How long after the switch is closed will it take for the current in the circuit

Question

How long after the switch is closed will it take for the current in the circuit to reach half of its maximum value ?

How long after the switch is closed will it take for the energy stored in the inductor to reach half of its maximum value?

A 2.30mH mH inductor is connected in series with a dc battery of negligible internal resistance, a 0.900k? k ohm resistor, and an open switch. How long after the switch is closed will it take for the current in the circuit to reach half of its maxi mu m value ? How long after the switch is closed will it take for the energy stored in the inductor to reach half of its maxi mu m value?

Explanation / Answer

current after t seconds is i=imax[1-exp^(-t/lamda)]....

here lamda =L/R=2.56*10^-6 sec......


i=imax/2......

(1/2)=[1-exp^(-t/lamda)].

exp^(-t/lamda)=(1/2)....

-t/lamda=ln(1/2)=-0.693.....

t=lamda*0.693....

t=1.774*10^-6 sec........

t=1.774us
B) energy sored is U=(1/2)*L*imax^2......

to get half value i must be imax/sqrt(2)....


i=imax/sqrt(2)......

[1/sqrt(2)]=[1-exp^(-t/lamda)].

exp^(-t/lamda)=[1/sqrt(2)]....


-t/lamda=ln[1/sqrt(2)]=-0.346.....

t=lamda*0.346....

t=0.886*10^-6 sec=0.886us

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