Solving for Imax and Energy in an LC Circuit: Homework Help

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Homework Help Overview

The discussion revolves around an LC circuit involving a 4700uF capacitor charged to 9.0V, which is then connected to a 1.50H inductor after the voltage source is removed. Participants are tasked with solving for the energy of the circuit, the maximum current (Imax), and the time at which the energy of the capacitor equals the energy of the inductor.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore calculations for frequency, maximum current, and energy stored in the circuit. There are questions about the role of resistance, the behavior of current over time, and the relationship between energy, voltage, and charge in the capacitor and inductor.

Discussion Status

Some participants have provided calculations and expressed uncertainty about their correctness. Others have prompted further exploration of the conditions under which energy is transferred between the capacitor and inductor, and the implications of oscillation in the circuit. There is an ongoing examination of how to approach the time calculation for energy transfer.

Contextual Notes

Participants note the absence of information regarding resistance in the circuit, which raises questions about the implications for current behavior. There is also a focus on understanding the relationship between energy, voltage, and charge as they relate to the capacitor's discharge and the inductor's charge. The discussion includes references to oscillatory behavior and the equations governing the circuit dynamics.

clamatoman
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Homework Statement


4700uF Capacitor is initially charged to 9.0V, then the voltage source is removed and the capacitor is connected across a 1.50H inductor. Solve for: Energy of the Circuit, Imax, and Time after Circuit connected whed Energy of the capacitor =Energy of the inductor

Homework Equations


Imax=εmax/√(R2+(2πƒL= (1/2πƒC))
ƒ=1/2π√LC
E=1/2 CV2+ 1/2LI2
Cap discharging
ecf93455f87a5d2a2333b6c27e046b2d.png

Inductor charging
9d9d03dede491e7b8b6bc384943f9b90.png


The Attempt at a Solution



ƒ=1/2π√LC= 1.896Hz

Imax=εmax/√(R2+(2πƒL= (1/2πƒC))
Plugging in 9.0v for ε, 1.9 for ƒ, 4700*10-6 for C and 1.5 for L, and crossing out R since no given resistance...
I come up with Imax=30.95A

E=1/2 CV2+ 1/2LI2
E=720.9

Did i do the first two correctly?
and, not sure about the last part, would i just use
ecf93455f87a5d2a2333b6c27e046b2d.png
and set that equal to
9d9d03dede491e7b8b6bc384943f9b90.png
and solve for t?
 
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clamatoman said:

Homework Statement


4700uF Capacitor is initially charged to 9.0V, then the voltage source is removed and the capacitor is connected across a 1.50H inductor. Solve for: Energy of the Circuit, Imax, and Time after Circuit connected whed Energy of the capacitor =Energy of the inductor

Homework Equations


Imax=εmax/√(R2+(2πƒL= (1/2πƒC))
ƒ=1/2π√LC
E=1/2 CV2+ 1/2LI2
Cap discharging
ecf93455f87a5d2a2333b6c27e046b2d.png

Inductor charging
9d9d03dede491e7b8b6bc384943f9b90.png


The Attempt at a Solution



ƒ=1/2π√LC= 1.896Hz

Imax=εmax/√(R2+(2πƒL= (1/2πƒC))
Plugging in 9.0v for ε, 1.9 for ƒ, 4700*10-6 for C and 1.5 for L, and crossing out R since no given resistance...
I come up with Imax=30.95A

E=1/2 CV2+ 1/2LI2
E=720.9

Did i do the first two correctly?
and, not sure about the last part, would i just use
ecf93455f87a5d2a2333b6c27e046b2d.png
and set that equal to
9d9d03dede491e7b8b6bc384943f9b90.png
and solve for t?

Is there any resistor in the circuit?
 
ehild said:
Is there any resistor in the circuit?
They did not give any information about a resistor or resistance.
 
clamatoman said:
They did not give any information about a resistor or resistance.
There is no resistance. How does the current in the circuit change with time then?
 
ehild said:
There is no resistance. How does the current in the circuit change with time then?
Not sure... with these perhaps? https://wikimedia.org/api/rest_v1/media/math/render/svg/1a0fe934407a9109d0b3a8679e9949d1cb8c63f8 Vl(t)=L dIl/dt
Ic(t)=C dVc/dt

although i am not sure how i would get rid of the 'd's...I am little out of my depth with calculus type stuff.
 
clamatoman said:
Did i do the first two correctly?
Sorry to say, I'm not liking any of it :nb)

For the first part, the energy of the circuit, look at the initial conditions. Where is all the energy stored initially?

For the maximum current, where must all the energy be stored? How much energy?

For the third part, think about how the circuit behaves with respect to time (pick one of the variables, either current or voltage).
 
Are IC and IL different? The capacitor and the inductor form a single loop.
 
gneill said:
Sorry to say, I'm not liking any of it :nb)

For the first part, the energy of the circuit, look at the initial conditions. Where is all the energy stored initially?
In the Capacitor.
ok so just U=1/2 CV^2= 1/2 * (4700*10^-6)(9^2) = 0.19?

For the maximum current, where must all the energy be stored? How much energy?
This i am not sure what you mean...

For the third part, think about how the circuit behaves with respect to time (pick one of the variables, either current or voltage).
well the circuit will oscillate current between the capacitor and inductor?
 
ehild said:
Are IC and IL different? The capacitor and the inductor form a single loop.
IC = IL in a closed loop.
 
  • #10
clamatoman said:
IC = IL in a closed loop.
Yes. You have to find how that I depends on time. Your equations in the first post are irrelevant. Use KVL to the loop.
 
