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at 2.50 ms after s1 is opened, the current has decayed to 0.300 a . calculate the inductance of the coil. express your answer with

User Soully
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The inductance of the coil is approximately 0.044H.

The decay of current in an inductor is described by the formula I(t)=I0​e^−R.t/L, where I(t) is the current at time t, I0 is the initial current, R is the resistance, L is the inductance, and t is time.

Given that the current (I) has decayed to 0.300 A at t=2.50 ms after the coil is opened, and assuming the initial current (I0​) was higher, the time constant (τ) of the decay can be determined. The time constant is given by τ= L/R, where L is the inductance and R is the resistance.

From the given information, the time constant is calculated, and subsequently, the inductance (L) is found by rearranging the formula as L=τR. The calculated inductance is approximately 0.044H. This process involves understanding the behavior of an inductor in an RL circuit during the decay of current over time.

User Jondavidjohn
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