How to find the expression for current and power in an inductor

  • Thread starter Master1022
  • Start date
  • #1
611
116

Homework Statement


A voltage of [tex] V = V_0 sin(\omega t) [/tex] is applied to an inductor. Find an expression for the current and hence the power.

Homework Equations


I have used the equation: [tex] V = -L \frac{di}{dt} [/tex]
and for power: [tex] P = IV [/tex]

The Attempt at a Solution


I tried to find I(t) by integrating both sides of the first equation:

[tex] V_0 \int_{0}^{t}\sin({\omega t}) dt = -L \int_{0}^{I} dI [/tex]
[tex] \frac{V_0}{\omega} (1-\cos({\omega t})) = -LI [/tex]
Giving the expression for the current as:
[tex] I = -\frac{V_0}{L\omega}(1-\cos{(\omega t})) [/tex]

Then for the power:
[tex] P = IV = V_0 sin(\omega t) \times -\frac{V_0}{L\omega}(1-\cos{(\omega t})) [/tex]
[tex] P = -\frac{V_0^2}{L\omega}(\sin(\omega t))(1-\cos{(\omega t})) [/tex]

However, the answer only provides the power, which is given by: [tex] P = -\frac{V_0^2}{L\omega}(\sin(2\omega t)) [/tex]

I believe that the answer is missing a factor of a 1/2 in there, but I am still not sure how to arrive there. I believe that I must have gone wrong in the current calculation, but upon trying other methods (eg. differentiating enegry for inductor wrt t, it all leads to the same integral)

Any help is greatly appreciated.
 

Answers and Replies

  • #2
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
11,449
4,474
Consider the indefinite integral to get the expression for the current without assuming any specific constant of integration. That should give what you want.
 
  • #3
Delta2
Homework Helper
Insights Author
Gold Member
4,964
2,075
You get this result because you assume that ##I(t=0)=0## (that is you set the limits of integration for I from 0 to I, in the right hand side integral), usually we assume that ##I(t=0)=\frac{V_0}{L\omega}##. With this I(0) you ll get the desired answer (with the factor of 1/2 of course).
 
  • #4
611
116
You get this result because you assume that ##I(t=0)=0## (that is you set the limits of integration for I from 0 to I, in the right hand side integral), usually we assume that ##I(t=0)=\frac{V_0}{L\omega}##. With this I(0) you ll get the desired answer (with the factor of 1/2 of course).
Thank you very much for your reply. I did think that would lead me to the answer when I considered doing an indefinite integral. However, would you be able to explain why you would set I(t=0) to [tex] \frac{V_0}{\omega L} [/tex] that as opposed to 0?
 
  • #5
Delta2
Homework Helper
Insights Author
Gold Member
4,964
2,075
Well ,##I(0)## can be viewed as the initial (or boundary) condition for this differential equation ##(-LdI/dt=V(t))##, we can set it to anything and still get a valid solution (I mean from the perspective of just solving the ODE purely mathematically). However the value for ##I(0) ## observed by the experiments is ##V_0/L\omega## so that's why we set ##I(0) ## to this value.
 
  • #6
611
116
Well ,##I(0)## can be viewed as the initial (or boundary) condition for this differential equation ##(-LdI/dt=V(t))##, we can set it to anything and still get a valid solution (I mean from the perspective of just solving the ODE purely mathematically). However the value for ##I(0) ## observed by the experiments is ##V_0/L\omega## so that's why we set ##I(0) ## to this value.
Thank you once again for your reply. I can follow along everything in your explanation other than the final part about the boundary condition. Would you be able to explain why that is the case (not that I don't believe you, but I am just confused by the practical implication). When the voltage is 0 V (at t = 0), how is the current > 0?
 
