Put simply, a circuit is a closed conducting wire loop with some elements in the mix, such as batteries, resistors, and capacitors. Only batteries and resistors are covered in AP® Physics 1; Physics 2 also covers capacitors. In this article, we’ll review the characteristics of circuits, resistors, and capacitors and do some example problems similar to those on the AP® Physics 1 and 2 exams.

Current

Electric current is defined as the charge passing through a cross-sectional area per time:

Electrical current = charge per time.

Charge is measured in Coulombs (C), so current has units of Coulombs per second, or Amperes: 1A=1C/s. For example, if 5.0C of charge passed through a resistor in 10s, the current would be (5.0C)/(10s)=0.5A. Physicists usually denote current by the letter I. Both lowercase and capital letters are commonly used.

Here’s a brief explanation of the microscopic origin of electrical current:

Electrical Current: a Brief Microscopic View

Electrical circuits are useful because they manipulate the flow of electrical charge. Imagine a metal wire. You might know from chemistry class that metallic bonds are formed when every atom shares its valence electron(s) with all other atoms – the electrons are not “attached” to any individual atom, and hence form an “electron sea.”

When the electron “sea” exhibits net movement in a particular direction along the wire, we say that the wire carries an electrical current. The electrical current moves in the opposite direction from the electrons because electrons are negatively charged; physicists use the direction positive charges would move (if they could move, as atomic nuclei are heavy) as a sign convention.

The electron “sea” moves very, very slowly (on the order of micrometers per second), but there are so many electrons that even very slow motion produces a measurable current.

Although the actual electrons may not travel all the way around the circuit, the current must do so. This is why circuits must be closed, or have no gaps. If the circuit has a gap, then there is no path for the current to make a round-trip, and we say the circuit is open.

In a steady-state circuit, there is no charge build-up in any node (a node is where two or more wires come together.) Therefore, the current coming into the node must equal the current leaving the node:

Current in = current out.

This is called Kirchhoff’s node rule, Kirchhoff’s junction rule, or Kirchhoff’s first law. It is a statement of the principle of charge conservation at a circuit node.

Voltage

Voltage difference is defined as the change in potential energy per unit charge:

Voltage difference = (change in potential energy) / (charge).

Since potential energy is measured in Joules (J) and charge in Coulombs (C), voltage is measured in Joules per Coulomb, or Volts: 1V=1J/C. Physicists denote voltage by the capital letter V. We say voltage “difference” because no point in a circuit has a specific, true voltage – only the difference in voltage between two points in a circuit has physical meaning.

When current passes through a circuit element, the charged particles may gain potential energy (positive voltage difference) or lose potential energy (negative voltage difference). There is a particular voltage difference across each circuit element, but when the current makes a round-trip, its energy must be the same as when it started:

The sum of voltage differences in a closed loop must be zero.

This is called Kirchhoff’s loop rule or Kirchhoff’s second law.

Keep in mind that the terms “voltage difference,” “potential difference,” “voltage drop,” “change in potential,” and other similar terms all mean the same thing.

Typically, batteries are required to keep the electric current in a circuit nonzero. More specifically, the resistance of a circuit causes energy loss, which is replenished by the chemical energy of a battery. The voltage across a battery is called the electromotive force, or emf:

Emf = electromotive force = voltage increase across a battery.

The electromotive “force” is not a true force; the name is a historical relic that nobody’s bothered to replace yet. You can read the emf of a battery on its label; for example, the small rectangular batteries usually have an emf of 9V.

The circuit symbol for battery on the AP® exams is:

The longer line on the battery symbol always represents the higher voltage. (You may see batteries with more lines; there should still be one short and one long line at the ends of the symbol.) If the emf of the battery is 9V, then we say the potential difference across the battery from left to right is 9V. The potential difference from right to left is -9V.

Resistance

Resistance is a quality of a circuit element that causes energy loss. For example, if the material of a wire is not perfectly conducting (nothing is, save superconductors), then energy will be lost when current flows through the wire. Often, circuit elements are designed that purposely cause the loss of energy; these circuit elements are called resistors. The circuit symbol for a resistor on the AP® exam is:

All the resistors and resistances you meet in AP® Physics follow a relation called Ohm’s Law:

V=IR.

In this equation, V is the potential difference across the circuit element, I is the current through the element, and R is its resistance. Circuit elements that follow this law are called Ohmic.

