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### Course: Electrical engineering>Unit 2

Lesson 1: Circuit elements

# Ideal elements and sources

Ideal models of the resistor, capacitor, and inductor.  Ideal voltage and current sources. Written by Willy McAllister.
An electric circuit is made of elements. Elements include at least one source. The source is connected to a bunch of components. We are going to describe sources and components with ideal mathematical abstractions. By the end of this article, we will have a nice collection of equations, which can be combined to generate lots of useful electronic functions. The next article describes real-world components that come close to the ideal abstractions we define here.
Elements are either sources or components.
Sources provide energy to a circuit. There are two basic types.
• Voltage source
• Current source
Components come in three basic types, each characterized by a different voltage-current relationship.
• Resistor
• Capacitor
• Inductor
These sources and components have two terminals or connection points. Not surprisingly, they are referred to as $2$-terminal elements.

## Ideal sources

### Constant voltage source

An ideal constant voltage source has a fixed output voltage, independent of the current drawn by the components connected to its terminals, as shown in this current versus voltage plot:
The equation for a constant voltage source is,
$v=\text{V}$
where $\text{V}$ is some constant output voltage, like $v=3\phantom{\rule{0.167em}{0ex}}\text{V}$.
You frequently see the variable name $e$ associated with voltage, derived from the term "electromotive force" or emf. This term is sometimes used when talking about the voltage from a source (battery or generator).
The two common symbols for constant voltage sources:
The symbol on the left is used for a battery. The longer horizontal line on the battery symbol represents the positive terminal of the battery, and the shorter horizontal line represents the negative terminal. The circle symbol represents some other source of voltage, often a power supply. It is a good practice to draw the $+$ and $-$ signs inside the circle.

### Variable voltage source

An ideal variable voltage source generates a known voltage as a function of time, independent of the current drawn by the components connected to its terminals, as shown in this $voltage$ versus $time$ plot:
The equation for a variable voltage source is,
$v=v\left(t\right)$
$v\left(t\right)$ can be a sine wave or any other time-varying voltage, for example, a single voltage step, or a repeating square wave.
The symbol for a variable voltage source:
The squiggle inside the circle suggests this particular symbol represents a sine wave generator. You will come across variations of this symbol for different waveform shapes.
These ideal mathematical abstractions of voltage sources can produce arbitrarily huge output current if the components they are connected to demand it. That doesn't happen in real life, of course. One place gigantic currents pop up is when you simulate a circuit. The computer doesn't mind a current of a zillion amperes, but it's probably not what you intended.

### Constant current source

An ideal constant current source has a fixed output current, independent of the voltage connected to its terminals, as shown in this $current$ versus $voltage$ plot:
The equation for a constant current source is,
$i=\text{I}$
where $\text{I}$ is a constant output current, like $i=2\phantom{\rule{0.167em}{0ex}}\text{mA}$.
The symbol for a constant current source:
The arrow indicates the direction of positive current flow.
The voltage at the terminals of an ideal current source becomes whatever is required to push out the constant current, even if that voltage is gigantic. When we build real current sources, of course, the range of operation is significantly restricted compared to the ideal current source abstraction.

## Resistor

The voltage across a resistor is directly proportional to the current flowing through it.
$v=\text{R}\phantom{\rule{0.167em}{0ex}}i\phantom{\rule{2em}{0ex}}\text{Ohm’s Law}$
This relationship is known as Ohm's law. You'll use this equation a lot in your work with circuits.
$\text{R}$ is a constant of proportionality, representing the resistance. Resistance has units of ohms, denoted by the Greek capital Omega symbol, $\mathrm{\Omega }$.
The $i$-$v$ graph for a resistor is shown below. The equation plotted is $i=v/\text{R}$, so the slope of the line is $1/\text{R}$.
The symbols for a resistor:
In the US and Japan the resistor symbol is a zig-zag. In the UK, Europe and other parts of the world, the resistor is often drawn as a box.
Ohm's Law can be written a number of ways, all of them useful,
$v=i\phantom{\rule{0.167em}{0ex}}\text{R}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}i=\frac{v}{\text{R}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\text{R}=\frac{v}{i}$
Ohm's Law is worth committing to memory.

