RL Circuits — Question 2

Question 2

A decay ratio measures inductance without knowing the starting current

An isolated series RL loop with R=5.00 ohms has current i1 at time t1 and current i2=i1/2 at time t1+.0200 s. Its L is constant and there is no source.

1. Infer the time constant and L.

2. Find the additional time required for the current to fall from i2 to i2/4.

Question 2 - Solution

A decay ratio measures inductance without knowing the starting current

(a) Method and calculation

$$i_2/i_1=e^{-\Delta t/\tau}=1/2\Rightarrow\boxed{\tau=.0200/\ln2\,s}.$$$$\boxed{L=R\tau=.100/\ln2\,H=.144270\,H}.$$

(b) Interpretation and further calculation

$$(i_2/4)/i_2=e^{-\Delta t_2/\tau}\Rightarrow\Delta t_2=\tau\ln4.$$$$\boxed{\Delta t_2=.0400\,s}.$$

Physical check

Ratios eliminate the unknown amplitude; the inferred time constant must remain the same throughout an ideal exponential decay.