Electrical Systems

Electrical systems involve the flow of electric charge through circuits of conducting material, usually a wire or metal cable. If there is a difference in voltage across a conductor, electrons will flow from the high voltage to low voltage creating a current. Current can be used to transfer energy between two points along a wire; this is the basis of all electronic devices. Unlike electrostatics which is mainly concerned with the activity of individual charges, electrical systems involve the movement of many electrons and the combined effects of the moving charges.

**Voltage Source: **A power source that provides a fixed voltage, usually a battery. While the voltage source has a certain amount of voltage available (also known as **emf**, or electromotive force), it tends to have some resistance, meaning the actual voltage (called the terminal voltage) it provides is less than the emf. SI unit: V

Voltage causes currents - think of voltage as how much force is pushing an electric current!

**Voltage: **Potential energy, measured as a difference between potential at two points in space. SI units: V

**Voltmeter: **Measures the potential difference (voltage) between two points in a circuit.

**Current: **The flow of electric charges through a wire. SI units: A

**Ammeter: **Measures current.

**Resistance: **The amount a device opposes the flow of a current. SI units: Ω

**Capacitance: **A device’s capacity to store charge. SI units: F

**Ohm’s Law: **The current between two points through a conductor is directly proportional to the voltage between those two points and inversely proportional to the resistance between those two points.

**Power: **The concept of power in electrical systems is exactly the same as in mechanics: the rate of energy transfer. However, power in electrical systems is calculated with different formulas. SI units: J

**JDC Circuits: **Also known as direct current circuits, electrons in DC circuits are constantly moving along a wire towards the positive charge.

**RC Circuits: **Circuit comprised of resistors and capacitors connected to a voltage source.

Can you guess what the R and C in RC circuit stand for now?

Circuits include a** voltage source** (usually a battery) and a conducting wire connecting opposite ends of the voltage source, providing a closed loop for charge to flow (**DC circuits**). Some other elements that may be included in a circuit are resistors and capacitors.

In circuit diagrams, we use these symbols:

When working with circuit diagrams, we are usually calculating one of the following:

**voltage****current****resistance****capacitance**(if capacitors are included)

**Ohm’s law** relates current (I), voltage (V), and resistance (R): V = IR.

Two other important rules are Kirchoff’s laws:

*Kirchoff’s Law for Voltage*(also known as the Loop Rule): The sum of all the potential differences in a given loop is equal to 0.*Kirchoff’s Law for Current*(also known as the Junction Rule): At any junction, the sum of the current flowing in equals the sum of the current flowing out.

Resistors are devices that resist the flow of current and are used to control the current flowing through a circuit. The **power **that a battery provides can be calculated by P = IV. This power flows through the circuit until it hits a resistor, which dissipates some energy. This energy goes into heating the resistor.

Resistance is low in a conductor and high in an insulator.

Electrical Systems cont.

The current passes through each resistor and the total voltage drop across the resistors is equal to their sum. The total resistance is equal to the sum of the resistance of each resistor.

\(V_{total} = V₁ + V₂ + V₃ + ...\)

\(R_{total} = R₁ + R₂ + R₃ + ...\)

If the current splits up (and can’t go through both resistors), then the resistors are in parallel.

The current is split up between the resistors, so the current is equal to the sum of the current through each resistor. The voltage drop across each resistor is the same. The total resistance is equal to the sum of the reciprocal of each resistor’s resistance.

\(I_{total} = I₁ + I₂ + I₃ + ...\)

\(V_{total} = V₁ = V₂ = V₃ = ...\)

\(\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...\)

Capacitors are devices that store energy. When the switch is closed, the voltage source forces electrons from one plate to the other, creating equal and opposite charges on the two plates. This creates a potential difference that increases until the potential difference is equal to the emf of the voltage source. At this point, the voltage source can no longer add charges to the plates. Thus, the flow of current stops and it is as if the switch had been opened again!

A large capacitance means more charge can be stored. Capacitance depends on the area of the plates, the distance between plates, and the dielectric constant (varies for different materials between the plates).

The total capacitance is equal to the sum of the reciprocal of each capacitor’s capacitance.

\(\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ...\)

The total capacitance is equal to the sum of the capacitance of each capacitor.

