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

• Master1022
In summary, the voltage V = V_0 sin(\omega t) is applied to an inductor, and the expression for the current and power are derived using the equations V = -L \frac{di}{dt} and P = IV. However, to arrive at the correct answer, it is necessary to consider the indefinite integral and set the initial condition for the current, I(0), to the observed experimental value of V_0/L\omega. This is because the initial condition represents the transient state of the circuit at t=0, where the current starts from 0 and quickly oscillates to reach its final value.f

## Homework Statement

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

## Homework Equations

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

## The Attempt at a Solution

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

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

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

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

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.

Consider the indefinite integral to get the expression for the current without assuming any specific constant of integration. That should give what you want.

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).

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 $$\frac{V_0}{\omega L}$$ that as opposed to 0?

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.

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?

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.

• Master1022
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:
• kuruman
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?

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.

• Master1022
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/

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?

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.

• Master1022