Chapter 31: Current & Resistance & Simple Circuits

Chapter 31: Current & Resistance & Simple Circuits

Simple Single Loop Circuits J E V = IR

What makes the wires Glow?

Electrical Power As the charge moves through the resistor the system loses this electric potential energy during collisions of the electrons with the atoms of the resistor This energy is transformed into internal energy in the resistor as increased vibrational motion of the atoms in the resistor The resistor emits thermal radiation which can make it glow. U q V du dq V dq P V I V dt dt dt P I V

POWER P P I V Energy J time s Watt P I V I( IR) I R 2 V P I V V R V R 2 You pay for ENERGY not for ELECTRONS! Kilowatt-hour is the energy consumed in one hour: [kwh]=j NOT TIME! Power x Time

QUESTION The voltage and power on a light bulb read 120 V, 60 W How much current will flow through the bulb? USE: P = I V I = P/ V = 60 W/120 V = 1/2 Amp

QUESTION The power and voltage on a light bulb read 120 V, 60 W What is the resistance of the filament? (I =.5 A) Hint: USE OHMS LAW: V = IR R = V/I = 120 V/.5 A = 240

QUESTION The power rating for two light bulbs read 30W and 60W. Which bulb has the greatest resistance at 120V? 2 2 P V / R R V / P R V W 2 (120 ) /30 480 R V W 2 (120 ) / 60 240 RULE: THE MORE POWER DISSIPATED IN A BULB, THE BRIGHTER IT IS, the LESS Resistance and greater current! P= IV!

Quick Quiz For the two lightbulbs shown in this figure, WHICH BULB BURNS BRIGTHEST? rank the current values at the points, from greatest to least. I a = I b > I c = I d > I e = I f. The 60 W bulb has the lowest resistance and therefore draws the most current and burns brightest!

Series Circuits The current is the same in each device. The equivalent resistance of the circuit is the sum of the individual resistances. Parallel Circuits The voltage of each device is the full voltage of the EMF source (the battery) The total current is divided between each path: 1 1 1 R R R total 1 2 R R R total 1 2

Series Circuits The current is the same in each device. The equivalent resistance of the circuit is the sum of the individual resistances. R=R 1 +R 2 V V + V, I I I 1 2 1 2 V total IR total I( R + R ) IR + IR V + V 1 2 1 2 1 2 R R R S 1 2

Series Circuits The current is the same in each device. The equivalent resistance of the circuit is the sum of the individual resistances. R=R 1 +R 2 V V + V, R R R R... 1 2 S 1 2 3 To find the current, use the total voltage and equivalent resistance: I V R S

Circuits Problem:3 Bulbs in Series If one more bulb is added to each circuit (3 bulbs total), how does the brightness of the bulbs change? Or not? In the series circuit, the bulbs DIM. WHY? V V1 V2 V3 In series, each of the three equal bulbs gets one third of the Voltage (V/3) that a single bulb would get. P 2 2 V /3 1 V P R R 9 R 9 V 2 singlebulb Note: P=VI but I is due to the equivalent Resistance: I = V/R s =V/3R So the Current through each is 1/3 the current through a single bulb and P=VI=V/3 x I/3 = VI/9 = P/9. The bulbs burn 1/9 as bright!

Parallel Circuits The voltage of each device is the full voltage of the EMF source (the battery) The total current is divided between each path: V V 1 1 V I I1 I2 V ( ) R R R R R Equivalent Resistance: 1 2 1 2 1 1 1 1... R R R R P 1 2 3 P

Circuits Problem:3 Bulbs in Parallel If one more bulb is added to each circuit (3 bulbs total), how does the brightness of the bulbs change? Or not? In the parallel circuit, the bulbs DO NOT DIM. WHY? In parallel, each of the three equal bulbs gets the full voltage of the battery source. P V R 2 P singlebulb Is this getting something for nothing? NO! Parallel circuits drain the battery faster!

Parallel Circuits As the number of branches is increased, the overall resistance of the circuit is DECREASED. Overall resistance is lowered with each added path between any two points of the circuit. This means the overall resistance of the circuit is less than the resistance of any one of the branches!!!! (Weird?) As overall resistance is lowered, more current is drawn. This is how you blow fuses!

Circuits Problem:Bulbs in Series vs Parallel A circuit contains a 48-V battery and two 240 light bulbs. In which circuit does each bulb burn brighter? RULE: THE MORE POWER DISSIPATED IN A BULB, THE BRIGHTER IT IS. P IV Find the power in each bulb when in series and in parallel.

