Current Electricity and Circuits V R = Chapter 22 and 23 I I = Q/t

Electric Circuits Now that we have the concept of voltage, we can use this concept to understand electric circuits Just like we can use pipes to carry water, we can use wires to carry electricity. The flow of water through pipes is caused by pressure differences, and the flow is measured by volume of water per time V = “Electrical pressure” - measured in volts. H2O High Pressure Low Pressure

1 Amp = 1 Coulomb per 1 second = 1 C/s Electric Current In electricity, the concept of voltage will be like pressure. Water flows from high pressure to low pressure ; electricity flows from high voltage to low voltage. But what flows in electricity? Charges! How do we measure this flow? Current I = DQ/Dt Electric current is measured in Amperes, in honor of Andre Marie Ampere. 1 Amp = 1 Coulomb per 1 second = 1 C/s

Direction of Current Flow For historical reasons, current is conventionally thought to flow from the positive to the negative potential in a circuit. Current flow is the rate of flow of “positive” charge (It’s really the electrons flowing in the opposite direction) Electron flow is from a lower potential (voltage) to a higher potential (voltage). Electrons carry negative charge Positive current flow is in opposite direction + V I - electron motion positive current direction

Conventional Current + - + - high V low V R1 R2 I I Conventional current flows from high V to low V and + to – across components like resistors. Conventional current flows from the positive side of the battery to the around to the negative side. Don’t be confused by diagrams that look like it is flowing across the battery from the negative to the positive. I I + - + - high V low V R1 R2

Power, Current and Voltage P= W/t Power = work/time For electric circuits, P = V .I Power = electric potential difference . current because, = (work/charge).(charge/time) Charge divides out leaving W/t, which is power Electrical energy is a measure of the amount of power used and the time of use. Electrical energy, kilowatt hours, is the product of the power and the time.

Battery or Cell – Voltage Source A battery in an electrical circuit plays the same role as a pump in a water system. Because of the “pumping” nature of voltage sources, we need to have a complete circuit before we have a current. Just like a pump needs water coming into it in order to pump water out, so the voltage source needs charges coming into it (into the negative terminal) in order to “pump” them out (of the positive terminal). Does the battery or power supply actually supply the charges that will flow through the circuit?

Voltage Sources Lab Power Supply A Battery Solar Cell Electric Power Plant Nerve Cell

Symbols for Voltage Sources In circuit diagrams, symbols are used to represent each component. The following symbols may be used for voltage sources: ~ Time-varying source + _ Battery . Generator (power plant) Solar Cell Note: The battery will be the voltage source used in our diagrams Remember: A voltage is measured between two points

“Ground” “Ground” refers to the reference terminal to which all other voltages are measured. The earth is really just one big ground node. Most people choose the earth as the reference ground when a connection to it is available. + _ V1 + _ V2 + _ V3 + _ V2

Ground Symbol Positive relative to ground + _ V1 + _ V2 + _ V3 + _ Negative relative to ground

Simple Circuit A simple circuit contains the minimum things needed to have a functioning electric circuit. A simple circuit requires three (3) things: A source of electrical potential difference or voltage. (typically a battery or electrical outlet) A conductive path which would allow for the movement of charges. (typically made of wire) An electrical resistance (resistor) which is loosely defined as any object that uses electricity to do work. (a light bulb, electric motor, heating element, speaker, etc.)

Ohm’s Law It takes pressure to make fluid flow due to the viscosity of the fluid and the size of the pipe. Voltage is required to make electric current flow due to the resistance in the circuit. By experiment we find that if we increase the voltage, we increase the current: V is proportional to I. The constant of proportionality we call the. resistance, R

The ratio of potential difference to current Ohm’s Law The ratio of potential difference to current is constant. If R = V/I is a constant value for a given resistor, then that resistor is said to obey Ohm’s Law. Ohm’s Law in equation form: V=IR The SI unit of resistance is the Ohm, W, named in honor of Georg Simon Ohm.

Ohm’s Law

Resistance The resistance depends on material and geometry (shape). For a wire: R = r L / A where where r is called the resistivity (in Ohm-m) and measures how hard it is for current to flow through the material, L is the length of the wire, and A is the cross-sectional area of the wire. In a circuit the symbol used for resistors is usually: In circuits, the resistance of resistors is often given or can be calculated. The resistance of the wire connecting the circuit is sometimes neglected.

Resistance and Temperature Many circuit elements do not obey Ohm’s Law. Resistors that get hot, like light bulbs and heating elements, do not keep a constant resistance. Resistance generally increases as objects become hotter.

Series and Parallel Circuits There are two ways devices can be connected in a circuit, series or parallel. Current flows differently through each type of circuit so each requires a different set of equations. R1 I V1 + R2 Vbat V2 Itotal - + Resistors connected in series I1 R1 Vbat I2 R2 - Resistors connected in parallel

Series Resistors - Equations Current is constant –There is only one path for the current to take so in a series circuit, resistance and voltage add, but current stays the same. 1. total resistance is the sum of the separate resistors RT = R1 + R2 + R3 + ... 2. current is the same through each resistor IT = I1 = I2 = I3 = ... 3. total potential difference is the sum of each VT = V1 + V2 + V3 + ...

