# How Does Power Calculation Translate to Energy Absorption in a Circuit?

• XodoX
In summary, a circuit element is shown below, with the voltage v(t) and the current i(t) defined. The voltage and current are given as v(t) = 5sin(t) [V] and i(t) = −10cos(t) [A], respectively. The power that the element is delivering is 50 sin(t) cos(t) [W], and the amount of energy that the circuit element absorbs between t = 0 and t = 1 [s] is 13.3 [J].
XodoX
1. A circuit element is shown below, with the voltage v(t) and the current i(t) defined. The
voltage and current are given as
v (t ) = 5sin (t ) [V]
i (t ) = −10cos (t ) [A] .
Calculate the power that the element is delivering.

b) For the same circuit in the previous problem, calculate the amount of energy that the circuit
element absorbs between t = 0 and t = 1 .

I don't get the b0 part. The solution for the first part is 50 sin(t) cos(t) [W]. How do I get the amount of energy for that time frame?

XodoX said:
1. A circuit element is shown below, with the voltage v(t) and the current i(t) defined. The
voltage and current are given as
v (t ) = 5sin (t ) [V]
i (t ) = −10cos (t ) [A] .
Calculate the power that the element is delivering.

b) For the same circuit in the previous problem, calculate the amount of energy that the circuit
element absorbs between t = 0 and t = 1 .

I don't get the b0 part. The solution for the first part is 50 sin(t) cos(t) [W]. How do I get the amount of energy for that time frame?

What kind of a circuit element is it? Why are they asking about delivering power in one part, and absorbing power in the other part?

To get energy over a time period, you just integrate the power multiplied by dt over the time interval.

I forgot how to take the integral of cos/sin

u= sin(t) du=cos(t) dt

Maybe you can explain the next step to me:) I think that once I got the result of it I am done.

XodoX said:
I forgot how to take the integral of cos/sin

u= sin(t) du=cos(t) dt

Maybe you can explain the next step to me:) I think that once I got the result of it I am done.

The wikipedia.org article on integration by parts is pretty reasonable:

http://en.wikipedia.org/wiki/Integration_by_parts

.

## 1. What is the equation V(t) = 5sin(t)?

The equation V(t) = 5sin(t) represents a sinusoidal function where the amplitude is 5 and the period is equal to 2π.

## 2. What does the variable t represent in the equation V(t) = 5sin(t)?

The variable t represents time in seconds. It is the independent variable used to calculate the voltage at a specific point in time.

## 3. How do you calculate the voltage at a specific time using the equation V(t) = 5sin(t)?

To calculate the voltage at a specific time, simply plug in the time value in seconds for t into the equation V(t) = 5sin(t) and solve for V. The resulting value will be the voltage at that particular time.

## 4. What is the maximum voltage in the function V(t) = 5sin(t)?

The maximum voltage in the function V(t) = 5sin(t) is 5 volts. This is because the amplitude of the function is 5, meaning that the voltage values will oscillate between -5 and 5.

## 5. How does the graph of V(t) = 5sin(t) look like?

The graph of V(t) = 5sin(t) is a sinusoidal curve with an amplitude of 5 and a period of 2π. The curve will start at the origin (0,0) and reach a maximum value of 5 at t = π/2 seconds. It will then decrease back to 0 at t = π seconds, reach a minimum value of -5 at t = 3π/2 seconds, and return to 0 at t = 2π seconds. The curve will continue to repeat this pattern for all values of t.

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