I have already answered a) & b) they are correct, I am stuck on part c) and the

Question

I have already answered a) & b) they are correct, I am stuck on part c) and the EXCERCISE portion, thanks and please write out each solution :)

Use the worked example above to help you solve this problem. A 12.0-V battery is connected in series to a 29.0- resistor and a 5.16-H inductor.

(a) Find the maximum current in the circuit. 0.413 A

(b) Find the energy stored in the inductor at this time.   0.440J

(c) How much energy is stored in the inductor when the current is changing at a rate of 1.82 A/s? ______ J

EXERCISE HINTS:  GETTING STARTED | I'M STUCK!

Use the values from PRACTICE IT to help you work this exercise. For the same circuit, find the energy stored in the inductor when the rate of change of current is 1.05 A/s.

_______J

Explanation / Answer

a) we know in series circuit the net current flow in the components are same. Now if total voltage is V=12V then

V=VR+VL where VR=RI , I is the same current flowing through the inductor but in case of inductor we have a phase angle between VL and I . Therefore the inductive reactance depends on the phase and frequency. So the maximum current can flow through only R as the VL leads I in inductor by 90 degree.

Therefore maximum current I=V/R=12/29=.413amp

b) energy stored in the inductor is given by W=LI^2/2

the circuit is series circuit therefore total current flow through R and L is same =.413amp

so W=5.16 * .413^2/2= .440Joules

c) Now the current is changing rate at 1.82A/s

the energy stored at canging rate of current in inductor is given by

P=LIdI/dt

dI/dt=1.82A/s

then P=5.16*.413*1.82watts=3.87watts

Exercise:

now dI/dt=1.05A/s

then P=5.16*.413*1.05Watts=2.237watts

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