Question

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average e.m.f. of 200 V induced, give an estimate of the self inductance of the circuit.

Answer 1

dl = l₂ – l₁ = 0.0 – 5.0 = -5.0A,

dt = 0.1s, e = 200V, L=?

Since

e = -L\(\frac{dI}{dt}\)

∴ L = -\(\frac{e}{\frac{dI}{dt}}=\frac{200}{\frac{-5.0}{0.1}}=\frac{200}{50}\)

or L = 4H