A magnetic circuit is applied with a current that changes at a rate of 5 A/sec. The circuit has an inductance of 2H, then the self-induced EMF is

• A. −10 V
• B. 0.4 V
• C. −2.5 V
• D. −4 V

Answer: (A)

Given
Inductance L = 2 H
Rate of change of current di/dt = 5 A/sec
Self induced EMF = − (Rate of change of current × Inductance) = −L(di/dt)
= −(5 × 2) = −10V