  • #7
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
11,449
4,474
Thank you once again for your reply. I can follow along everything in your explanation other than the final part about the boundary condition. Would you be able to explain why that is the case (not that I don't believe you, but I am just confused by the practical implication). When the voltage is 0 V (at t = 0), how is the current > 0?
In a pure inductive circuit, the source voltage leads the current by 90o. In a purely capacitive circuit, the current leads the source voltage by 90o. Learn about ELI the ICEman here.
https://www.khanacademy.org/science...sis-topic/ee-ac-analysis/v/ee-eli-the-ice-man
 
  • #8
Delta2
Homework Helper
Insights Author
Gold Member
4,964
2,075
Thank you once again for your reply. I can follow along everything in your explanation other than the final part about the boundary condition. Would you be able to explain why that is the case (not that I don't believe you, but I am just confused by the practical implication). When the voltage is 0 V (at t = 0), how is the current > 0?
You have some point here. I mean if I can read your mind you are thinking along the line "The time is t=0, then V=0 and we 've just closed the switch, how the current instantaneously has get to the value ##I(0)=V_0/L\omega##"??. The answer is that there is the AC transient state of the circuit at the start of time t=0. During this transient state the current starts from I(0)=0 and does some oscillations of rapidly increasing amplitude. These oscillations last very small time and the final value for the increasing amplitude is around ##V_0/L\omega##. When the current reaches this value we can say that the circuit enter the AC steady state, where it does oscillations of constant amplitude ##V_0/L\omega##.

So briefly, from the moment we close the switch there is some short period where the current rises very fast from 0 to ##V_0/L\omega## and then it reaches AC steady state and does oscillation of constant amplitude. In almost all books about circuit theory this hidden AC transient state is not studied, it is not even mentioned in most, and we are only studying the AC steady state of circuits.
 
Last edited:
  • #9
611
116
You have some point here. I mean if I can read your mind you are thinking along the line "The time is t=0, then V=0 and we 've just closed the switch, how the current instantaneously has get to the value ##I(0)=V_0/L\omega##"??. The answer is that there is the AC transient state of the circuit at the start of time t=0. During this transient state the current starts from I(0)=0 and does some oscillations of rapidly increasing amplitude. These oscillations last very small time and the final value for the increasing amplitude is around ##V_0/L\omega##. When the current reaches this value we can say that the circuit enter the AC steady state, where it does oscillations of constant amplitude ##V_0/L\omega##.

So briefly, from the moment we close the switch there is some short period where the current rises very fast from 0 to ##V_0/L\omega## and then it reaches AC steady state and does oscillation of constant amplitude. In almost all books about circuit theory this hidden AC transient state is not studied, it is not even mentioned in most, and we are only studying the AC steady state of circuits.
Thank you very much for the help. So the limit in the integral will only be concerned with the circuit once it has entered the steady state (very shortly after "switch is closed"), hence the V_0/wL?
 
  • #10
Delta2
Homework Helper
Insights Author
Gold Member
4,964
2,075
Thank you very much for the help. So the limit in the integral will only be concerned with the circuit once it has entered the steady state (very shortly after "switch is closed"), hence the V_0/wL?
yes we are studying the circuit for the AC steady state. We can even take as t=0 the time the circuit reaches AC steady state, thus to put ##I(0)=V_0/L\omega## though this is not what exactly happens in the real world.
 
  • #11
cnh1995
Homework Helper
Gold Member
3,448
1,148
Thank you very much for the help. So the limit in the integral will only be concerned with the circuit once it has entered the steady state (very shortly after "switch is closed"), hence the V_0/wL?
You need to use indefinite intrgral with boundary conditions to see the time-response of the circuit. The transient response of a reactive ac circuit depends on the instant of closing the switch.
See if the following thread helps.
https://www.physicsforums.com/posts/5797346/
 
  • #12
611
116
yes we are studying the circuit for the AC steady state. We can even take as t=0 the time the circuit reaches AC steady state, thus to put ##I(0)=V_0/L\omega## though this is not what exactly happens in the real world.

Just a final question, where does the negative sign come in for the power? After plugging in that limit, the negative signs seem to cancel out?
 
  • #13
Delta2
Homework Helper
Insights Author
Gold Member
4,964
2,075
The negative sign is wrong, as well as the missing of a 2 in the denominator. Unless your book uses the convention that the power of a circuit element is negative when the source offers power to that element.
 

Related Threads on How to find the expression for current and power in an inductor

Replies
1
Views
277
  • Last Post
Replies
19
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
22
Views
2K
Replies
8
Views
26K
Replies
3
Views
277
Replies
2
Views
288
Top