CHAPTER 18 Direct Current (DC) Circuits Symbols: Resistor Battery (long line is positive side) Flow of conventional (positive) current Open Switch Closed.

CHAPTER 18CHAPTER 18Direct Current (DC) CircuitsDirect Current (DC) Circuits

Symbols:Symbols:

Resistor

Battery (long line is positive side)

Flow of conventional (positive) current

Open Switch

Closed Switch

Capacitor

VoltmeterAmmeter

I

V

A

Resistors in SeriesResistors in SeriesThe same amount of current flows thru each resistor.The same amount of current flows thru each resistor.

RT = Req = R1 + R2 when resistors are in seriesCircuit AnalysisCircuit AnalysisFind the voltage at every point (a, b, and c) in Fig 18.2 if the battery is 12V and R1 and R2 are 2.0 and 4.0 ohm respectively.StrategyStrategyUsually: 1. Find total resistance (RT)

2. Use total resistance to find total current (IT) 3. Use IT and individual resistance to calculate

various V’s.

1.Find total resistance (RT)2. Use total resistance to find total current (IT)3. Use IT and individual resistance to calculate

various V’s.RT = 2.0 + 4.0 = 6.0 (RT = R1 + R2)I = 12 Volt/6.0 = 2.0 Amp (A) (V = I R)V at battery discharge (+ end) = 12 VV across resistors = IRV1 = 2.0 A x 2.0 = 4.0 VV2 = 2.0 A x 4.0 = 8.0 VVVaa = 12 V = 12 V VVcc = 0 V (8V-8V) = 0 V (8V-8V)VVbb = 8 V (12V-4V) = 8 V (12V-4V)

Series CircuitSeries Circuit

RT = R1 + R2

IT = I1 = I2

V = I R R = VI

VT

IT

V1

I1

V2

I2= +

VT = V1 + V2

VT

IT

V1

IT

V2

IT= +

IT = I1 + I2

V = I R I = VR

VT

RT

V1

R1

V2

R2

= +

VT = V1 = V2

Resistors in ParallelResistors in ParallelTotal current splits and flows partially thru Total current splits and flows partially thru

each resistor before recombining.each resistor before recombining.

= + 1R2

1R1

1RT

VT

RT

VT

R1

VT

R2

= +

Circuit AnalysisCircuit AnalysisFind the current flowing through each resistor in Fig 18.4 if the battery is 12V and R1 and R2 are 2.0 and 4.0ohm respectively.

1RT

1R2

1R1

= + 12.0

14.0

= +

1RT

= 3.04.0

= .75 -1

RT = 1.3 V1 = 12 Volts = I1 x R1

12 V = I1 x 2.0I1 = 6.0 A

V2 = 12 Volts = I2 x R2

12 V = I2 x 4.0I2 = 3.0 A

V = I R

IT = I1 + I2 = 9.0A

VT = IT x RT

VT = 9.0A x 1.3VT = 12V

Combination CircuitsCombination Circuits

Reading the Circuit DiagramReading the Circuit Diagram

The 4.0 resistor is in series with the 8.0 resistor in front of it because all of the current that passes through the 8.0 resistor must also pass through the 4.0 resistor.

The 6.0 resistor is not in series with the 4.0 resistor because all the current passing through the 4.0 resistor does not pass through the 6.0 resistor. Some of it passes through the 3.0 resistor.

The 6.0 resistor and the 3.0 resistor are in parallel because all the current entering point b passes through point c.

Combination Circuit ExamplesCombination Circuit ExamplesCalculate Total Resistance

P18.6

P18.8

P18.7

• Trace Current Flow• 18, 9.0, 6.0 parallel RC = 3.0• Combo 3.0 in series with 12• RT = 12 + 3.0 = 15 IT = 2.0A

• Trace current flow a to b• 2 resistors in parallel• Horizontal R,R, and RC are in series

RC = .5R

RT = 2.5R

• Vertical R is immaterial

• Trace current flow• Two 5.0 in series RC1 = 10.0• RC1 in parallel with vertical 5.0

RC2 = 3.3• Horizontal 5.0, RC2 and horizontal 1.5 are in series

RT = 9.8

Test YourselfTest Yourself

Working from top right of circuit:

RP1 = 3.0

RS1 =

RP2 =

RS2 =

RPC =

RT =

6.0

3.0

2.7

5.0

5.7

Quiz YourselfQuiz Yourself

RP1 = 3.3

RS1 = 7.3

RP2 = 2.1

RT = 5.1

Real BatteriesReal Batteries

= emf = electromotive force (not a true force) = 12 volts in a 12Volt battery = “gross voltage”All batteries have internal resistance (especially as they grow old)r = internal resistanceVoltage drop within the battery is IrV = Terminal Voltage (“net voltage”) = - IrTerminal Voltage is measured at the terminals of the battery with current flowing.

Example ProblemExample Problem (Circuit Analysis)Find a) the power dissipated across each resistor, b) the current through each resistor, and c) the voltage between all the resistors.StrategyStrategyFind the total resistance and then the total current.RT = 3.9IT = 3.1ATrace the circuit starting with the positive side of the battery and determine V at exit of each resistor (V-V)VB = 12.0VV2 = 5.8V (12.0-2.0x3.1)V4 = V6 = V10 = 0 Volts all directly connected to

` negative terminal of battery

I4 = 5.8V/4.0 = 1.5 AI6 = 5.8V/6.0 = 1.0 AI10 = 5.8V/10.0 = .6 AIT = 3.1 A

With all current and resistance known, calculate power dissipation.

PT = (3.1A)2 (3.9) = 37.5 Watts

P4 = (I4)2 R4= 9.0 WattsP6 = 6.0 WattsP10 = 3.6 WattsP2 = 19.2 WattsPT = 37.8 Watts