Consider an ideal inductor (that is, an inductor with no internal resistance). A
ID: 3895610 • Letter: C
Question
Consider an ideal inductor (that is, an inductor with no internal resistance). An EMF of 150 V is measured across the inductor during a 2.00 ms span when the current through the inductor increases steadily from 1.00 A to 4.00 A. Consider an ideal inductor (that is, an inductor with no internal resistance). An EMF of 150 V is measured across the inductor during a 2.00 ms span when the current through the inductor increases steadily from 1.00 A to 4.00 A.
a. What is the inductance of this inductor? How much energy is stored in the inductor at the end of the time span, when the current has reached 4.00 A?
b. Immediately after the switch is closed, what is the current through the inductor?
c. Immediately after the switch is closed, which end of the inductor, the top or the bottom, will be at the higher voltage
d. Immediately after the switch is closed, what is the voltage across the inductor? [Hint: Use part (b) and Kirchhoff
Explanation / Answer
a) V = L (di/dt)
150 = L (4-1)/(2*10^(-3))
L = 100 milli henry = 0.1 henry
b) when switch is open current is
I(t) = I(0) e^(-t/R2L)
as current in inductor cant change immediately .....
current after closing the switch is also
I(t) = I(0) e^(-t/R2L)
c) the top end will be at higher potential as current flows from top to bottom
d) voltage across the inductor after closing the switch is
V = L (di/dt)
V = L (d(i(0)e^(-t/R2L))/dt)
= L I(0) (-1/R2 L) e^(-t/R2L)
= -I(0)e^(-t/R2L) / R2