 = 4.6 L\\L = 0.0864/4.6\\L = 0.0187 \: H[/tex]
Therefore, an inductance of 0.0187 H is needed to satisfy the requirement.