APC – UNIT 9 DC Circuits

Whenever electric charges move, an electric current is said to exist. The current is the rate at which the charge flows through a certain cross-section A. We look at the charges flowing perpendicularly to a surface of area A +-

Atomic View of Current Consider a wire connected to a potential difference… + - E Existence of E inside wire (conductor) does not contradict our previous results for E = 0 inside conductor. Why?

Current Density (j) Current density is a vector field within a wire. The vector at each point points in the direction of the E-field The current density and the electric field are established IN a conductor whereas a potential difference is maintained ACROSS a conductor.

Drift Velocity Current was originally thought to be positive charge carriers (Franklin) and therefore that became conventional current flow. However, it is the free electrons (valence) that move but they encounter many collisions with atoms in the wire. (non-conventional) Overall speed of electrons is VERY slow. It is called the drift velocity, v d. 5.5hr to move 1m. The thermal motion of electrons is very random but fast (10 6 m/s). When E-field is established, the e- ‘drift’.

In a time Δt, electrons travel a distance Δx = v d Δt. Volume of electrons in Δt pass through area A is given as V = A Δ x = A v d Δ t If there are n free electrons per unit volume (n= N/V) where N = # of electrons then the total charge through area A in time Δt is given by dQ = (# of charges)x(charge per e-) dQ = nV(-e) = -n A v d (dt) e Minus means dirn of + current opposes dirn of v d Also… We can relate the current to the motion of the charges

Ohm’s Law

Conductivity Within a wire of length L, I = jA and V = EL, substituting into Ohm’s Law we get High resistivity produces less current density for same E-field Conductivity is the reciprocal of resistivity. + - E VbVb VaVa

Electrical Power Consider the simple circuit below. Imagine a positive quantity of charge moving around the circuit from point A through an ideal battery, through the resistor, and back to A again. A B As charge moves from A to B through battery, its electrical energy increases by an amount QΔV while the chemical PE of battery decreases by that amount. When the charge moves through the resistor, it loses EPE as it undergoes collisions with atoms in R and produces thermal energy.

A B The rate at which charge loses PE in resistor is given by From this we get power lost in the resistor: Using Ohm’s Law we can also get

Series Circuit Characteristics

Parallel Circuit Characteristics

Short Circuit

Ammeter and Voltmeter AMMETERS have a very small resistance to limit their effect on introducing resistance into the circuit being measured. Connected in SERIES. VOLTMETERS (  V) have a very large resistance to reduce the amount of current drawn from the circuit being measured (short). Connected in PARALLEL.

Compound Circuit 10V a) Find the potential difference across R 4 b) If the wire before R 2 is cut (inoperable) what happens to the total current? c) If R 2 is replaced by a wire what happens to the total current?

Potentiometer or Variable Resistor Device that allows for you to vary the resistance by changing the effective length of wire symbol

EMF (electromotive force), ε A ideal battery has no internal resistance (friction). However, a real battery has some internal resistance where there is a voltage drop within battery leaving less ΔV for external circuit.

Different sized batteries (AAA vs D) have different amp-hour ratings. The larger the battery, the higher the amp-hour rating for the same V. Larger-sized batteries have more charge to supply The battery capacity that battery manufacturers print on a battery is usually the product of 20 hours multiplied by the maximum constant current that a new battery can supply for 20 hours at 68 F° (20 C°), down to a predetermined terminal voltage per cell. A battery rated at 100 A·h will deliver 5 A over a 20 hour period at room temperature.

Series and Parallel EMFs Battery Charging

EMFs in parallel can produce more current than a single emf while maintaining voltage. When connecting in parallel you are doubling the capacity (amp hours) (60amphrs means if the load drew 10A, it would last 6hrs) of the battery while maintaining the voltage of the individual batteries Batteries MUST be the same, If not, there will be relatively large currents circulating from one battery through another, the higher-voltage batteries overpowering the lower-voltage batteries.

Kirchoff’s Rules 1) Junction Rule (  I j = 0) Circuits that are complex in that they cannot be reduced to series or parallel combinations require a different approach.

2) Loop Rule  (  V j ) = 0 (conservation of energy)

Kirchoff Example Calculate the current in each branch of the circuit. I1I1 I3I3 I2I2

If a voltmeter was connected between points c and f, what would be the reading (V cf )? V cf means V c – V f.

RC CIRCUITS Initially, at t=0, at the instant a switch closes there is a potential difference of 0 across an uncharged capacitor (it acts like a wire…short circuit). Often RC circuits are used to control timing. Some examples include windshield wipers, strobe lights, and flashbulbs in a camera, some pacemakers. Ultimately, the capacitor reaches its maximum charge and there is no current flow through the capacitor (it acts like an open circuit as t goes to infinity (R=∞)). At this point, ΔV C = ε. Note that C does not charge instantaneously. Current (i) decays over time and is not steady.

ξ A closer look at current during charging process 1) At instant switch is connected to ‘a’, 2) As C charges,

ξ Applying Kirchoff’s Loop Rule to above circuit (CW):

Current, i, as function of time

Time Constant, τ There is a quantity referred to as the time constant of the RC circuit. This is the time required for the capacitor to reach 63% of its charge capacity and maximum voltage. It also represents the time needed for the current to drop to 37% of its original value. It can be shown that after 1 time constant (RC), V C is 63% of its maximum voltage, V o.

0.37 I max 0.63 V max 1 time constant, RC

Discharging ξ After a very long time, the capacitor would be fully charged. If the switch was then moved to ‘b’… The capacitor would discharge through the resistor as a function of time similar to the previous derivation. i Voltage across resistor equals voltage across capacitor at all times in above circuit.

Example Both switches are initially open, and the capacitor is uncharged. What is the current through the battery just after switch S 1 is closed? Both switches are initially open, and the capacitor is uncharged. What is the current through the battery after switch 1 has been closed a long time? a) I b = 0 b) I b = ε / (3R) c) I b = ε /(2R)d) I b = ε / R b) I b = V/(3R) a) I b = 0 c) I b = V/(2R) d) I b = V/R 2R R C S1S1 S2S2 ε

Both switches are initially open, and the capacitor is uncharged. What is the current through the battery just after switch S 1 & S 2 are closed? After a long time what is the current through the battery? a) I b = 0 b) I b = ε / (3R) c) I b = ε /(2R) d) I b = ε / R b) I b = ε /(3R) a) I b = 0 c) I b = ε /(2R) d) I b = ε /R After a long time S 1 is opened. What is the voltage across R and 2R after 2τ? 2R R C S1S1 S2S2 ε

Example S R2R2 R1R1 C Find V R2 & V R1 after S has been closed for 1τ. 12V Find total current at this time if R 1 = 10 Ω and R 2 = 20Ω

Each circuit below has a 1.0F capacitor charged to 100 Volts. When the switch is closed: c) Which lights consumes more energy assuming we wait until both can’t be seen? Example a) Which system will be brightest? b) Which lights will stay on longest?