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• • • • Day 1 slides 1-11, hw 8-1 webassign There are 4 webassigns for unit 8-2 = circuits 8-3 = kirchoff 8-4 RC etc Copy DC circuits wkshts 1, 2, 3, + bulb demo (saved as word doc) for use on day 2 and 3 APC – UNIT 8 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 + - The time rate of the charge flow through A defines the current (=charges per time): Units: C/s=A (Ampere) I = dQ/dt Scalar quantity Current moving from + to – is called conventional current flow. 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? E = 0 inside conductor was only for electrostatic conditions. No longer dealing with that…charges are now free to move. Only when charges are at rest, does E = 0. Current Density (j) Current per cross sectional area I j dA where dA is the element of surface & I is the current through the surface over which integration takes place Current density is a vector field within a wire. The vector at each point points in the direction of the E-field I j dA 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. (nonconventional). The thermal motion of electrons is very random but fast (106 m/s). When Efield 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 = vd Δt. Volume of electrons in Δt pass through area A is given as V = A Δ x = A vd 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-) Also… I neAvd dQ = nV(-e) = -n A vd (dt) e j nevd A A dQ I neAvd dt Minus means dirn of + current opposes dirn of vd Ohm’s Law V IR Rwire L A Units for resistance are ohms (Ω) ρ = resistivity… a parameter that depends on the properties of material Units are Ωm Materials that obey Ohm’s Law are said to be ohmic. Incandescent light bulbs are non-ohmic. Conductivity + E Vb - Within a wire of length L, I = jA and V = EL, substituting into Ohm’s Law we get EL ( jA)( L A Conductivity is the reciprocal of resistivity. ) E j 1 j E Va High resistivity produces less current density for same E-field 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. The rate at which charge loses PE in resistor is given by U Q V IV t t From this we get power lost in the resistor: P IV Using Ohm’s Law we can also get 2 V PI R R 2 Day 2 • • • • Get out circuit board, batteries Wire for short Potentiometer rheostat Use drycell for more stable current flow. Series Circuit Characteristics RT REQ R1 R2 ... IT I1 I 2 ... VT V1 V2 ... Parallel Circuit Characteristics VT V1 V2 ... IT I1 I 2 ... 1 1 1 ... RT R1 R2 Short Circuit EXAMPLE 1 a) Determine the equivalent resistance of the circuit shown if R1 = 860Ω, R2 = 640Ω, and R3 = 470Ω. REQ = 836.9 Ω b) Determine the voltage drop (potential difference) across R3 . V =6.73V 3 c) Determine the current through R2 . I2 =8.23mA 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. EXAMPLE 2 a) If V = 10V, find the potential difference across R4. 4.08V D C b) If the wire at ‘C’ is cut, what happens to the total current in circuit? RT goes from 13.5Ω to 60Ω, so IT decreases. Less pathways means more total resistance. EXAMPLE 2 RESET problem c) If wire is cut at ‘D’ between R3 & R4 what happens to VA? To VX? D C Vx = 0 . Originally VA = VR2 = 6.89V. VA still = VR2 , however, that value is now 7.5V. d) RESET…If R2 is “shorted out” what happens to the total current? The short leaves out R2, R3, & R4 . Total R goes down to 6Ω from 19.5Ω, total current goes up. R1 S V R3 R4 R2 Describe what happens to the voltage across each resistor when the switch S is closed? V1 and V2 increase; V3 and V4 decrease Work on DC circuits #2 worksheet S1 + 1 - S2 2 S3 3 4 5 S4 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 An ideal battery has no internal resistance (friction). However, a real battery has some internal resistance where there is a voltage drop(EPE loss/charge) within the battery leaving less ΔV for external circuit. The voltage available for external circuit is called the terminal voltage, Vab . The internal resistance is r. Therefore, Vab = – Ir Vab = Open circuit voltage (ideal) The current