Self Inductance

Question

The current in a coil changes from 5A to 1A in 0.1s. If the induced EMF is 8V, the coefficient of self-inductance is:

RankGuru · Accepted Answer

1. **Self-Inductance Formula**: The magnitude of induced EMF is $|e| = L \left| \frac{di}{dt} ight|$.
2. **Given Values**: $e = 8	ext{ V}$, $di = 1 - 5 = -4	ext{ A}$, and $dt = 0.1	ext{ s}$.
3. **Calculation**: 
   $$8 = L 	imes \left| \frac{-4}{0.1} ight| = L 	imes 40$$
4. **Result**: $L = \frac{8}{40} = 0.2	ext{ H}$.

Rank Guru

Self Inductance

Select the correct option:

Solution

🔒 Solution Hidden from View

More self inductance Practice Questions

Self-induced emf in coil ε = -L (dI/dt). L depends on?

Induced EMF in inductor L with changing current di/dt is:

About This Question