Question

If current drops from 4 A to 0 A in 0.5 s, and voltage is 120 V, find the self-inductance.

Hint:

The formula for self-inductance is derived from the relationship between voltage, current, and time in an inductor.

Answer 1

Step 1: Understanding the formula.
The self-inductance \( L \) can be calculated using the formula: \[ L = \frac{V \Delta t}{\Delta I} \] where: - \( V \) is the voltage (120 V), - \( \Delta t \) is the time duration (0.5 s), - \( \Delta I \) is the change in current (from 4 A to 0 A, so \( \Delta I = 4 \, \text{A} \)). Step 2: Substituting the values.
Substituting the known values into the formula: \[ L = \frac{120 \times 0.5}{4} = \frac{60}{4} = 15 \, \text{H} \] Step 3: Conclusion.
Thus, the self-inductance is \( 15 \, \text{H} \).

Answer 2

Step 1: Self-inductance is found using L = (V × Δt) ÷ ΔI.

Step 2: Here, V = 120 V, Δt = 0.5 s, and the current changes from 4 A to 0 A, so ΔI = 4 A.

Step 3: Plugging in: L = (120 × 0.5) ÷ 4 = 60 ÷ 4 = 15.

Step 4: So the self-inductance is 15 H.

Answer 3

Via Lenz's law reasoning.
As the current falls from 4 A to 0 A, the coil develops a self-induced back-emf that opposes this change; its magnitude equals the given voltage of 120 V.
The rate at which current changes is \( \dfrac{\Delta I}{\Delta t} = \dfrac{4}{0.5} = 8 \, \text{A/s} \).
Since the induced emf is \( V = L \dfrac{dI}{dt} \), the self-inductance is \[ L = \frac{V}{dI/dt} = \frac{120}{8} = 15 \, \text{H} \]