The current in a 50 mH inducator is known to be i=120mA,t≤0; i=A1e−500t+A2e−2000tA,t≥0. The voltage across the inducator (passive sign convention) is 3 V at t=0. a) Find the expression for the voltage across the inducator for t&gt;0. b) Find the time, greater then zero, when the power at the terminals of the inductor is zero.

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The current in a 50 mH inducator is known to be    i=120mA,t≤0; i=A1e−500t+A2e−2000tA,t≥0.    The voltage across the inducator (passive sign convention) is 3 V at t=0.     a) Find the expression for the voltage across the inducator for t&amp;gt;0.    b) Find the time, greater then zero, when the power at the terminals of the inductor is zero.

Huses1969 · Accepted Answer

a) We know that i(0)=120mA so we can write    i(0)=A1e0+A2e0=A1+A2=0.12A    We also know that v(0)=3V so we write     didt=−500A1e−500t−2000A2e−2000t    v=Ldidt=−25A1e−500t−100A2e−2000t    v(0)=−25A1e0−100A2e0=−25A1−100A2=3V    Now we have two equations with two unknowns:    A1+A2=0.12    −25A1−100A2=3    We solve this and get    A1=0.2 and A2=−0.08    Now we can write the expression for the voltage    v(t)=−25(0.2)e−500t−100(−0.08)e−2000t    v(t)=−5e−500t+8e−2000tV

Coon2000 · Accepted Answer

b) p=vi so we search for when the voltage current are zero i(t)=0.2e−500t−0.08e−2000t i(t)=0⇒ 0.2e−500t=0.08e−2000t 0.20.08=e−2000te−500t ln⁡(2.5)=ln⁡(e−1500t) t=−ln⁡(2.5)1500 t=−6.1⋅10−4s Whitch is not greater then zero. Let`s look at the voltage now: v(t)=−5e−500t+8e−2000t v(t)=0⇒ 5e−500t=8e−2000t 85=e−500te−2000t ln⁡(1.6)=ln⁡(e1500t) t=ln⁡(1.6)1500 t=313.33μs

The current in a 50 mH inducator is known to be i=120mA,t\le 0;

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