PHY 102: Lecture Voltage 5.2 Current 5.3 Resistance 5.4 Ohm’s Law 5.5 Electric Power 5.6 Series Circuits 5.7 Parallel Circuits 5.8 Combined Series/Parallel Circuits 5.9 Capacitors in Series / Parallel

PHY 102: Lecture 5 Electric Circuits 5.1 Voltage

Electromotive Force In a battery, a chemical reaction occurs that transfers electrons from one terminal (leaving it positively charged) to another terminal (leaving it negatively charged) Because of the positive and negative charges on the battery terminals, an electric potential difference exists between them The maximum potential difference is called the electromotive force (emf) of the battery The symbol for electromotive force is E In a typical car battery, the chemical reaction maintains the potential of the positive terminal at a maximum of 12 volts higher than the potential of the negative terminal

Battery Illustrations

PHY 102: Lecture 5 Electric Circuits 5.2 Current

Electron Current Battery creates an electric field within and parallel to wire The electric field is directed from the positive toward the negative terminal The electric field exerts a force on the free electrons in the wire, and they respond by moving Charges move inside a wire and cross an imaginary surface that is perpendicular to their motion This flow of charge is known as a electric current The symbol for electric current is I The electric current is defined as the amount of charge per unit time that crosses the imaginary surface If the rate of flow is constant, I =  q/  t SI unit of current: ampere = 1 Coulomb / s

Current Illustration

Direct / Alternating Current If the charges move around a circuit in the same direction at all times, the current is said to be direct current (dc) The current is said to be alternating current (ac) when the charges move first one way and then the opposite way, change direction from moment to moment

Problem 1 The battery pack of a pocket calculator has a voltage of 3.0 V and delivers a current of 0.17 mA ( A) In one hour of operation (a) how much charge flows in the circuit (b) how much energy does the battery deliver to the calculator circuit?  (a) Charge that flows in one hour   q = I(  T) = (0.17 x A)(3600 s) = 0.61 C  (b) Energy delivered in one hour  E=charge x (energy/charge)=(0.61 C)(3.0V) = 1.8 J

Conventional Current Electrons flow in metal wires It is customary not to use the flow of electrons when discussing circuits Instead, a so-called conventional current is used Conventional current is the hypothetical flow of positive charges that would have the same effect in the circuit as the movement of negative charges

Conventional Current Illustration

PHY 102: Lecture 5 Electric Circuits 5.3 Resistance

Resistance and Resistivity R =  (L/A )   is the resistivity of the material   =  0 [1 +  (T – T 0 )]   is temperature coefficient of resistivity R = R 0 [1 +  (T – T 0 )] There is a class of materials whose resistivity suddenly goes to zero below a certain temperature Tc, which is called the critical temperature

Resistivity of Various Materials

Problem The instructions for an electric lawn mower suggest that a 20-gauge extension cord can be used for distances up to 35 m A thicker 16-gauge cord should be used for longer distances to keep the resistance of the wire as small as possible Cross-sectional area of 20-gauge wire is 5.2 x m 2 Cross-sectional area of 16-gauge wire is 13 x m 2 Determine the resistance of (a) 35 m of 20-gauge copper wire and (b) 75 m of 16-gauge copper wire

Problem Resistivity of copper is 1.72 x 10 -8

Problem A heating element on an electric stove contains a wire (length 1.1 m, cross-sectional area = 3. x m 2 ) This wire is embedded within an electrically insulating material that is contained within a metal casing The wire becomes hot in response to the flowing charge and heats the casing The material of the wire has a resistivity of  0 =6.5x10 -5 at T 0 = C The temperature coefficient of resistivity of  =2.0x10 -3 Determine the resistance of the heater wire at an operating temperature of T = C

Problem 4 - 2

PHY 102: Lecture 5 Electric Circuits 5.4 Ohm’s Law

Georg Ohm (1787 – 1854) 1827 Publishes Ohm’s Law German Work was ignored by his colleagues He lived in poverty for much of his life –“sole effort is to detract from the dignity of nature” book critic –“a professor who preached such heresies was unworthy to teach science” German Minister of Education 1841 work recognized by Royal Society 1852 appointed to the chair of physics at University of Munich

Ohm’s Law V = IR I = V/R R = V/I V is the voltage applied across a piece of material I is the current through the material R is the resistant of the piece of material SI Unit of Resistance: ohm (  )

Elements of Ohm’s Law

Problem 2 The filament in a light bulb is a resistor in the form of a thin piece of wire The wire becomes hot enough to emit light Flashlight uses two 1.5-V batteries (3.0 V) The current is 0.40 A in the filament Determine resistance of glowing filament  R = V/I = 3.0 V / 0.40 A = 7.5 

PHY 102: Lecture 5 Electric Circuits 5.5 Electric Power

One of the most important functions of the current in an electric circuit is to transfer energy from a source to an electrical device When an amount of positive charge  q moves from the higher potential (+) to the lower potential (-), its electric potential energy decreases This decrease is (  q)V P = Change in energy / Time interval = (  q)V/  t = IV P = IV P = V 2 / R P = I 2 R SI Unit of Power: watt