Remember we said that V is the change in potential energy per charge. When we multiply by current, we get the change in potential energy per time:

\dfrac{\Delta energy}{charge}\times\dfrac{charge}{time} = \dfrac{\Delta energy}{time}

Since power is energy per time, we now have a formula for the power dissipated by a resistor:

P=VI=I^2R=V^2/R,

The last two expressions come from substituting V=IR in various forms in the leftmost equality. The function of the battery is to compensate for this energy loss by transforming its chemical energy into electrical energy, which is eventually dissipated in the form of heat by the resistor(s) according to the formula above. This is why batteries get used up!

Combining Resistances

On the AP® exam, you’ll see resistors connected to each other in different combinations. All traditional combinations can be broken down into simpler combinations of two types: series and parallel.

You’ll almost certainly be asked to find the “equivalent resistance” of a combination of resistors. “Equivalent resistance” stands for the resistance of one resistor which would give the same current under the same voltage difference as the original combination does, or equivalently, the resistance of one resistor which would give the same voltage difference under the same current. All we need to derive the formula for resistor combinations are Ohm’s Law, V=IR, and Kirchhoff’s laws.

Resistors in Series

Resistors in series are connected in the same current pathway; you could think of them as two waterwheels on the same river. Below are two series resistors and the desired equivalent resistance.

The same current I passes through both { R }_{ 1 }and { R }_{ 2 }, so the voltages across these resistors are

{ V }_{ 1 }={ IR }_{ 1 }, { V }_{ 2 }={ IR }_{ 2 }.

The total voltage across the combination is

{ V }_{ 1 }+{ V }_{ 2 }=I ({ R }_{ 1 }+{ R }_{ 2 }).

We want the equivalent resistance to give the same voltage as the total voltage across the combination:

{ V }_{ eq }={ V }_{ 1 }+{ V }_{ 2 }= I ({ R }_{ 1 }+{ R }_{ 2 }).

Since { V }_{ eq }={ IR }_{ eq }, we have { R }_{ eq }={ R }_{ 1 }+{ R }_{ 2 }. Extending this to an arbitrary number of resistors in series, the equivalent resistance of series resistors is

{R}_{eq}={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }

Resistors in Parallel

Resistors in parallel are connected in separate branches that join together before and after the current passes through the parallel resistors:

In the figure above, a total current I separates into smaller currents I₁ and I₂ which pass through resistors R₁ and R₂, respectively. By Kirchhoff’s junction rule,

I ={ I }_{ 1 }+{ I }_{ 2 }.

By Kirchhoff’s loop rule on the loop consisting of resistors 1 and 2,

{ V }_{ 1 }={ V }_{ 2 }, or

{ R }_{ 1 }{ I }_{ 1 }={ R }_{ 12}{ I }_{ 2 }

{ I }_{ 2 }={ R }_{ 1 }{ I }_{ 1 }/{ R }_{ 2 }.

Substituting this into Kirchhoff’s junction rule, we obtain

I ={ I }_{ 1 }(1+{ R }_{ 1 }/{ R }_{ 2 }).

The voltage across the equivalent resistance is by definition equal to either the voltage across resistor 1 or resistor 2 (they’re equal):

{ R }_{ eq }I={ R }_{ 1 }{ I }_{ 1 }.

Combining this with the previous equation, we have \dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}. Extending to an arbitrary number of resistors, the equivalent resistance of parallel resistors is given by:

{\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} }.

To solve for the equivalent resistance algebraically, this equation tells you to add the reciprocals of the parallel resistances and take the reciprocal of the sum. This is the desired equivalent resistance.

Here’s an example problem that involves both parallel and series resistors.

Equivalent Resistance

A 2A current enters the resistor combination below:

(a) Find the current through each resistor.

(b) Find the power dissipated by each resistor.

(c) Find the equivalent resistance of the combination.

(d) Find the total power dissipated by this combination.

Solution:

(a) All of the current must pass through the 15\Omega resistor, so the current through it is 2.0A. The parallel combination is composed of two identical resistances, so the current through either of them is 1.0A. If these resistances were not identical, you would have to use Kirchhoff’s laws to set up the equations

I={ I }_{ 1 }+{ I }_{ 2 }, { R }_{ 1 }{ I }_{ 1 }={ R }_{ 2 }{ I }_{ 2 },

and then solve for { I }_{ 1 } and { I }_{ 2 }.

(b) Using P = I^2R for each of the three resistors, we find

{ P }_{ 10\Omega }={ (1.0A) }^{ 2 }(10\Omega)=10W,

{ P }_{ 15\Omega }={ (2.0A) }^{ 2 }(15\Omega)=60W.

(c) The equivalent resistance of the parallel combination is given by

\dfrac{1}{R_{parallel}} = \dfrac{1}{10 \Omega} + \dfrac{1}{10 \Omega}, or R_{parallel} = 5 \Omega.