Circuits Problem:Bulbs in Series vs Parallel In Series: The Voltage is divided in series: each bulb gets half: V = 24V. 2 2 V V (24 V ) P IV V 2.4W R R 240 In Parallel: Voltage is the same in each bulb: 48V. P 2 2 V (48 V) R 240 9.6W Parallel Bulbs Burn Brighter!

Circuits Problem:Bulbs in Series vs Parallel If a bulb burns out - what happens to the other bulb in each circuit? Does it go out? Is it brighter? Dimmer? Or? In the series circuit, the burned out bulb will short the circuit and the other bulb will go out. In the parallel circuit the other bulb will have the same brightness.

QuickCheck 31.13 The three bulbs are identical and the two batteries are identical. Compare the brightnesses of the bulbs. A. A B C. B. A C B. C. A B C. D. A B C. E. A B C. Slide 31-84

QuickCheck 31.13 The three bulbs are identical and the two batteries are identical. Compare the brightnesses of the bulbs. A. A B C. B. A C B. C. A B C. D. A B C. E. A B C. Slide 31-85

QuickCheck 31.14 The three bulbs are identical and the two batteries are identical. Compare the brightnesses of the bulbs. A. A B C. B. A C B. C. A B C. D. A B C. E. A B C. Slide 31-86

QuickCheck 31.14 The three bulbs are identical and the two batteries are identical. Compare the brightnesses of the bulbs. A. A B C. B. A C B. C. A B C. D. A B C. E. A B C. Slide 31-87

Find Everything V 15.0V, R 10.0, R 20.0, R 30.0, 1 2 2

Single Loops: When a circuit can be reduced to one equivalent Resister. Reduce and Use Ohm s Law For the circuit shown, find the equivalent resistance of the circuit, the total current drawn by the battery, and the current through and the potential difference across each resistor. Place your results in a table for ease of reading. 6.0

Multi-Loop Circuits When resistors are connected so that the circuits formed cannot be reduced to a single equivalent resistor you can use two rules, called Kirchhoff s rules, to solve the problem by generating systems of equations to find the unknowns!!

1824 1887 German physicist Worked with Robert Bunsen They Gustav Kirchhoff Invented the spectroscope and founded the science of spectroscopy Discovered the elements cesium and rubidium Invented astronomical spectroscopy

Kirchhoff s Rules for Spectra: 1859 German physicist who developed the spectroscope and the science of emission spectroscopy with Bunsen. Kirkoff Bunsen * Rule 1 : A hot and opaque solid, liquid or highly compressed gas emits a continuous spectrum. * Rule 2 : A hot, transparent gas produces an emission spectrum with bright lines. * Rule 3 : If a continuous spectrum passes through a gas at a lower temperature, the transparent cooler gas generates dark absorption lines.

Kirchhoff s Rules Equations In order to solve a particular circuit problem, the number of independent equations you need to obtain from the two rules equals the number of unknown currents Any capacitor acts as an open branch in a circuit The current in the branch containing the capacitor is zero under steady-state conditions

Kirchhoff s Junction Rule Junction Rule The sum of the currents at any junction must equal zero Currents directed into the junction are entered into the -equation as +I and those leaving as -I A statement of Conservation of Charge Mathematically, junction I 0

Kirchhoff s Loop Rule Loop Rule The sum of the potential differences across all elements around any closed circuit loop must be zero A statement of Conservation of Energy Mathematically, V closed loop 0

Using the Loop Rule Traveling around the loop from a to b In (a), the resistor is traversed in the direction of the current, the potential across the resistor is IR In (b), the resistor is traversed in the direction opposite of the current, the potential across the resistor is is + IR

Loop Rule: Polarity In (c), the source of emf is traversed in the direction of the emf (from to +), and the change in the electric potential is +ε Discharging I In (d), the source of emf is traversed in the direction opposite of the emf (from + to -), and the change in the electric potential is ε which means you are CHARGING the battery! I Charging

Kirchhoff s Rules Junction Rule: junction Loop Rule: I 0 V closed loop 0 First use junction rule and assign values to the current - guess the directions! Then use the loop rule CONSISTENTLY (e.g. Clockwise) on each loop. Any capacitor acts as an open branch in a circuit once under steady state conditions

Single Circuit: Multiple Batts What is the direction of current? If you choose wrong Current Directions you get negative currents. It still works out but try to pick the correct direction! I The 12 V WINS! When polarities of the batteries are opposed, one gets CHARGED.

