Energy in inductors A constant current I exists in a solenoid whose inductance i
ID: 1668023 • Letter: E
Question
Energy in inductorsA constant current I exists in a solenoid whose inductance is L. The current is then reduced to zero in a certain amount of time.
(a) If the wire from which the solenoid is made has no resistance, is there a voltage across the solenoid during the time when the current is constant?
A voltage does not exist across the solenoid.
A voltage exists across the solenoid.
(b) If the wire from which the solenoid is made has no resistance, is there an emf across the solenoid during the time that the current is being reduced to zero?
An emf exists across the solenoid.
An emf does not exist across the solenoid.
(c) Does the solenoid store electrical energy when the current is constant?
No, because the solenoid stores electrical energy only when the current is changing.
No, because a solenoid cannot store electrical energy under any condition.
Yes, the solenoid stores electrical energy.
(d) When the current is reduced from its constant value to zero, what is the rate P at which energy is removed from the solenoid?
P = (1/2LI2) / t
P = (1/2LI2)t2
P = (1/2LI2)t
P = t / (1/2LI2)
A solenoid has an inductance of L = 2.1 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 is the emf induced across the solenoid?
Explanation / Answer
a. If the wire has no resistance, there is no voltage across thesolenoid. V = IR. If R = 0 , then V = 0 V. b. Yes. According to Faraday’s law of electromagneticinduction, expressed as = L(I /t ) , an emf is induced in thesolenoid as long as the current is changing in time. c. Yes. the amount of electrical energy E stored by an inductor isE=1/2(LI)2, where L is the inductance and I is thecurrent. d. The power is equal to the energy removed divided by the time,P=E/t=[1/2(LI)2]/t. e. The emf induced in the solenoid is Emf= L(I /t)=-(2.1H)[(0A-15A)/(75x10-3s)]=+420V f. The energy stored in the solenoid is E=1/2(LI)2=1/2(2.1H)(15 A)2=236.25J g. The rate (or power, P) at which the energy is removed is P=E/t=[1/2(LI)2]/t=[1/2(2.1H)(15A)2]/(75x10-3s)=3150W
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