Stored Magnetic Energy - Inductor Circuit

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SUMMARY

The discussion focuses on calculating the energy delivered by a battery in an inductor circuit with a 12 V EMF and a 6.6 Ω resistor. The inductance of the solenoid is determined to be 14.56171141 H, and the current through the battery is calculated as 0.8542201 A. The energy delivered by the battery after 1.4 seconds is derived using the formula U = 0.5LI^2, resulting in 10.62556402 J, but further integration of the current over time is necessary to accurately compute the total energy delivered, leading to a suggestion to integrate the power over time.

PREREQUISITES
  • Understanding of inductance and its calculation in solenoids
  • Familiarity with the formula for energy stored in an inductor: U = 0.5LI^2
  • Knowledge of integrating functions over time to find total energy
  • Basic principles of electric circuits, including Ohm's Law and power calculations
NEXT STEPS
  • Learn about the derivation of the current in an RL circuit: I(t) = (V/R)(1 - e^(-Rt/L))
  • Study the integration of power to find energy delivered in electrical circuits
  • Explore the concept of instantaneous power and its relation to energy over time
  • Review the principles of electromagnetic induction and energy storage in inductors
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone involved in circuit analysis and energy calculations in inductive systems.

Bryon
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Homework Statement


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?


Homework Equations


U = 0.5LI^2

∫Udt from 0 to 1.4s

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?
 
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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
 
Do I just inetrate (V/R)*(1-e^(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.
 
Bryon said:
Do I just inetrate (V/R)*(1-e^(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.

You have the expression for the current with respect to time. That's the current that is being delivered by the battery to the circuit. Do you know what the instantaneous power (watts) delivered by the battery is?
 
Yes, it is P=V*I(t). So, when I find the total current over time I can just plug it into that!
 
Bryon said:
Yes, it is P=V*I(t). So, when I find the total current over time I can just plug it into that!

Yup. Integrate the power to find the total energy delivered.
 
gneill said:
Yup. Integrate the power to find the total energy delivered.

But same is integrate current and then multiply by V

both methods seems different but are same, right?
 
Thanks for the help! I see if i can get it in a bit!
 
Yes; in this case the voltage is constant, so it can be "pulled out of" the integral.
 

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