CHAPTER 18 Direct Current (DC) Circuits Symbols: Resistor Battery (long line is positive side) Flow of conventional (positive) current Open Switch Closed Switch Capacitor Voltmeter Ammeter I V A

Resistors in Series The same amount of current flows thru each resistor. R T = R eq = R 1 + R 2 when resistors are in series Circuit Analysis Find the voltage at every point (a, b, and c) in Fig 18.2 if the battery is 12V and R 1 and R 2 are 2.0 and 4.0 ohm respectively.Strategy Usually: 1. Find total resistance (R T ) 2. Use total resistance to find total current (I T ) 3. Use I T and individual resistance to calculate various V’s.

1.Find total resistance (R T ) 2. Use total resistance to find total current (I T ) 3. Use I T and individual resistance to calculate various V’s. R T = 2.0   = 6.0  (R T = R 1 + R 2 ) I = 12 Volt/6.0  = 2.0 Amp (A)(V = I R) V at battery discharge (+ end) = 12 V V across resistors = IR V 1 = 2.0 A x 2.0  = 4.0 V V 2 = 2.0 A x 4.0  = 8.0 V V a = 12 V V c = 0 V (8V-8V) V b = 8 V (12V-4V)

Series Circuit R T = R 1 + R 2 I T = I 1 = I 2 V = I R R = VIVI VTITVTIT V1I1V1I1 V2I2V2I2 =+ V T = V 1 + V 2 VTITVTIT V1ITV1IT V2ITV2IT =+

I T = I 1 + I 2 V = I R I = VRVR VTRTVTRT V1R1V1R1 V2R2V2R2 =+ V T = V 1 = V 2 Resistors in Parallel Total current splits and flows partially thru each resistor before recombining. = + 1R21R2 1R11R1 1RT1RT VTRTVTRT VTR1VTR1 VTR2VTR2 =+

Circuit Analysis Find the current flowing through each resistor in Fig 18.4 if the battery is 12V and R 1 and R 2 are 2.0 and 4.0ohm respectively. 1RT1RT 1R21R2 1R11R1 =  1 4.0 =+ 1RT1RT =  =.75  -1 R T = 1.3  V 1 = 12 Volts = I 1 x R 1 12 V = I 1 x 2.0 I 1 = 6.0 A V 2 = 12 Volts = I 2 x R 2 12 V = I 2 x 4.0 I 2 = 3.0 A V = I R I T = I 1 + I 2 = 9.0A V T = I T x R T V T = 9.0A x 1.3 V T = 12V

Combination Circuits Reading 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 Examples Calculate Total Resistance P18.6 P18.8 P18.7 Trace Current Flow 18, 9.0, 6.0 parallel R C = 3.0 Combo 3.0 in series with 12 R T = 12 + 3.0 = 15 I T = 2.0A Trace current flow a to b 2 resistors in parallel Horizontal R,R, and R C are in series R C =.5R R T = 2.5R Vertical R is immaterial Trace current flow Two 5.0 in series R C1 = 10.0 R C1 in parallel with vertical 5.0 R C2 = 3.3 Horizontal 5.0, R C2 and horizontal 1.5 are in series R T = 9.8

Test Yourself Working from top right of circuit: R P1 = 3.0  R S1 = R P2 = R S2 = R PC = R T = 6.0  3.0  2.7  5.0  5.7 

Quiz Yourself R P1 = 3.3  R S1 = 7.3  R P2 = 2.1  R T = 5.1 

Real 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 resistance Voltage drop within the battery is Ir V = Terminal Voltage (“net voltage”) =  - Ir Terminal Voltage is measured at the terminals of the battery with current flowing.

Example Problem Example 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. Strategy Find the total resistance and then the total current. R T = 3.9 I T = 3.1A Trace the circuit starting with the positive side of the battery and determine V at exit of each resistor (V-V) V B = 12.0V V 2 = 5.8V ( x3.1) V 4 = V 6 = V 10 = 0 Volts all directly connected to `negative terminal of battery

I 4 = 5.8V/4.0 = 1.5 A I 6 = 5.8V/6.0 = 1.0 A I 10 = 5.8V/10.0 =.6 A I T = 3.1 A With all current and resistance known, calculate power dissipation. P T = (3.1A) 2 (3.9) = 37.5 Watts P 4 = (I 4 ) 2 R 4 = 9.0 Watts P 6 = 6.0 Watts P 10 = 3.6 Watts P 2 = 19.2 Watts P T = 37.8 Watts