RL Circuits — Question 8

Question 8

A fractional rise-time requirement gives an inductance bound with resistor tolerance

An initially unenergized series RL circuit is stepped to constant voltage. At time T its current must be no greater than fraction f of its own final current, with $0<f<1$. Resistance can be as large as Rmax.

1. Derive the minimum L that guarantees the requirement.

2. Evaluate for T=.0100 s, f=.200 and Rmax=12.0 ohms.

Question 8 - Solution

A fractional rise-time requirement gives an inductance bound with resistor tolerance

(a) Method and calculation

$$i(T)/i_\infty=1-e^{-RT/L}\le f\Rightarrow RT/L\le-\ln(1-f).$$$$\boxed{L\ge\frac{R_{\max}T}{-\ln(1-f)}}.$$

(b) Interpretation and further calculation

$$L_{\min}=\frac{(12.0)(.0100)}{-\ln(.800)}.$$$$\boxed{L_{\min}=.537770\,H}.$$

Physical check

The largest resistance produces the fastest fractional rise. The criterion is relative to each circuit final current, not a fixed absolute current.