Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
Initial current, I1 = 5.0 A
Final current, I2 = 0.0 A
Change in current`dI=I_1-I_2=5A`,
Time taken for the change, t = 0.1 s
Average emf, e = 200 V
For self-inductance (L) of the coil, we have the relation for average emf as:
e = L `"di"/"dt"`
Hence, the self induction of the coil is 4 H.
Video Tutorials For All Subjects
- Inductance - Self-Inductance