Multiple Loop Circuit How Many Currents? If you choose wrong Current Directions you get negative currents. It still works out but try to pick the correct direction! I 1 I 2 I 1 I 3 I 3

QuickCheck 31.4 The current through the 3 resistor is A. 9 A. B. 6 A. C. 5 A. D. 3 A. E. 1 A. Slide 31-34

QuickCheck 31.4 The current through the 3 resistor is A. 9 A. B. 6 A. C. 5 A. D. 3 A. E. 1 A. Slide 31-35

Kirchhoff Problem

What is the current in the resistors?

Many-Loop Many Equations! You can solve systems of equations with simple algebra or by using linear algebra (Cramer s Rule) but I can t teach you that! I ve put a document on the website if you want to teach yourself. BUT YOU DON T NEED TO! Algebra will work and I won t put a circuit with more than two loops on the exam! Also, you calculator can solve systems of equations!

Which Circuit(s) Requires Kirchoff? Rank net Power Supply Figure P28-50, p. 824

Voltage in a single-resistor circuit with an Ideal 1.5 Volt battery

Electromotive Force: EMF e = V+ Ir The emf e is the voltage labeled on the battery. It is greater than the terminal voltage. V=IR The actual potential difference V between the terminals of the battery depends on the current in the circuit and the internal resistance of the battery: V = e Ir If the internal resistance of the battery is zero, the terminal voltage equals the emf : V = e R is the load resistance: V = IR

Power with real batts The total power output of the battery is I V Ie This power is delivered to the external resistor (I 2 R) and to the internal resistor (I 2 r) is maximum when R = r!! (see nice proof in text) I 2 R I 2 r

HW: What is the emf and internal resistance of the battery?

Next Time: RC Circuits: Charging The capacitor continues to charge until it reaches its maximum charge (Q = Cε). Once the capacitor is fully charged, the current in the circuit is zero and the The potential difference across the capacitor matches that supplied by the battery.

RC Simulator http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=31

RC Circuit: Charging The charge on the capacitor varies with time q(t) = Ce(1 e -t/rc ) = Q(1 e -t/rc ) t is the time constant t = RC The current through R is: ε t RC I( t) e R The time constant t has units of time represents the time required for the charge to increase from zero to 63.2% of its maximum The energy stored in the charged capacitor is ½ Qe = ½ Ce 2

RC Circuit: Charging t RC Qt ( ) εc( 1 e ) t RC Capacitor: VC ( t ) V0 ( 1 e ) C C Resistor: ε VR ( t ) I( t ) R R e εe V 0 e R t RC t RC t RC

Charging an RC Circuit http://faculty.wwu.edu/vawter/physicsnet/topics/dc-current/rcseries.html

Discharging

An RC Problem The switch in the figure has been in position a for a long time. It is changed to position b at t = 0s. What are the charge on the capacitor and the current I through the resistor a)immediately after the switch is closed? b)what is the time constant t? c)at t = 50 s? d)at t = 200 s?

RC Circuit Charging q(t) = Q(1 e -t/rc ) I( t) ε R e t RC

HO RC Problem

Discharging a Capacitor in an RC Circuit When a charged capacitor is placed in the circuit, it can be discharged q = Qe -t/rc The charge decreases exponentiallyat t = t = RC, the charge decreases to 0.368 Q max In other words, in one time constant, the capacitor loses 63.2% of its initial charge The current is: I t dq dt Q RC e t RC

QuickCheck 31.16 What does the voltmeter read? A. 6 V. B. 3 V. C. 2 V. D. Some other value. E. Nothing because this will fry the meter. Slide 31-91

QuickCheck 31.16 What does the voltmeter read? A. 6 V. B. 3 V. C. 2 V. D. Some other value. E. Nothing because this will fry the meter. Slide 31-92

QuickCheck 31.17 What does the ammeter read? A. 6 A. B. 3 A. C. 2 A. D. Some other value. E. Nothing because this will fry the meter. Slide 31-93

QuickCheck 31.17 What does the ammeter read? A. 6 A. B. 3 A. C. 2 A. D. Some other value. E. Nothing because this will fry the meter. Slide 31-94

Current: Dead or Alive DEATH: NEUROLOGIC CRITERIA: An individual with irreversible cessation of all brain function, including the brain stem, is dead. CARDIOPULMONARY CRITERIA: An individual with irreversible cessation of circulatory and respiratory function is dead.

