The 5.05 A current through a 1.50 H inductor is dissipated by a 1.89 resistor in
ID: 1401470 • Letter: T
Question
The 5.05 A current through a 1.50 H inductor is dissipated by a 1.89 resistor in a circuit like that in the figure below, with the switch in position 2.
(a) What is the initial energy in the inductor?
19.127 J This answer is correct
(b) How long will it take the current to decline to 5.00% of its initial value?
2.378 s This answer is correct
(c) Calculate the average power dissipated, and compare it with the initial power dissipated by the resistor.
Pavg
2.52 IS NOTT THE CORRECT ANSWER FOR C!!
What is the energy when the current is 5.00% of the initial current? How is average power defined in terms of initial and final energies?
What is the energy when the current is 5.00% of the initial current? How is average power defined in terms of initial and final energies? W Pi
Pavg
2.52 IS NOTT THE CORRECT ANSWER FOR C!!
=
What is the energy when the current is 5.00% of the initial current? How is average power defined in terms of initial and final energies?
Explanation / Answer
initial power dissipated by the resistor is Pi = i^2*R = 5.05^2*1.89 = 48.2 W
Avg power dissipated is Pavg = irms^2*R = (5.05/1.414)^2*1.89 = 24.1 W
Pavg/pi = 24.1/48.2 = 1/2 = 0.5
-----------------------------------------------
Energy when current is 5% of the initial current = 0.5*L*i^2 = 0.5*1.5*(0.05*5.05)^2 = 0.0478 J
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