### Power in a resistor

Power is dissipated by a resistor when current flows through it.
Energy in flowing electrons becomes bulk heat as electrons collide with atoms in the resistor material. Power can be expressed a few ways using Ohm's Law. These are all equivalent,
$p=v\phantom{\rule{0.167em}{0ex}}i$
$p=\left(\text{i}\phantom{\rule{0.167em}{0ex}}\text{R}\right)\phantom{\rule{0.167em}{0ex}}i={i}^{2}\phantom{\rule{0.167em}{0ex}}\text{R}$
$p=v\left(\frac{v}{\text{R}}\right)=\frac{{v}^{2}}{\text{R}}$
The last two expressions reveal that power in a resistor goes up (or down) proportional to the square of voltage or current.
• Increase either voltage or current by a factor of $2$, the power consumed goes up by a factor of $4$.
• Reduce either voltage or current by half, and you reduce the power by
• Aaron finds a way to cut the voltage across a resistor by a factor of two. When Beth looks at Aaron's new design, she figures out how to reduce the current in the resistor by a factor of two.

## Capacitor

The basic equation describing a capacitor relates charge on the capacitor to the voltage across the capacitor.
$\text{Q}=\text{C}\phantom{\rule{0.167em}{0ex}}\text{V}$
The constant of proportionality $\text{C}$ is the capacitance. Capacitance has units of farads, symbolized by the capital letter $\text{F}$. The unit of capacitance is the farad, and from the equation above we see that, $1\phantom{\rule{0.167em}{0ex}}\text{farad}=1\phantom{\rule{0.167em}{0ex}}\text{coulomb}/\text{volt}$
If the charge can move, we have a term for this; moving charge is called current. Current is the time rate of change of charge,
$i=\frac{dq}{dt}$
Using this idea, let's take the derivative of both sides of $\text{Q}=\text{C}\phantom{\rule{0.167em}{0ex}}\text{V}$ with respect to time and see what we get,
$\frac{dq}{dt}=\text{C}\phantom{\rule{0.167em}{0ex}}\frac{dv}{dt}$
and we end up with an equation saying the current in a capacitor is directly proportional to the time rate of change of the voltage across the capacitor,
$i=\text{C}\phantom{\rule{0.167em}{0ex}}\frac{dv}{dt}$
This capacitor equation captures the $i$-$v$ relationship for capacitors. It also tells us that electric circuits can be affected by time.
The symbols for a capacitor:
The version with the curved line is used for capacitors manufactured in a way that requires one terminal to have a positive voltage with respect to the other terminal. The curved line indicates the terminal that needs to be kept at the more negative voltage.
We can flip the capacitor equation around to solve for $v$ in terms of $i$ by integrating both sides, resulting in the integral form of the capacitor equation,
$v=\frac{1}{\text{C}}\phantom{\rule{0.167em}{0ex}}{\int }_{-\mathrm{\infty }}^{\phantom{\rule{0.167em}{0ex}}T}i\phantom{\rule{0.167em}{0ex}}dt$
The $-\mathrm{\infty }$ lower limit on the integral suggests that the capacitor's voltage at time $T$ depends not just on the capacitor current right now, but on the entire past history of the current. That's a long time ago, so we often write this integral starting at some known voltage ${v}_{0}$ at some known time like $t=0$, and then keeping track of the changes from there.
$v=\frac{1}{\text{C}}\phantom{\rule{0.167em}{0ex}}{\int }_{\phantom{\rule{0.167em}{0ex}}0}^{\phantom{\rule{0.167em}{0ex}}T}i\phantom{\rule{0.167em}{0ex}}dt+{v}_{0}$

