Question 7.21: The coil of a certain relay is operated by a 12-V battery. I......

The current through the coil is given by

i(t) = i(∞) + [ i(0) – i(∞) ] e^{-t/\tau}

where

i(0) = 0 ,    i(∞) = \frac{12}{150} =    80 mA

\tau = \frac{L}{R} = \frac{30 × 10^{-3} }{150 } = 0.2 ms

Thus,

i(t) = 80 [ 1 – e^{-t/\tau } ] m A

If  i(t_{d} ) = 50 mA then

50 = 80 [ 1 – e^{-t_{d} / \tau } ]    ⇒   \frac{5}{8} = 1 – e^{-t_{d} / \tau }

or

e^{-t_{d} / \tau } = \frac{3}{8}    ⇒    e^{t_{d} / \tau } = \frac{8}{3}

By taking the natural logarithm of both sides, we get

t_{d} = \tau  ln \frac{8}{3} = 0.2  ln \frac{8}{3} ms = 0.1962 ms

Alternatively, we may find t_{d}   using

t_{d} = \tau  ln \frac{i(0) – i(∞)}{i(t_{d}) – i(∞)}