
#1
Mar2811, 12:09 PM

P: 99

1. The problem statement, all variables and given/known data
In the above circuit, the EMF from the battery is 12 V and the resistor has a resistance is 6.6 Ω. The inductor consists of a long, thin cylindrical coil of wire with 30000 turns, a radius of 5 cm and a length of 61 cm. Answer the following questions for a time 1.4 seconds after the battery has been connected. (c) How much energy has been delivered by the battery up to this point? 2. Relevant equations U = 0.5LI^2 ∫Udt from 0 to 1.4s 3. The attempt at a solution I found the inductance and the current for the 1st two parts of the problem: (a) What is the inductance of the solenoid? 14.56171141 H (b) What is the current through the battery? 0.8542201 A I plugged in the numbers to find the energy: U =0.5*14.56171141*(0.8542201^2) = 10.62556402 J Then I think that you would have to integrate over time from 0 to 1.4s which gave me a result of 7.4378948 J. It did not like the answer. Any suggestions? 



#2
Mar2811, 12:34 PM

P: 1,135

you know the eqn of current in circuit at some time t
use it to find the charge flown through battery ... q = ∫i dt where t changes from 0 to 1.4 now work done by battery = Q(EMF) = energy delivered 



#3
Mar2911, 07:52 AM

P: 99

Do I just inetrate (V/R)*(1e^(Rt/L))? Its the only thing i can think of now that will be of any help. I tried to figure this one out from the hints you gave me but no luck.




#4
Mar2911, 08:27 AM

Mentor
P: 11,416

Stored Magnetic Energy  Inductor Circuit 



#5
Mar2911, 08:39 AM

P: 99

Yes, it is P=V*I(t). So, when I find the total current over time I can just plug it in to that!




#6
Mar2911, 08:54 AM

Mentor
P: 11,416





#7
Mar2911, 08:58 AM

P: 1,135

both methods seems different but are same, right? 



#8
Mar2911, 09:01 AM

P: 99

Thanks for the help! I see if i can get it in a bit!




#9
Mar2911, 09:02 AM

Mentor
P: 11,416

Yes; in this case the voltage is constant, so it can be "pulled out of" the integral.



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