### Power and energy in a capacitor

The instantaneous power in watts associated with a capacitor is,
$p=v\phantom{\rule{0.167em}{0ex}}i$
$p=v\phantom{\rule{0.167em}{0ex}}\text{C}\phantom{\rule{0.167em}{0ex}}\frac{dv}{dt}$
The energy $\left(U\right)$ stored in a capacitor is power integrated over time,
$U=\int p\phantom{\rule{0.167em}{0ex}}dt=\int v\phantom{\rule{0.167em}{0ex}}\text{C}\phantom{\rule{0.167em}{0ex}}\frac{dv}{dt}\phantom{\rule{0.167em}{0ex}}dt=\text{C}\int v\phantom{\rule{0.167em}{0ex}}dv$
If we assume the capacitor voltage was $0\phantom{\rule{0.167em}{0ex}}\text{V}$ at the beginning of the integration, then the integral evaluates to:
$U=\frac{1}{2}\phantom{\rule{0.167em}{0ex}}\text{C}\phantom{\rule{0.167em}{0ex}}{v}^{2}$
Unlike a resistor, where the energy is lost to heat, the energy in an ideal capacitor does not dissipate. Instead, energy in the capacitor, in the form of stored charge, is recovered when the charge flows back out of the capacitor.

## Inductor

The voltage across an inductor is directly proportional to the time rate of change of current through the inductor,
$v=\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}$
This property arises from the inductor's ability to store energy in a surrounding magnetic field. The stored magnetic energy can return to the circuit by generating an electric current.
The constant of proportionality $\text{L}$ is the called the inductance. The unit of inductance is the henry, denoted by the capital letter H.
The reason this property of inductance arises in coils of wire is a complex topic involving the intimate relationship between electricity and magnetism, which is beyond the scope of this article. For now, please trust that the voltage across an inductor is proportional to the rate of change of current.
The symbol for an inductor:
It looks like a wire wrapped in a coil, since that is the usual way to make an inductor.
Similar to the capacitor equation, we can write the inductor equation in integral form to get $i$ in terms of $v$. Notice the kinship between the capacitor and inductor equations.
$i=\frac{1}{\text{L}}{\int }_{-\mathrm{\infty }}^{\phantom{\rule{0.167em}{0ex}}T}v\phantom{\rule{0.167em}{0ex}}dt$
$v=\frac{1}{\text{C}}\phantom{\rule{0.167em}{0ex}}{\int }_{-\mathrm{\infty }}^{\phantom{\rule{0.167em}{0ex}}T}i\phantom{\rule{0.167em}{0ex}}dt$
The $-\mathrm{\infty }$ lower limit on the integral means the inductor's current at time $T$ depends on the entire past history of the inductor voltage. We usually write this integral starting from some known current ${i}_{0}$ at some known time like $t=0$, and then keeping track of the changes from there.
$i=\frac{1}{\text{L}}\phantom{\rule{0.167em}{0ex}}{\int }_{\phantom{\rule{0.167em}{0ex}}0}^{\phantom{\rule{0.167em}{0ex}}T}v\phantom{\rule{0.167em}{0ex}}dt+{i}_{0}$

### Power and energy in an inductor

The instantaneous power in watts associated with an inductor is
$p=i\phantom{\rule{0.167em}{0ex}}v$
$p=i\phantom{\rule{0.167em}{0ex}}\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}$
The energy $\left(U\right)$ stored in the magnetic field of an inductor is power integrated over time,
$U=\int p\phantom{\rule{0.167em}{0ex}}dt=\int i\phantom{\rule{0.167em}{0ex}}\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}\phantom{\rule{0.167em}{0ex}}dt=\text{L}\int i\phantom{\rule{0.167em}{0ex}}di$
$U=\frac{1}{2}\phantom{\rule{0.167em}{0ex}}\text{L}\phantom{\rule{0.167em}{0ex}}{i}^{2}$
Unlike a resistor, where the energy is lost to heat, the energy in an ideal inductor does not dissipate. Instead, the energy stored in the inductor's magnetic field can be fully recovered when the energy in the magnetic field gets converted back into an electric current in the wire.

