Electromagnetism A battery with negligible internal resistance and with electrom

Question

Electromagnetism

A battery with negligible internal resistance and with electromotive force 12.0 Vis connected through a switch S to an inductor with inductance L 400 HH and two resistors with resistances RI 1.20 k and R 20.0 k2, see figure 3.1. The switch has been closed for a long time 3.1 Calculate the power dissipations PI and P2 in the two resistors and the stored energy UL in the inductor. At time t to the switch is opened, see figure 3.2 3.2 Calculate the voltage across R2 right after the switch is opened 3.3 Calculate how long time 4t goes until the current has decayed to half of the value at t t

Explanation / Answer

after long time ,inductor act as short circuit

i=current across R1=12/R1=0.01 A

P1=power =current^2*resistance=0.12 W

current across R2=12/R2=6*10^(-4) A

P2=7.2*10^(-3) W

UL=power stored in inductor=0.5*L*i^2=2*10^(-8) W

voltage across R2=12 V

after switch is open

there will no path for current to flow as circuit is broken so current will be 0 which is agains the rules of inductor .

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