Parallel Resistors - Equations In a parallel circuit, resistance adds as reciprocals, voltage stays the same, and current splits 1. reciprocal of the total resistance is the sum of the reciprocals of the separate resistors 1/RT = 1/R1 + 1/R2 +1/R3 + ... 2. total current is the sum of the current through each resistor IT = I1 + I2 + I3 + ... 3. potential difference is the same across each resistor VT = V1 = V2 = V3 = ...

Kirchhoff’s Current Law: Junction Rule Kirchhoff’s current law says: Current into junction = Current leaving junction OR The sum of all the currents entering a node is zero I1 = I2 + I3 OR I1 – I2 – I3 = 0 I1 node I2 I3 Io How much is the current Io , if I2 is 2.5 mA and I3 is 4 mA?

Kirchhoff’s Current Law applies to all types of networks Fiber optic network (I is light intensity) I1 I2 “KCL” for light: I1 = I2 + I3 I3

Kirchhoff’s Current Law applies to all types of networks Human Blood Vessels (f is blood flow rate) Organ f2 f1 f1 “KCL” for blood flow: f1 = f2 + f3 f3

Kirchhoff’s Voltage Law: Loop Rule Kirchhoff’s Voltage Law: Sum of all voltage drops and voltage rises in a circuit (a closed loop) equals zero OR The voltage measured between any two nodes does not depend of the path taken. V1= V2 + V3 V1= V2 + V4 voltage V3= V4 + _ + V2 – V3 – V4 V1 voltage voltage

Kirchhoff’s Voltage Law: Loop Rule + _ + V2 – V3 – V4 V1 Kirchhoff’s Voltage Law: V1 = V2 + V3 OR –V1 + V2 + V3 = 0

Using the Loop Rule Assign symbols and directions of currents in the loop If the direction is chosen wrong, the current will come out with a right magnitude, but a negative sign (it’s ok). Choose a direction (cw or ccw) for going around the loop. Record drops and rises of voltage according to this: If a resistor is traversed in the direction of the current: -V = -IR If a resistor is traversed in the direction opposite to the current: +V=+IR If EMF is traversed “from – to + ”: +E If EMF is traversed “from + to – ”: -E

Using the Loop Rule Loops can be chosen arbitrarily. For example, the circuit below contains a number of closed paths. Three have been selected for discussion. Suppose that for each element, respective current flows from + to - signs. + - v2 - v5 + - - - Path 1 v1 v4 v6 + + + Path 2 v3 v7 - + + - Path 3 - + + v8 v12 v10 + - - + v11 - - v9 +

Using the Loop Rule • • - v7 + v10 – v9 + v8 = 0 “b” Using sum of the drops = 0 • + - v2 - v5 + - - - Blue path, starting at “a” - v7 + v10 – v9 + v8 = 0 v1 v4 v6 + + + v3 v7 - + + - • “a” Red path, starting at “b” +v2 – v5 – v6 – v8 + v9 – v11 – v12 + v1 = 0 - + + v8 v12 v10 + - - Yellow path, starting at “b” + v2 – v5 – v6 – v7 + v10 – v11 - v12 + v1 = 0 + v11 - - v9 +

Using the Loop Rule Example: For the circuit below find I, V1, V2, V3, V4 and the power supplied by the 10 volt source. For convenience, we start at point “a” and sum voltage drops =0 in the direction of the current I. 2. We note that: V1 = - 20I, V2 = 40I, V3 = - 15I, V4 = 5I (2) 3. We substitute the above into Eq. 1 to obtain Eq. 3 below. 10 + 20I – 30 + 15I + 5I – 20 + 40I = 0 (3) Solving this equation gives, I = 0.5 A.

Using the Loop Rule Using this value of I in Eq. 2 gives: V1 = - 10 V P10(supplied) = -10I = - 5 W (We use the minus sign in –10I because the current is entering the + terminal) In this case, power is being absorbed by the 10 volt supply.

8.0 2.0 5.0 RT = VT = IT = PT = E = 12 V R1 R1 R3 R2 R3 R2 R, W V, V A P, W E = 12 V R1 8.0 R1 R3 R2 2.0 R3 5.0 RT = R2 VT = IT = PT =

8.0 2.0 5.0 RT = 15 Ω VT = 12 V IT = 0.80 A PT = 9.6 W E = 12 V R1 6.4 5.1 R1 R3 R2 2.0 1.6 0.80 1.3 R3 5.0 4.0 0.80 3.2 RT = 15 Ω R2 VT = 12 V IT = 0.80 A PT = 9.6 W

12 8.0 12 RT = VT = IT = PT = E = 12 V R1 R2 R1 R3 R2 R3 R, W V, V I, A P, W E = 12 V R1 12 R2 8.0 R1 R3 12 R2 RT = R3 VT = IT = PT =

R, W V, V I, A P, W E = 12 V R1 12 12 1.0 12 8.0 12 1.5 R2 18 R1 12 12 1.0 12 R3 R2 RT = 3.42 Ω R3 VT = 12 V IT = 3.50 A PT = 42 W

Toll Road – Circuit Analogy Adding toll booths in series increases resistance and slows the current flow. Adding toll booths in parallel lowers resistance and increases the current flow.