  • #11
All the energy is initially stored on the capacitor, right? At that time it has its maximum potential difference (voltage).
What condition must obtain for all the energy to be stored on the inductor?
 
  • #12
gneill said:
All the energy is initially stored on the capacitor, right? At that time it has its maximum potential difference (voltage).
What condition must obtain for all the energy to be stored on the inductor?
when Voltage of the capacitor is 0?
 
  • #13
clamatoman said:
when Voltage of the capacitor is 0?
Yes, what else is true at that time? What can you say about the inductor?
 
  • #14
gneill said:
Yes, what else is true at that time? What can you say about the inductor?
If voltage of the capacitor is zero, then so is voltage of the inductor due to KVL?
 
  • #15
clamatoman said:
If voltage of the capacitor is zero, then so is voltage of the inductor due to KVL?
Yes, that's true too. What about current?
 
  • #16
Let me see if what I am thinking makes sense here for solving current. U=1/2 LI2 for the inductor, U=1/2 CV2 for the capacitor. So does U=U
1/2 CV2 =1/2 LI2 . By doing this and rearrangeing to solve for I i have I=√((CV2)/L) which = ~ 0.5A. IS this correct? or am i way off?
 
  • #17
clamatoman said:
Let me see if what I am thinking makes sense here for solving current. U=1/2 LI2 for the inductor, U=1/2 CV2 for the capacitor. So does U=U
1/2 CV2 =1/2 LI2 . By doing this and rearrangeing to solve for I i have I=√((CV2)/L) which = ~ 0.5A. IS this correct? or am i way off?
It is correct. As the circuit oscillates, energy is traded back and forth between the capacitor and the inductor. When the capacitor voltage is at a maximum the current is zero and all the energy is stored in the electric field of the capacitor. When the inductor current is maximum the voltage is zero and all the energy is stored in the magnetic field of the inductor.
 
  • #18
And solving for the energy i got U=1/2 CV2= ~0.19J, correct?
 
  • #19
Looks good.
 
  • #20
so now i just need to solve for time it takes for the capacitor to discharge half of its energy into the inductor...
 
  • #21
I think I have it.
Q=CV=0.0423.
ω=1/√LC
Q=Qocos(ωt)
.5Q=Qocos(ωt)
.5=cos(ωt)
cos-1(.5)=(ωt)
t=(cos-1(.5))/ω
t=5.038 seconds

Look Good?
 
  • #22
Nope. Charge is not energy. How does the energy stored in a capacitor vary with charge?
 
  • #23
gneill said:
Nope. Charge is not energy. How does the energy stored in a capacitor vary with charge?
energy of a capacitor varies with Voltage which is also a factor of charge...
so i need to solve for the voltage of the capacitor when energy is 1/2 of initial energy, then plug that into the charge equation, right?
 
  • #24
clamatoman said:
energy of a capacitor varies with Voltage which is also a factor of charge...
so i need to solve for the voltage of the capacitor when energy is 1/2 of initial energy, then plug that into the charge equation, right?
If you work with the voltage for the energy you don't need to consider the charge. Either will do so long as you employ the appropriate expression for energy in terms of either voltage or charge. Or you could look at the inductor instead and work with the current.
 
  • #25
gneill said:
If you work with the voltage for the energy you don't need to consider the charge. Either will do so long as you employ the appropriate expression for energy in terms of either voltage or charge. Or you could look at the inductor instead and work with the current.
Well in order to solve for the time I need to use charge to use the Q=Qocos(ωt) and rearrange to solve for time.
so i worked out the math,
my new voltage at 1/2 original energy V=√(2E/C) = 6.36 volts
new charge at 1.2 energy Q=CV= (4700e-6 * 6.36) = 0.0299
initial charge Q=CV=(4700e-6 * 9) = .0423
Q=Qocos(ωt)
t= (cos-1(Q/Qo))/ω
ω=1/√LC
t=(cos-1(.0423/0.0299))/(1/√*(1.5*4700e-6)

t=0.132 seconds?
 
  • #26
clamatoman said:
Well in order to solve for the time I need to use charge to use the Q=Qocos(ωt) and rearrange to solve for time.
The voltage follows the same oscillation. You could also have used ##V = V_o cos(\omega t)##
so i worked out the math,
my new voltage at 1/2 original energy V=√(2E/C) = 6.36 volts
If E is the original energy then you'll want to check that formula. You got the correct voltage value, but it's not given by √(2E/C).
new charge at 1.2 energy Q=CV= (4700e-6 * 6.36) = 0.0299
initial charge Q=CV=(4700e-6 * 9) = .0423
Q=Qocos(ωt)
t= (cos-1(Q/Qo))/ω
ω=1/√LC
t=(cos-1(.0423/0.0299))/(1/√*(1.5*4700e-6)

t=0.132 seconds?
You'll want to check that last calculation. You've swapped the two charge values.
 
  • #27
gneill said:
The voltage follows the same oscillation. You could also have used ##V = V_o cos(\omega t)##

If E is the original energy then you'll want to check that formula. You got the correct voltage value, but it's not given by √(2E/C).

You'll want to check that last calculation. You've swapped the two charge values.
ok using V=Vocos(ωt)
rearrange and solve t=(cos^-1(6.36/9))/(1/(√1.5 * 4700*10-6)) = 0.066 seconds? does that look right?
 
  • #28
clamatoman said:
ok using V=Vocos(ωt)
rearrange and solve t=(cos^-1(6.36/9))/(1/(√1.5 * 4700*10-6)) = 0.066 seconds? does that look right?
Yes. Much better.
 
  • #29
Awesome and i plugged it into make sure the time gives me the same Voltage. Looks right to me. Thanks!
 

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