chapter 5 & 6 dr. farid farahmand cet 236. outline identify a series/parallel circuit determine...

15
Chapter 5 & 6 Dr. Farid Farahmand CET 236

Post on 22-Dec-2015

220 views

Category:

Documents


1 download

TRANSCRIPT

Chapter 5 & 6Dr. Farid Farahmand

CET 236

Outline

Identify a series/parallel circuitDetermine the current and voltage in a

circuitDetermine total resistanceApply Ohm’s lawApply Kirchhoff’s voltage law

Series Circuit

A series circuit provides only one path for current between two points current is the same through each

series resistor. Find the total resistance in a series

circuit sum of the resistances of each

individual series resistor.

RT = R1 + R2 + R3 + . . . + Rn

Ohm’s Law in Series Circuits

If you know the total current, you can find the voltage drop across any of the series resistors by using: VR = ITR

The polarity of a voltage drop across a resistor is positive at the end of the resistor that is closest to the positive terminal of the voltage source

An open in a series circuit prevents current; and, there is zero voltage drop across each series resistor. The total voltage appears across the points between which there is

an open

+

-

Voltage Sources in Series

When two or more voltage sources are in series, the total voltage is equal to the the algebraic sum (including polarities of the sources) of the individual source voltages.

Kirchhoff’s Voltage Law (KVL)

The sum of all the voltage drops around a single closed loop in a circuit is equal to the total source voltage in that loop.

VS = V1 + V2 + V3 + … + Vn

Or

VS - V1 - V2 - V3 = 0

Since each resistor has the same current, the voltage drops are proportional to the resistance values.

Vx = (Rx/RT)VS

Power in a Resistive Circuit

The total amount of power in a series/parallel resistive circuit is equal to the sum of the powers in each resistor in series.

PT = P1 + P2 + P3 + . . . + Pn

Remember

Px = Vx.Ix

The amount of power in a resistor is important the power rating of the resistor must be high

enough to handle the expected power in the circuit.

Open and Short Circuit

When an open occurs in a series circuit, all of the source voltage appears across the open. The most common failure

in a series circuit is an open.

When a short occurs a portion of the series resistance is bypassed, thus reducing the total resistance. A short in a series circuit

results in more current than normal.

What is the current when there is no short?

What is the current when the circuit is shorted?

Open and Short Circuit

When an open occurs in a series circuit, all of the source voltage appears across the open. The most common failure

in a series circuit is an open.

When a short occurs a portion of the series resistance is bypassed, thus reducing the total resistance. A short in a series circuit

results in more current than normal.

Let’s Examine Parallel Circuits!

Resistors in Parallel

Each current path is called a branch. A parallel circuit is one that has more than one branch.

The voltage across any given branch of a parallel circuit is equal to the voltage across each of the other branches in parallel

Two Branches

Kirchhoff’s Current Law (KCL)

Two ways of stating it: The sum of the currents into a

junction (total current in) is equal to the sum of the currents out of that junction (total current out).

The algebraic sum of all the currents entering and leaving a junction is equal to zero.

IIN(1) + IIN(2) + . . . + IIN(n) = IOUT(1) + IOUT(2) + . . . +IOUT(m)

Total Parallel Resistance

When resistors are connected in parallel, the total resistance of the circuit decreases.

The total resistance of a parallel circuit is always less than the value of the smallest resistor.

1/RT = 1/R1 + 1/R2 + 1/R3 + . . . + 1/Rnor

RT = R1||R2||R3||R4||R5….Rn

What is the total resistance in a parallel circuit with 2 resistors?

RT = R1R2/(R1 + R2)

Current in Parallel Circuits

The total current produced by all current sources is equal to the algebraic sum of the individual current sources.

A parallel circuit acts as a current divider because the current entering the junction of parallel branches “divides” up into several individual branch currents.

I1=Vs/R1 = IT.RT/R1

I1 = (R2/(R1 + R2))IT

I2 = (R1/(R1 + R2))IT

Given IT, find I1 and I2!

Circuit Example: How does it work?