Problem 5 In a flashlight, the current is 0.40 A The voltage is 3.0 V Find (a) the power delivered to the bulb Find (b) the electrical energy dissipated in the bulb in 5.5 minutes of operation Power is P = IV = (0.40 A)(3.0 V) = 1.2 W Energy is E = P  t = (1.2 W)(330 s) = 400 J

PHY 102: Lecture 5 Electric Circuits 5.6 Series Circuits

Series Wiring Series wiring means that the devices are connected in such a way that there is the same electric current through each device The equivalent resistance is:  R S = R 1 + R 2 + R 3 + … The voltage across all the resistors in series is the sum of the individual voltages across each resistor The total power delivered to any number of resistors in series is equal to the power delivered to the equivalent resistance

Problem R 1 = 47 ohms, R 2 = 86 ohms, V = 24 V (a) Find equivalent resistance (b) Find total current (c) Find total power

Problem R 1 = 47 ohms, R 2 = 86 ohms, V = 24 V (d) Find V 1 and V 2 (e) Find power dissipated by R 1 (f) Find power dissipated by R 2

Problem (a) Equivalent (total) resistance  R s = R 1 +R 2 = = 133 ohms Ohm’s Law applies to the entire circuit and each part of a circuit (b) Total current (use Ohm’s Law)  I = V / R s = 24 / 133 = 0.18 A (c) Total power (use a power formula)  P = IV = 0.18 x 24 = 4.32 W

Problem (d) Find V 1 and V 2 (use Ohm’s Law)  V 1 = IR 1 = 0.18 x 47 = 8.46 volts  V 2 = IR 2 = 0.18 x 86 = volts  Check = volts (e) Power dissipated by R1  P 1 = IV 1 = 0.18 x 8.46 = 1.52 watts  P 2 = IV 2 = 0.18 x = 2.79 watts  Check = 4.31 watts

PHY 102: Lecture 5 Electric Circuits 5.7 Parallel Circuits

Parallel Wiring Parallel wiring means that the devices are connected in such a way that the same voltage is applied across each device The equivalent resistance is:  1/R P = 1/R 1 + 1/R 2 + 1/R 3 + … The total current from the voltage source is the sum of the currents in the individual resistors The total power delivered to any number of resistors in parallel is equal to the power delivered to the equivalent resistor

Problem R 1 = 8 ohms, R 2 = 4 ohms, V = 6 V (a) Find equivalent resistance (b) Find total current (c) Find total power

Problem R 1 = 8 ohms, R 2 = 4 ohms, V = 6 V (d) Find I1 and I2 (e) Find power dissipated by R 1 (f) Find power dissipated by R 2

Problem (a) Equivalent (total) resistance  1/R P = 1/R 1 + 1/R 2  1/R P = 1/8 + ¼ = =  R p = 1 / = 2.67 ohms (b) Total current (use Ohm’s Law)  I = V / R P = 6 / 2.67 = 2.25 A (c) Total power (use a power formula)  P = IV = 2.25 x 6 = 13.5 W

Problem (d) Find I 1 and I 2 (use Ohm’s Law)  I 1 = V/R 1 = 6 / 8 = 0.75 amps  I 2 = V/R 2 = 6 / 4 = 1.50 amps  Check = 2.25 volts (e) Power dissipated by R1  P 1 = IV 1 = 0.75 x 6 = 4.5 watts  P 2 = IV 2 = 1.50 x 6 = 9.0 watts  Check = 13.5 watts

PHY 102: Lecture 5 Electric Circuits 5.7 Combined Series/Parallel Circuits

Problem 8 - 1

Problem Sum the two series resistors 220 ohms and 250 ohms to get 470 ohms Sum the two parallel resistors 180 ohms and 470 ohms to get 130 ohms This is R AB Sum the two series resistors 110 ohms and 130 ohms to get 240 ohms R E = 240 ohms

Problem (a) Find the total current  I = V / R E = 24 / 240 = 0.10 amps (b) Find the current through R 110  R 110 = total current = 0.10 amps (c) Find the voltage across R 110  V 110 = IR 110 = 0.10 x 110 = 11 volts (d) Find the voltage across R AB  V AB = IR AB = 0.10 x 130 = 13 volts  Check = 24 volts

Problem (e) Find the current through R 180  I 180 = V 180 / R 180 = 13 / 180 = amps (f) Find the current through R 470  I 470 = V 470 / R 470 = 13 / 470 = amps  Check = 0.10 amps

PHY 102: Lecture 5 Electric Circuits 5.8 Capacitors in Serial / Parallel

Parallel Capacitors C P = C 1 + C 2 + C 3 + … Total energy = ½ C P V 2

Series Capacitors All capacitors in series contain charges of the same magnitude on their plates 1/C S = 1/C 1 + 1/C 2 + 1/C 3 + … Total energy = ½ C S V 2