## Summary of ideal component equations

Here are the three important circuit component $i$-$v$ equations,
$v=i\phantom{\rule{0.167em}{0ex}}\text{R}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}$ Ohm's Law
$i=\text{C}\phantom{\rule{0.167em}{0ex}}\frac{dv}{dt}\phantom{\rule{2em}{0ex}}$ capacitor equation
$v=\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}\phantom{\rule{2em}{0ex}}$ inductor equation
These three expression are your tools of the trade for circuit analysis.
In addition, we also developed these expressions for power and energy.
The power in a resistor is
$p=i\phantom{\rule{0.167em}{0ex}}v\phantom{\rule{1em}{0ex}}$ or$\phantom{\rule{1em}{0ex}}{i}^{2}\phantom{\rule{0.167em}{0ex}}r\phantom{\rule{1em}{0ex}}$or $\phantom{\rule{1em}{0ex}}{v}^{2}/r$
The energy in a capacitor is $\frac{1}{2}\phantom{\rule{0.167em}{0ex}}\text{C}\phantom{\rule{0.167em}{0ex}}{v}^{2}$
The energy in an inductor is $\frac{1}{2}\phantom{\rule{0.167em}{0ex}}\text{L}\phantom{\rule{0.167em}{0ex}}{i}^{2}$
The next article describes how real-world physical components come close to the mathematical ideal.

## Want to join the conversation?

• The text articles are very interesting, however, they would be even more useful if they could be printed out. It there any way to make this possible?
• I have an add-on for my Firefox browser called "Print Edit" which lets me delete portions of the page before printing.
• what is power? is energy always dissipated as heat?
• In an enclosed system (in this case, an electrical circuit), the amount of Energy is constant.

Energy consumed over a period of time, is power.
E = P x t , Unit : E = joule or Watt-hrs , P = Watt or VA.

Or we can say, we have a circuit with 5V battery, and a 5 ohms bulb connected, how much energy needed to run (on) this bulb for 1 minute?

V=IR, P=VI, E=Pt.
| solving for E, we get E = ( V x ( V / R ) ) x t = 5V x 5V / 5Ohm * 60s = 300 Joule. :)

-----
Is energy always dissipated as heat?

IMHO, for an ideal circuit, no.
Heat is the rate of change of temperature, or measure of a structure lattice vibration. If we say it dissipate as heat, then we need to consider the whole physics surrounding it. :)

A water pipe that flows, doesn't just deliver water current at a constant volume per second, it also vibrate the pipe. it have friction with the pipe wall, it flows through pipes that have different diameters, the water also have friction with it's own water molecules. So, does the water circuit have a loss of energy? yes.
Is it just heat/vibration? not really.

Back to our concern, the electrical circuit. It really have a lot of similarities with the water pipe example. But instead of water flow circuit, it is a charge flow circuit.
If the physical circuit is not built with superconductor, then there is almost always have internal resistance (from the wire/load/connections/change of conductor diameters).
The energy loss can be in the form of insulation capacitance, or conductors inductions, and ionization (charge exposed to atmosphere will charge the air). The rate is very minimal for small circuits, but in electrical transmission business, the energy loss can be very significant. :)
• What is the voltage across a ideal current source? Is it just the sum of voltage drops across every passive element in the circuit?
• Whatever it needs to be to make the current line up, as per Ohm's Law. An ideal current source adjusts its voltage constantly, as per Ohm's Law, in order to keep its current locked to the predetermined value despite whatever changes occur in the load's resistance or impedance. Likewise, an ideal voltage source will constantly match how much current it supplies, holding the voltage steady.
• What is the difference between potential difference, voltage and electromotive force?
• The term "voltage" is the honorary name given to "potential difference". They mean the same thing. I sometimes forget that "volt" and "voltage" are in honor of a person, and they only get their technical meaning because we say so. (Just like the physics term the newton means kg-m/s^2.) One interesting twist about the definition of voltage... it is defined to be a difference. So if someone asks, "What's the voltage difference between these to points?", it's kind of redundant (the same as saying "What is the potential difference difference?"). So the precise phrase would be, "What is the voltage between these two points?"

The term "electromotive force" (emf) is also a measure of potential difference. This term emphasizes what potential difference does: it impresses a "motive force" on an electron. EMF is the term used when talking about electric force generated by a moving magnetic field. The units of EMF are also volts (joules/coulomb), and the symbol is "E" or "e". You will see "e" all the time in circuits, used interchangeably with "v". Sal has a nice video on where the term EMF comes from: https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-12-induced-current-in-a-wire.