Electric Shock What causes electric Shock in the human body, Voltage or Current? Electric Shock occurs when current is produced in the body, which is caused by an impressed voltage. Voltage is the CAUSE Current does the DAMAGE

Electric Shock Current (A) Effect 0.001 Can be felt 0.005 Painful 0.010 Causes involuntary muscle spasms 0.015 Causes loss of muscle control 0.070 If through heart, serious! If current lasts for 1 s - FATAL! Dry Skin Body Resistance: 500,000 Wet Skin Body Resistance: 1000

Electric Shock Therapy ELECTRO CONVULSIVE THERAPY An electric shock is applied to produce a convulsive seizure. The shock is typically between 140-170 volts and lasts between 0.5 and 1 seconds. No explanation of how it works. Used in the treatment of: 1.Chronic endogenous depression 2.Bipolar disorder. 3.Acute mania. 4.Certain types of schizophrenia In the U.S. 33,000-50,000 people receive ECT each year

Question What current would you draw if you were unfortunate to short-circuit a 120 V line with dry hands? Wet hands? Dry Skin Body Resistance: 500,000 Wet Skin Body Resistance: 1000 DRY: I = V/R = 120 V/500,000 =.00024 (live!) WET: I = V/R = 120 V/1000 =.12 (dead!)

Electrical equipment manufacturers use electrical cords that have a third wire, called a ground This safety ground normally carries no current and is both grounded and connected to the appliance Ground Wire

Ground-Fault Interrupters (GFI) Special power outlets Used in hazardous areas Designed to protect people from electrical shock Senses currents (< 5 ma) leaking to ground Quickly shuts off the current when above this level

Is AC Deadlier than DC? Low frequency (50-60 Hz) AC currents can be more dangerous than similar levels of DC current since the alternating fluctuations can cause the heart to lose coordination, inducing ventricular fibrillation, which then rapidly leads to death. High voltage DC power can be more dangerous than AC, however, since it tends to cause muscles to lock in position, stopping the victim from releasing the energized conductor once grasped. Frequency MATTERS.

Is AC Deadlier than DC? They are BOTH Deadly! Any practical distribution system will use voltage levels quite sufficient for a dangerous amount of current to flow, whether it uses alternating or direct current. Ultimately, the advantages of AC power transmission outweighed this theoretical risk, and it was eventually adopted as the standard worldwide after Nikola Tesla designed the first AC hydroelectric power plant at Niagara Falls, New York which started producing electrical power in 1895.

War of Currents 1880 s Thomas Edison, American inventor and businessman, pushed for the development of a DC power network. George Westinghouse, American entrepreneur and engineer, backed financially the development of a practical AC power network. Nikola Tesla, Serbian inventor, physicist, and electro-mechanical engineer, was instrumental in developing AC networks.

Edison's Publicity Campaign Edison wired NYC with DC. He carried out a campaign to discourage the use of AC, including spreading information on fatal AC accidents, killing animals, and lobbying against the use of AC in state legislatures. Edison opposed capital punishment, but his desire to disparage the system of alternating current led to the invention of the electric chair. Harold P. Brown, who was at this time being secretly paid by Edison, constructed the first electric chair for the state of New York in order to promote the idea that alternating current was deadlier than DC. The first electric chair, which was used to execute William Kemmler in 1890

GE & Edison: We bring good things to light. More than a 1000 killed since 1890! The first electric chair, which was used to execute William Kemmler in 1890 Nebraska: Only state that requires it. 15-second-long jolt of 2,450 volts of electricity (~ 8 Amps)

Limitations of DC Transmission Large currents in wires produce heat and energy losses by P = I 2 R. Large expensive conductors would be needed or else very high voltage drops (and efficiency losses) would result. High loads of direct current could rarely be transmitted for distances greater than one mile without introducing excessive voltage drops. Direct current can not easily be changed to higher or lower voltages. Separate electrical lines are needed to distribute power to appliances that used different voltages, for example, lighting and electric motors.

Advantages of AC Transmission Alternating Current can be transformed to step the voltage up or down with transformers. Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V). Convert AC to DC with a rectifier in appliances. AC is more efficient for Transmission & Distribution of electrical power than DC over long distances! But what if the power generation is LOCAL?

In The Future. Long Distance AC Power Transmission may not be needed!!! New DC technologies and Solar Power will make AC obsolete!

Quick Quiz For the two lightbulbs shown in this figure, rank the current values at the points, from greatest to least. I a = I b > I c = I d > I e = I f. The 60 W bulb has the lowest resistance and therefore draws the most current and burns brightest!