\(C_{total} = C_1 + C_2 + C_3 + ...\)

Transistors are semiconductors that are used to amplify and switch electrical signals. They have three or more terminals that can connect to outside circuits. When a voltage or current passes through one terminal, the current flowing through the other terminals changes. Oftentimes, the output power is significantly greater than the input making transistors very effective signal amplifiers.

An** RC circuit **is a circuit comprised of resistors, capacitors, and a voltage source. Initially, the capacitor is uncharged and the potential difference across the resistor is the emf. When the switch is closed, the current and the potential difference across the resistor start to decrease as charge accumulates on the capacitor.

At time = 0, the capacitor is uncharged so it acts like a wire.

\(I = \frac{emf}{r}\) I - initial/maximum current

As time \(\rightarrow \alpha\), the capacitor is fully charged and there is no current running through the capacitor, making the system resemble an open circuit.

Electrical System Problem Guide

\(V = IR\)

V - voltage

I - current

R - resistance

\(P = IV = I^2R = \frac{V^2}{R}\)

P - power

\(C = \frac{Q}{V}\)

C - capacitance

Q - charge

\(C = \varepsilon_0 \frac{A}{d}\)

\(ε_0\) - dielectric constant

A - area of plate

d - distance between plates

\(E= \frac{1}{2}QV = \frac{1}{2}CV^2 = Pt\)

E - energy

\(I = \frac{\Delta Q}{\Delta t}\)

The general strategy for any circuit problem is to first find the total equivalent resistance or capacitance, depending on what kind of circuit you are dealing with. Using that, we can find out whatever information on the circuit as a whole that has been left out. Then, work your way back down using Ohm’s law in in order to find the characteristics of smaller sections of the circuit until you get to your answer. When finding total resistance or capacitance, it is usually easier to work starting with the most buried components. The problem below will illustrate this concept.

In the circuit below, find the voltage drop and the power dissipated by the resistor labeled R₂.

Before we can even think about the individual resistors, we need to find out how much current is running through the whole circuit. So, to do that, we will first need to find the total resistance. We’ll start by summing the resistance of resistors 3 and 4.

\(R_{3,4} = R_3 + R_4 = 2\Omega + 1\Omega = 3\Omega\)

Next we’ll combine \(R_{3,4}\) with R₂.

\(\frac{1}{R_{2,3,4}} = \frac{1}{R_2} + \frac{1}{R_{2,3}} = \frac{1}{2\Omega} + \frac{1}{3\Omega}\)

\(\frac{1}{R_{2,3,4}} = \frac{5}{6}\Omega \rightarrow R_{2,3,4} = \frac{6}{5}\Omega\)

Finally we can find \(R_t\) by combining \(R_{2,3,4}\) with \(R_1\).

\(R_t = R_1 + R_{2,3,4} = 5\Omega + \frac{6}{5}\Omega = 6.2\Omega\)

Note that the three steps above can be combined into one expression: \(R_t = R_1 + \frac{1}{\frac{1}{R_2} + \frac{1}{R_3 + R_4}}\)

Now, we can find the amount of current that runs through the whole circuit using Ohm’s Law.

\(V= IR_t \rightarrow I = \frac{V}{R_t} = \frac{12v}{6.2\Omega} = 1.93A\)

Electrical Systems cont.

Now, since \(R_1\) and the group \(R_{2,3,5}\) are in series, we know that the same amount of current is going through both of these resistors. So, we can use Ohm’s Law to find the voltage drop across \(R_{2,3,4}\).

\(V_{2,3,4} = IR_{2,3,4} = 1.93A . 1.2 \Omega = 2.31V\)

Since, R₂ and the group \(R_{3,4}\) are in parallel, we know that the voltage applied across them is equal. So, know the answer to the first part of the problem is \(V_2 = 2.31 V\). We can use this information to find the current through the resistor and then the power it dissipates.

\(V_2 = I_2 R_2 \rightarrow I_2 = \frac{V_2}{R_2} = \frac{2.31V}{2\Omega} = 1.15A\)

Now we can find the power the resistor dissipates.

\(P = I^2_2 R_2 = 1.15^2 A . 2\Omega = 2.64W\)