CHAPTER 18 Direct Current (DC) Circuits Symbols: Resistor Battery (long line is positive side) Flow...
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Transcript of CHAPTER 18 Direct Current (DC) Circuits Symbols: Resistor Battery (long line is positive side) Flow...
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