Power and Energy

Introduction: This article introduces power and energy in the context of voltage and current.

The Essentials

Along with voltage and current, the output of the circuit is one important characteristic. The output can described as the power or energy. When it comes to design, having the units voltage and current calculated alone are not sufficient.

In physics, power is described as work being done over time. Power can be described as the following:

\[ p = \frac{dw}{dt} \]

where

p = the power in watts
w = the energy in joules
t = the time in seconds

So, in this case, 1 Watt is the same as 1 Joule per second.

Remembering our previous definitions of current and voltage, we can rewrite the equation as the following:

\[ p = \frac{dw}{dq} \frac{dq}{dt} = vi = i^{2}r \]

where

p = the power in watts
v = the voltage in volts
i = the current in amps
r = the resistance in ohms

So power at a terminal can be described as the voltage across the terminal multiplied by the current through the terminal.

Example

Let's do a brief example of what you might go through.

Let's say that the voltage across the device and current through it is described as \(v(t) = 20 sin(4 t) \) and \( i(t) = 10(1+ e^{-2t} \)) mA. Calculate (1) the total charge in the device at \( t = 1 \) s and \( q(0) = 0 \) and (2) calculate the power consumed by the device at \( t = 1 s \).

Firstly, to find the charge we remember that the current is the flow of charge over time, and since we are curious of the charge we ought to take its integral:

\[ Q = \int_{0}^{1} 10(1+e^{-2t})\, dt = 10(t-\frac{1}{2}e^{-2t}) \big|_{0}^{1} \]

Evaluate this integral from 0 to 1 and you should end up with this result:

\[ 10(1-\frac{1}{2}e^{-2}+\frac{1}{2}) = 14.32mC \]

This result is the charge in the device at t = 1 s, answering the first question.

Now for the second question. We are interested in the power consumed by the device at t = 1 s. Looking at equation 2 above, remember that power is voltage times current, so we do that in this problem. As we multiply, we get this equation:

\[ p(t) = v(t) \cdot p(t) = 20sin(4 t) \cdot 10(1+e^{-2t}) = 200(1+e^{-2t})sin(4 t) \]

With this equation, we solve for p(1). When we do this, we get:

\[ p(1) = 200(1+e^{-2})sin(4) = -171.84 mW \]

So in this scenario, the device appears to be supplying energy at \( t = 1 \) s since the power is negative at that time.

Practice

Problem 1.) The charge entering the positive terminal of an element is described as:

\[ q(t) = 5sin(4\pi t)\, \text{mC} \]

The voltage across the same terminal is described as:

\[ v(t) = 3cos(4\pi t) \,\text{V} \]

a) Find the power delivered to terminal at \( t = 0.3 \) s and (b) Calculate the energy delivered to the element between 0 and 0.6 s

Problem 2.) The current entering the positive terminal of a device is

\[ i(t) = 6e^{-2t} \text{mA} \]

and the voltage across the device is

\[ v(t) = 10 \frac{di}{dt}\, \text{mV} \]

(a) Find the charge delivered to the device between \( t = 0 \) and \( t = 2 \)s. (b) Calculate the power absorbed. (c) Determine the energy absorbed in 3 s.

Problem 3.) Figure 1. shows a given schematic for a circuit with numerous elements:

Polarity of voltage vabD

Figure 1: Practice Problem 3 Schematic

Say \( p_{1} = 60 \) W absorbed, \( p_{3} = -145 \) W absorbed, and \( p_{4} = 75 \) W absorbed. How many watts does element 2 absorb?

Solutions:

Problem 1 Solution.) (a) \( p = 123 mJ \). We are given the equations for charge and voltage and we are interested in the power equation. We know that current is change of charge with respect to time, so we need to take a derivative of the charge equation

\[ i(t) = \frac{dq}{dt} = 20\pi cos(4\pi t) \]

From this, we multiply the voltage and current equation to get the power equation

\[ p(t) = v(t) \cdot i(t) = 60\pi cos^{2}(4\pi t) \]

We then evaluate the function at \( t = 0.3 \) s, and we get \( p(0.3) = 123.37 \) mJ.

(b) \( W = 58.75 mJ \). The energy of the terminal is represented in equation 1. In order to get the energy equation, integrating equation 1 is necessary

\[ W = \int p(t) \, dt = \int_{0}^{0.6} 60\pi cos^{2}(4\pi t)\, dt \]

Then we need to evaluate this integral. A trig identity will help with evaluating the \( cos^{2} \) term.

\[ W = \int_{0}^{0.6} 30\pi(1+cos(8\pi t) \, dt = 30\pi(t + \frac{1}{8\pi}sin(8\pi t))\big|^{0.6}_0 \]

After evaluating, you should get W = 58.75 mJ.

Problem 2 Solution.) (a) \( Q = 2.95 mC \). To solve for charge in device from \( t = 0 \) s to \( t = 2 \) s, we need to take the integral of the current equation.

\[ Q = \int_{0}^{2}6e^{-2t} \, dt = -3e^{-2t} \big|^{2}_0 \]

Evaluating this integral should result in 2.95 mC.

(b) \( p(t) = -720e^{-4t} \mu \text{W} \). The energy of the terminal is represented in equation 1. In order to get the energy equation, integrating equation 1 is necessary

(c) \( w(3) = -180 \mu J \). We'll need to use the equation for power found in part (b). Looking at equation 1, we recognize that power is the change of energy over time, so if we take the integral of power, we can get energy. Let's integrate the power equation with respect to time

\[ W = \int_{0}^{3}-720e^{-4t} \, dt = 180e^{-4t} \big|^{3}_0 \]

Evaluating the integral results in \( w(3) = = -180 \mu J \).

Problem 3 Solution.) \( p_{2} \) = 10 W. Given from the values we have, we notice that \( p_{3} \) is supplying power while \( p_{1} \) and \( p_{4} \) are absorbing power. We know from the law of conservation of energy, the power within the circuit must sum to 0. So let's write that out

\[ P_{1} + P_{2} + P_{3} + P_{4} = 0 \]

We know some of these values, let plug these in and solve for \( p_{2} \)

\[ -10 + P_{2} = 0 \]

In order for this statement to be true, \( p_{2} \) must be 10 W. Thus, power absorbed by \( p_{2} \) is 10 W.

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