Energy dissipated in a resistor

[Bob jefferson]

Homework Statement

A resistor of resistance 10Kohms has voltage across it described by the function
V(t) = 6sin(10t+pi/4)*exp(-2t) Volts
Calculate the energy dissipated in the resistor between t=0 and t=0.5

Homework Equations

The Attempt at a Solution

I have done the definite integral and have got 0.324 but i am not too sure if i am supposed use this answer in P=V^2/R .

 

[DrDu]

Definite integral sounds good, the question is which integral exactly?

 

[phyzguy]

Also, the units of the answer should be supplied.

 

[Bob jefferson]

Definite integral sounds good, the question is which integral exactly?

I integrated this V(t) = 6sin(10t+pi/4)*exp(-2t)

 

[DrDu]

Does the integral $\int V(t) dt$ have the dimension of an energy?

 

[Bob jefferson]

Does the integral $\int V(t) dt$ have the dimension of an energy?

No

 

[Bob jefferson]

Also, the units of the answer should be supplied.

I think the answer should be in joules as its asking for energy dissipated.

 

[phyzguy]

I think the answer should be in joules as its asking for energy dissipated.

That's correct. As DrDu pointed out, the integral you did doesn't have these units. Why did you go away from P = V^2 * R? What are the units of power? How do you go from power to energy?

 

[gneill]

Hint: Energy is the time integral of power.

 

[Bob jefferson]

Hint: Energy is the time integral of power.

so i just multiply 0.5s with the value i obtained from the integration.

 

[gneill]

so i just multiply 0.5s with the value i obtained from the integration.

No, you need to perform another integration where what you are integrating is the instantaneous power.

 

[Bob jefferson]

No, you need to perform another integration where what you are integrating is the instantaneous power.

Oh so actually putting V^2/r in the integral.

 

[phyzguy]

Oh so actually putting V^2/r in the integral.

Yes. Try this and tell us what you get.

 

[Bob jefferson]

No, you need to perform another integration where what you are integrating is the instantaneous power.

so would it be some thing like $\frac 1 R \int_0^{0.5} V^2(t) dt$

 

[phyzguy]

so would it be some thing like $\frac 1 R \int_0^{0.5} V^2(t) dt$

Yes.

 

[Bob jefferson]

Yes.

so i need a bit of guidance, would $sin^2(10t+\frac \pi 4)$ equal to $\frac 1 2(1-cos(2*10t+\frac \pi 4))$

 

[phyzguy]

$\sin^2(x) = \frac{1}{2}(1-\cos(2x))$, but that's not exactly what you wrote. The 2 should multiply everything inside the cos argument.