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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 . W e look at the charges flowing perpendicularly to a surface of area A. +. -. Atomic View of Current.
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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) The thermal motion of electrons is very random but fast (106 m/s). When E-field is established, the e- ‘drift’. Overall speed of electrons is VERY slow. It is called the drift velocity, vd. 5.5hr to move 1m.
We can relate the current to the motion of the charges In a time Δt, electrons travel a distance Δx = vdΔt. Volume of electrons in Δt pass through area A is given as V = A Δ x = A vdΔ 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-) Also… dQ = nV(-e) = -n A vd (dt) e Minus means dirn of + current opposes dirn of vd
Conductivity Vb Va Within a wire of length L, I = jA and V = EL, substituting into Ohm’s Law we get E - + High resistivity produces less current density for same E-field Conductivity is the reciprocal of resistivity.
B A 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. 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.
B A 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
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 (DV) have a very largeresistance to reduce the amount of current drawn from the circuit being measured (short). Connected in PARALLEL.
Compound Circuit a) Find the potential difference across R4 b) If the wire before R2 is cut (inoperable) what happens to the total current? 10V c) If R2 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 Circuits that are complex in that they cannot be reduced to series or parallel combinations require a different approach. 1) Junction Rule (S Ij = 0)
Calculate the current in each branch of the circuit. Kirchoff Example I1 I3 I2
If a voltmeter was connected between points c and f, what would be the reading (Vcf)? Vcf means Vc – Vf.
Often RC circuits are used to control timing. Some examples include windshield wipers, strobe lights, and flashbulbs in a camera, some pacemakers. 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). 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, ΔVC =ε. Note that C does not charge instantaneously. Current (i) decays over time and is not steady.
A closer look at current during charging process 1) At instant switch is connected to ‘a’, ξ 2) As C charges,
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), VC is 63% of its maximum voltage, Vo.
0.63 Vmax 0.37 Imax 1 time constant, RC
After a very long time, the capacitor would be fully charged. If the switch was then moved to ‘b’… Discharging The capacitor would discharge through the resistor as a function of time similar to the previous derivation. i ξ Voltage across resistor equals voltage across capacitor at all times in above circuit.
2R C R S1 S2 Example Both switches are initially open, and the capacitor is uncharged. What is the current through the battery just after switch S1 is closed? ε a) Ib = 0 b) Ib = ε/ (3R) c) Ib = ε /(2R) d) Ib = ε / R 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) Ib = 0 b) Ib = V/(3R) c) Ib = V/(2R) d) Ib = V/R
2R C R S1 S2 Both switches are initially open, and the capacitor is uncharged. What is the current through the battery just after switch S1 & S2 are closed? ε a) Ib = 0 b) Ib = ε/ (3R) c) Ib = ε /(2R) d) Ib = ε / R After a long time what is the current through the battery? a) Ib = 0 b) Ib = ε /(3R) After a long time S1 is opened. What is the voltage across R and 2R after 2τ? c) Ib = ε /(2R) d) Ib = ε /R
S Example Find VR2 & VR1 after S has been closed for 1τ. 12V R2 R1 C Find total current at this time if R1 = 10Ω and R2 = 20Ω
Example Each circuit below has a 1.0F capacitor charged to 100 Volts. When the switch is closed: a) Which system will be brightest? b) Which lights will stay on longest? c)Which lights consumes more energy assuming we wait until both can’t be seen?