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Given: Initial current (I₁) = 10 A, Final current (I₂) = 0, Time interval (t) = 0.5 seconds, and induced e.m.f. (ε) = 220 V. What is the self-inductance (L) of the coil?

1. Induced e.m.f. (ε) is calculated using the formula ε = -L (ΔI/Δt) = -L ((I₂ - I₁)/t).
2. Substituting values: -L (0 - 10)/0.5 = 20L.
3. From the equation: 220 = 20L.
4. Solving for L: L = 220/20 = 11 H (self-inductance of the coil).

Correct answer: Solving for L: L = 220/20 = 11 H (self-inductance of the coil).

Solution

The self-inductance (L) is calculated using the formula for induced emf: ε = -L (dI/dt). Substituting the given values, we find L = 11 H.

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