Combined Series parallel circuits Rt = R1 + 1/(1/R2 + 1/R3) add R2 and R3 as parallel then add the result to R1 Vt = V1 + (V2 = V3) It = I1 = I2 + I3 total current goes through R1 but splits when it go through R2 and R3

Internal Resistance Real batteries are constructed from materials which possess non-zero resistance. It follows that real batteries are not just pure voltage sources. They also possess internal resistances. A battery can be modeled as an emf connected in series with a resistor , which represents its internal resistance.

Capacitor /Water Tower A water tower holds water. A capacitor holds charge. While we normally define the capacity of a water tank by the TOTAL AMOUNT of water it can hold, we define the capacitance of an electric capacitor as the AMOUNT OF CHARGE PER VOLT instead. C = Q/V There is a TOTAL AMOUNT of charge a capacitor can hold, and this corresponds to a MAXIMUM VOLTAGE that can be placed across the capacitor. Each capacitor DOES HAVE A MAXIMUM VOLTAGE. If an electric capacitor is “over-filled” or equivalently a higher voltage is placed across the capacitor than the listed maximum voltage It will “break” by having the charge “escape”. This escaping charge is like lightning - a spark that usually destroys the capacitor

Capacitors – In Series C1 + V C2 There is only one way around the circuit, and you have to jump BOTH capacitors in making the circuit The positive charge on the left plate of C1 will attract a negative charge on the right plate, and the negative charge on the bottom plate of C2 will attract a positive charge on the top plate - just what is needed to give the negative charge on the right plate of C1. C1 (+Q1  ) + +Q1 -Q 1 +Q2 V C2 -Q2 (  +Qtotal) Vtotal = (V1 + V2) 1/C1 + 1/C2 = 1/Ceffective Qtotal = Q1 = Q2

Capacitors – Parallel For parallel, both plates are across the same voltage, so Vtotal = V1 = V2 . The charge can accumulate on either plate, so: Qtotal = (Q1 + Q2). Since the Q’s are in the numerator, we have: Ceff = C1 + C2. + +Q1 +Q2 C1 C2 V -Q1 -Q2 +Q1   +Qtotal = (Q1+Q2)  +Q2

The Light Bulb and its Components Has two metal contacts at the base which connect to the ends of an electrical circuit The metal contacts are attached to two stiff wires, which are attached to a thin metal filament. The filament is in the middle of the bulb, held up by a glass mount. The wires and the filament are housed in a glass bulb, which is filled with an inert gas, such as argon.

Light bulbs and Power Power dissipated by a bulb relates to the brightness of the bulb. The higher the power, the brighter the bulb. Power is measured in Watts [W] Power, P = I*V, can also be written P = I2*R or P = V2 / R by using Ohm’s law and doing some substitutions. If we wanted a higher power light bulb, should we have a bigger resistance or a smaller resistance for the light bulb?

Light bulbs and Power - Answer Answer: In this case, the voltage is being held constant due to the nature of the batteries. This means that the current will change as we change the resistance. Thus, the P = V2 / R would be the most straight-forward equation to use. This means that as R goes down, P goes up. (If we had used the P = I2*R formula, as R goes up, I would decrease – so it would not be clear what happened to power.) The answer: for more power, lower the resistance. This will allow more current to flow at the same voltage, and hence allow more power!

Light bulbs and Power The three light bulbs in the circuit all have the same resistance. Given that brightness is proportional to power dissipated, the brightness of bulbs B and C together, compared with the brightness of bulb A, is 1. twice as much. 2. the same. 3. half as much.

RC circuits When switch is closed, current flows because capacitor is charging As capacitor becomes charged, the current slows because the voltage across the resistor is  - Vc and Vc gradually approaches . Once capacitor is charged the current is zero q q Charge across capacitor RC is called the time constant

RC circuits If a capacitor is charged and the switch is closed, then current flows and the voltage on the capacitor gradually decreases. This leads to decreasing charge q Charge across capacitor 0.37q

Sources www.physics.ubc.ca/outreach/phys420/p420_04/mitsuko/OhmsLaw-Gr10-Science.ppt http://physics.bu.edu/~duffy/PY106 Physics by Zitzewitz www.glenbrook.k12.il.us/GBSSCI/PHYS/Class www.cbu.edu/~jholmes/P202/P1elec2.ppt