Question

An e.m.f. of 2 volt is produced in a coil when the
current changes at a steady rate from 3 to 2
amperes in 1 milli-second. The value of self inductance is
(a) zero (b) 2 mH
(b) 2 H (d) 200 mH

Answer 1

Given:

• Emf produced in coil = 2V
• Current change = 3A to 2A.
• Time = 1 m-sec.

To find:

• Value of self inductance

Solution:

• We know that       $e = -L \frac{dI}{dt}$  
• where e=Induced emf ; dI=Change in current; dt=Time taken; L = Self inductance coefficient.
• Given that e=2V; dI=3-2=1A ; dt= 1 × 10⁻³ sec.
• Substituting in the formula we get, $2 =L * \frac{1}{1*10^{-3}}$
• $L = 2 * 10^{-3} H$  
• Note that minus sign is ignored since it only indicates direction and that is not needed.
• L= 2 mH

Answer:

Option (b) 2 mH is the right value of self inductance.