The equivalent resistance of the entire combination is given by summing the resistance of the first combination with the remaining resistance in series:

R_{eq} = 5 \Omega + 15 \Omega = {20 \Omega}.

(d) Since voltage and current are identical in equivalent resistances and resistance combinations, power is identical as well. We can just use the power formula

P = I^2R_{eq} = (2.0 A)^2(20 \Omega) = {80 W}.

Of course, this is equal to the sum of the powers dissipated by the three resistors in part (b).

Capacitance (Physics 2 Only)

A capacitor is a device which consists of two metal plates separated by a small distance, which might contain an insulator such as paper or ceramic.

Below is the symbol for capacitance on the AP® Physics 2 Exam; you can see the symbolic representation of the two plates:

A capacitor has the capacity to store charge: one plate is positively charged and is at a higher potential; the other plate is negatively charged and is at a lower potential. The higher the capacitance, the more charge it can store for a given voltage difference. We call this ratio the capacitance, which is denoted by the capital letter C.

Q=CV.

Q represents the magnitude of charge on either plate and V represents the potential difference between the two plates. Capacitance is measured in Coulombs per Volt; physicists call this unit the Farad:

1F=1C/V.

Think about the equation Q=CV. If V is constant, then so is Q. Since no change in charge means no flow of current,

No current flows in or out of a fully charged capacitor.

Current flows only if the capacitor is charging or discharging. You won’t have to worry about this for the AP® Physics 2 Exam: the questions cover only the steady-state behavior of capacitors – when they are fully charged or not charged at all.

Similarly, to how resistors dissipate energy, capacitors store energy. The formula for stored energy in a capacitor are

U = \dfrac{CV^2}{2} = \dfrac{Q^2}{2C} = \dfrac{QV}{2}.

If you memorize one formula, you can get the other two by substituting various forms of the definition of capacitance, Q=CV.

Combining Capacitance’s (Physics 2 Only)

The laws for finding equivalent capacitance’s are the reverse of those for resistances. The derivations are quite similar so I won’t derive the formulas for equivalent capacitance’s, but here they are:

Capacitors in Series

{\dfrac{1}{C_{eq, series}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} }.

Capacitors in Parallel

{C_{eq, parallel} = C_1 + C_2 + C_3 }

Let’s do an example problem similar to one you might encounter on the AP® Physics 2 Exam.

Equivalent Capacitance

In the diagram below, each capacitor has a capacitance of 10mF, and the voltage across the combination is 5.0V.

(a) Find the voltage across each capacitor.

(b) Find the charge on each capacitor.

(c) Find the equivalent capacitance.

(d) Find the total energy stored in the circuit.

Solution:

(a) The voltage across both branches must be the same, by Kirchhoff’s loop rule. Since the capacitance’s of the two capacitors in the top branch are equal, both have a potential difference of 2.5V. The potential difference across the bottom capacitor is 5.0V.

(b) Using the equation Q=CV, we find

{ Q }_{ top }=(10mF)(2.5V)=25mC,

{ Q }_{ bottom }= (10mF)(5V)=50mC.

(c) Using the series addition rule, the equivalent capacitance of the top branch is given by

\dfrac{1}{C_{top}} = \dfrac{1}{10 mF} + \dfrac{1}{10 mF}, or C_{top} = 5 mF.

Using the parallel addition rule on both branches,

C_{eq} = 5 mF + 10 mF = {15mF}.

(d) We know the equivalent capacitance of the combination and the total voltage across it, so by the energy formula,

U = \dfrac{CV^2}{2} = \dfrac{(15 mF) (5.0 V)^2}{2} = {0.19 J}.

Wrapping Up Circuits for the AP® Physics 1 and 2 Exams

Here are important equations you should know for the AP® Physics 1 and 2 Exams.

Kirchhoff’s Junction Rule

Current in = current out

Kirchhoff’s Loop Rule

Sum of potential differences around a loop = 0

Ohm’s Law

{ V }_{ R }=IR

Power Dissipated by a Resistor

P = VI = I^2R = V^2/R

Series Equivalent Resistance

{R_{eq} = R_1 + R_2 + R_3 }

Parallel Equivalent Resistance

{\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}

Definition of Capacitance

Q=CV

Energy Stored in a Capacitor

U = \dfrac{CV^2}{2} = \dfrac{Q^2}{2C} = \dfrac{QV}{2}

Series Equivalent Capacitance

  {\dfrac{1}{C_{eq, series}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3}}

Parallel Equivalent Capacitance

{C_{eq, parallel} = C_1 + C_2 + C_3}

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