P = t / (1/2 LI 2 ) P =(1/2 LI 2 ) t P =(1/2 LI 2 ) t 2 P = (1/2 LI 2 ) / t A co

Question

P = t /(1/2LI2)
P =(1/2LI2)t
P =(1/2LI2)t2
P = (1/2LI2)/ t A constant current I exists in a solenoid whose inductanceis L. The current is then reduced to zero in a certainamount of time. (d) When the current is reduced from its constant value to zero,what is the rate P at which energy is removed from thesolenoid? P = t /(1/2LI2) P =(1/2LI2)t P =(1/2LI2)t2 P = (1/2LI2)/ t A solenoid has an inductance of L = 3.6 H and carries a current of I = 15 A. (e) If the current goes from 15 to 0 A in a time of 75 ms, what isthe emf varepsilon induced across the solenoid? (f) How much electrical energy E is stored in thesolenoid? (g) At what rate P must the electrical energy be removedfrom the solenoid when the current is reduced to 0 A in a time of75 ms? Note that the rate at which energy is removed is the power.

Explanation / Answer

Is that 75e-8? I cant tell what the exponent is.

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