Power in a circuit with an inductor and integrals

In summary, the conversation discusses how to calculate the net energy into an inductor using the voltage and current equations. They also mention using calculus to take the integral over one cycle, where one cycle is defined as a change in omega*t from zero to 2*pi. The final equation for power is P=(V_p^2/2omega*L)*sin(2omega*t).
  • #1
bokonon
20
0
1. The power into a circuit element in the product of the voltage across the element and the current through the element. Assuming a voltage v(t) = V_p cos(omega*t) across inductor L, integrate the power over one cycle and show that the net energy into the inductor is zero.

V_p is the peak voltage

2. Homework Equations :
P=IV
I=(V_p/(omega*L))sin(omega*t)
Ok, so I'm really bad at calculus. I combined the three above equations to solve for Power, P and got: P=((V_p)^2/(omega*L))cos(omega*t)sin(omega*t). I assume I have to take the integral of this, but I'm not even sure what one cycle means. Can anybody point me in the right direction?

My best guess of the integral would be something like (since integral(sinxcosx)=.5sinx^2):

.5sin(omega*t)^2*V_p/(omega*L)

Thanks
 
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  • #2
In one complete cycle omega*t changes from zero to 2*pi. Take the integration between these limits.
 
  • #3
Can you help me out a little bit more about how to take that integral?
 
  • #4
P= Vp^2/wL*sinwt*coswt = (Vp^2/2wL)*sin2wt.
Now find the integration with limits.
 

Related to Power in a circuit with an inductor and integrals

1. How does an inductor affect power in a circuit?

An inductor is a passive electrical component that stores energy in the form of a magnetic field. When current flows through an inductor, it creates a back EMF (electromotive force) that opposes the flow of current. This results in a delay in the flow of current, which leads to a decrease in power in the circuit. Therefore, an inductor has the ability to reduce power in a circuit.

2. What is the relationship between inductance and power in a circuit?

The amount of inductance in a circuit is directly proportional to the amount of power that can be stored in the inductor. This means that as the inductance increases, the power in the circuit also increases. Similarly, a decrease in inductance results in a decrease in power.

3. How do integrals play a role in power calculations for circuits with inductors?

Integrals are mathematical tools used to calculate the total power in a circuit. In circuits with inductors, integrals are used to calculate the energy stored in the inductor, which is a crucial component in determining the overall power in the circuit. The integral of the voltage across the inductor gives the energy stored in the magnetic field, which can then be used to calculate power.

4. Can power be completely lost in a circuit with an inductor?

No, power cannot be completely lost in a circuit with an inductor. The energy stored in the magnetic field of the inductor will eventually be released back into the circuit, resulting in a transfer of power. However, the delay caused by the inductor can lead to a decrease in power in the short term.

5. How can the power loss in a circuit with an inductor be minimized?

The power loss in a circuit with an inductor can be minimized by using components with low resistance and high inductance values. This will reduce the delay in current flow and result in less power loss. Additionally, using capacitors in conjunction with inductors can help to balance out the effects of inductance and minimize power loss in the circuit.

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