At the beginning of EE it can get confusing to see all these terms thrown around. It's hard to tell if they mean the same thing or if there is some important difference that you don't quite get, yet. For PE, V, and EMF, they are different ways to say "electric push".
• At the end of capacitor and inductor, what is the long curly symbol right after " V= "
and after "i = " ?
• The loopy symbol is from calculus. It is the "integral" sign.

In the capacitor equation above, v = ..., the integral sign tells you to add up the product of current, i, times a tiny interval of time, dt, for every t starting at time t = -infty and stopping at time t = T. Integration is the opposite of taking the derivative. You study integrals in integral calculus, https://www.khanacademy.org/math/integral-calculus.

Thanks for pointing this out. I added this response to the end of the capacitor section in the revised version of this article at http://spinningnumbers.org/a/ideal-elements-and-sources.html#capacitor.

P = (I^2)*R but also P = (V^2)/R

So is power Directly or Indirectly proportional to current?
• Hello Abhishek,

There are a few relationships to consider.

From Ohm’s Law:

voltage = current x resistance

Also power is defined as energy per unit time or if you prefer:

P = voltage x current

As to your question, there are two scenarios to consider:

1) In the first case we do not know or care about the resistance. And so we say:

P = voltage x current

2) In the second case we are given a fixed resistance. We then change the operating conditions of the circuit. Here is the important part - if we change the voltage the current must also change. For example if we assume a resistance of 2 Ohms and an initial voltage of 4 volts then we know the current is 2A. Consequently the power is 4 * 2 = 8 W. Now if we double the voltage to 8 volts the current will rise to 8 / 2 = 4A. The power is now 8 * 4 = 32 W.

When the voltage was 4 volts P = (V^2)/R = (4^2)/2 = 8 W also P = (I^2)R = (2^2)2 = 8W

When the voltage was 8 volts P = (V^2)/R = (8^2)/2 = 32 W also P = (I^2)R = (4^2)2 = 32W

Recommend you Google “Ohm’s Law pie chart.” You will find a wheel with solutions for V, I, R, and I. Work the equations voltage = current x resistance and P = voltage x current to solve for all items on the wheel.

Regards,

APD
• How should I understand current sources? Let's say that I have a 1 A ampere current source: that source adds (reiforces, gives) a current of 1 A to the circuit or it means that the current in the branches is exactly 1 A? Thank you for support!
• Gustavo - You asked: for "a 1 A ampere current source: that source adds a current of 1 A to the circuit or it means that the current in the branches is exactly 1 A? The best answer is your second choice: The current source makes the current in its branch exactly 1A. The current source does not "add" 1 A to another current.

In your comment to APD you described a 2-mesh circuit with a current source on the shared branch. If you try to use the Mesh Current Method, this configuration is a special case that requires extra care to solve. A current source in this position constrains two mesh currents to be the same value. You need to create another degree of freedom to solve the circuit. Search the web for the term "supermesh" to see what this means.
• what is a Ideal circult elements
• All the components above are ideal circuit elements. Remember that there are 2 types real and ideal. Real components are the physical components we can test and apply actual power to the circuits. where as your ideal circuit components are in layman's terms "theory". all these formulas above are of ideal conditions where as in real the conditions are much more wide spread to cause voltage variation or current spikes or all sorts of things that may change the end result.
• In the section "Power as a resistor" - Aaron and Beth part, I have a question. What I understand is that voltage AND current are reduced by half each.
Let's say
new current(I) = I'
new voltage(V) = V'

I' = I/2
V' = V/2

Then the new power is
P'=I'V'=(I/2)*(V/2)= IV/4 = P/4
Hence I got that power is reduced by factor of 4.

Where is the flaw in my solving process?
I used the equation P=IV, which involves both the changed variables in the question. I don't understand how it is 4*4=16.
Thank you.
• since the voltage cut by Aaron is cut across the resistor so,
p=v^2/r
i.e, v=1/2
v^2=1/4
p1=v^2/r
or, p1=v'r
where, v'=1/4
now Beth also saw that she could also cut the current across the resistor by a factor of 2,
p=i^2r
since i=1/2,
i^2= 1/4
so, p2=i^2*r or p=i'*r
here, i'=1/4
now on replacing the equations;
p=p1*p2
where r will cancelled out
p=v'*i'
p=i^2*v^2
p=1/16