Electric current and Ohm’s Law AP Physics Section 18-1 to 18-5 Electric current and Ohm’s Law

When a conductor connects a region of high potential to a region of low potential an electric current will flow through the conductor. The flow of positive charge from high to low potential (from + to –) is called conventional current. Current is the movement of any charged particles. These may be electrons in a metal, or ions in an electrolyte solution. ∆q I = The symbol for electric current is I. ∆t The unit of current is one of the base SI units, the ampere (A). 1A = 1 coulomb/second (1 C/s).

Production of electric current conventional current I Conventional current is the flow of positive charges from positive to negative (high to low potential). In metals, direct current is actually electron flow from negative to positive. + + – – + + – – + + + – – – + + flow of electrons in metal conductor – – + – Production of electric current In the diagram above, the current flow would eventually stop when the spheres reach equilibrium. However, a device may be used to “pump” the current back to high potential again, just like a water pump raising water to a higher elevation.

Sources of electromotive force Electromotive force (EMF, symbol: ℰ) is the potential difference created by any source of electrical energy including electrochemical cells and batteries, dynamos (DC generators), photovoltaic cells, thermoelectric generators, etc. photovoltaic cell DC power supply cell battery hand dynamo

Resistance I ∆V = IR ∆V R = Resistance is a property of materials that measures how much current flows compared to the potential difference across the material. Symbol: R Units: ohm (Ω) ∆V = IR ∆V often written, Ohm’s Law R = ∆V I AP sheet: I = R Properties of conductor that affect the resistance length (ℓ) — higher R for longer length. cross-sectional area (A) — higher R for smaller area. temperature — higher R at higher temperatures. material resistivity (ρ) — the material used affects R

Resistors resistivity, ρ materials from most to least conductive In electronics, special components with known resistances are used to control the flow of current. These are called resistors. The values in ohms are printed on the resistor, or a color code is used. silver copper gold aluminum tungsten iron platinum lead nichrome brown = 1 carbon black = 0 orange = 000 germanium gold = 5% silicon For given T, R = ρ ℓ A 10kΩ resistor symbol

Various types of resistors with different values and tolerances (possible degree of error). Resistor construction.

Measuring devices Ω V A Ammeter Voltmeter Ohmmeter Multimeter Various measuring devices are used to measure potential difference (voltage), current, and resistance. Ammeter Voltmeter V A Ω Ohmmeter Multimeter

Other common circuit symbols switch (single throw) variable resistor (trimmer) ground (earth) capacitor bulb potentiometer diode

Simple circuit problems Calculate the current in the circuits in milliamperes. 1) 2) 15kΩ 330Ω I I 9V 3V I = 0.6 mA I = 9.1 mA Calculate the resistance of the resistors. RV is very high. Why? 12V 6V V V 3) 4) R1 R2 I I R1 = 62Ω R2 = 470Ω 12V 6V A A 0.194 A RA is very low. Why? 12.8 mA

Electric Power I = ∆V2 UE = PEe UE ∆UE = W = q ∆V ∆V = q = It P = I∆V We learned that: UE ∆UE = W = q ∆V ∆V = so, q0 ∆q q = It I = also, so, ∆t Substitute q: ∆UE = W = It∆V Since power is energy or work per time in seconds: P = I∆V Units: watt (W) ∆V2 P = I2R P = and Also, since ∆V = IR: R

André-Marie Ampère Alessandro Volta Georg Simon Ohm James Watt

Calculating Power 1) 2) 3) 4) Calculate the power of all four previous circuits, 1–4. 1) 2) P = 0.0054 W P = 0.027 W 3) P = 2.3 W 4) P = 0.077 W If all four circuits run for 5.0 minutes, how much energy is used by each circuit. 1) energy = 1.6 J 2) energy = 8.1 J 3) energy = 690 J 4) energy = 23 J

I = ∆V ∆V2 ∆V2 P = energy/time energy = P•t = 100 W (3600 s) energy = A car has two 50 watt fog lights running, for a total of 100 W of power. The electrical system runs off a 12V lead acid battery. 1. How much energy do the fog lights use in one hour? P = energy/time energy = P•t = 100 W (3600 s) energy = 360000 J = 360 kJ 2. What is the current in the circuit? ∆V P 12 V 100 W P = I∆V I = = = 8.3 A 3. What is the total resistance of the two lights? P = R ∆V2 P ∆V2 100 W (12 V)2 R = = = 1.4 Ω