provided to the circuit depends on both the internal, r, & external resistance, R. When R >> r, most of the power that is delivered by the battery is transferred to the load resistor, R. As a battery ages, its internal resistance increases (the chemical processes slows) . This affects the battery’s ability to supply energy. Different sized batteries (AAA vs D) have different amp-hour (charge) ratings or capacities. The larger the battery, the higher the amp-hour rating for the same V. Larger-sized batteries have more charge to supply. More charge carriers can deliver more energy. A 12V car battery is more dangerous than 9, 1.5V D batteries. Series and Parallel EMFs Batteries in series in the same direction where signs alternate: total voltage is the sum of the separate voltages. Battery Charging Batteries in series, opposite direction where signs are like signs: total voltage is the difference, where the lower voltage battery is being charged. Hooking up car batteries in series to charge. Like signs must be connected together. Batteries in parallel are not used to increase voltage or charge one another, but serve to provide more energy when large currents are needed. Each cell only produces a fraction of total current so loss due to internal resistance is less than for single cell. Batteries will last longer. In this case, VR = 12V and if R=1, IT = 12A with each battery providing only 6A each. When connecting in parallel you are doubling the capacity (charge) 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. Vab worksheet Power delivered to Load (R) When is the power delivered to the load (R) a maximum when battery is NOT ideal? Meaning, how is R related to r in order to achieve maximum power output, assuming there is a load resistor? PI R 2 Determine I that flows through R, then plug into P. P( rR I /( r R) 2 2 ) R r 2rR R 2 Clearly, PMAX is achieved when denominator is a minimum. 2 R P 2 r 2rR R 2 2 R Turn R in numerator to 1/(1/R) and simplify P 2 1 2 1 1 2 ( )r ( )2rR ( ) R R R R P 2 2 r 2r R R We know P will be maxed out when denominator is a minimum. So, take the derivative of denominator and set equal to zero to see when R is a minimum. 2 d r ( R 2r ) 0 dR R d 2 1 ( r R R 2r ) 0 dR 2 r R 1 0 0 2 r / R 1 2 R r 2 2 2 R r Since resistance is + , then Rr 2nd derivative test to prove minimum or maximum for denominator, 2 d 2 2 2 r (r R 1) 3 dR R Since answer is positive, solution refers to a minimum. If denominator is minimum, then P is maximum when R=r. 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) (conservation of charge) The sum of the currents entering any junction must equal the sum of the currents leaving that junction. 2) Loop Rule S (Vj ) = 0 (conservation of energy) The sum of the potential differences across all the elements around any closed circuit loop must be zero. (valid for any closed loop); Kirchoff Example Arbitrarily assign currents and their directions in each branch abiding by I1 junction rule. If you are wrong about direction, your answer will I2 yield a minus sign. Calculate the current in each branch of the circuit. I3 Junction Rule: I3 = I1 + I2 Assign loops (direction is arbitrary) When moving against current ,it is a positive change in ΔV Top Loop (CW): {ahdcba} -I1 (30) + 45 – I3 (1 + 40) = 0 Outside Loop (CW) {ahdefga} -I1 (30) + I2 (20 + 1) – 80 = 0 When moving from – terminal to + terminal of battery, it’s a positive change in ΔV 3 equations, 3 unknowns, solve! I3 = I1 + I2 -I1 (30) + 45 – I3 (1 + 40) = 0 -I1 (30) + I2 (20 + 1) – 80 = 0 I2 = 2.6A I1 = -0.86A I3 = 1.7A Negative indicates current is opposite direction and that battery was charging instead of discharging. Don’t disregard minus sign when figuring out other currents. If a voltmeter was connected between points ‘c’ and ‘f’, what would be the reading (Vcf)? Vcf = Vc – Vf. If current was a negative, don’t switch the direction. Just keep as is. Voltmeter completes loop! Use loop rule for left side moving CW (fgabcf) -80 + 41(1.7) + Vcf = 0 -10.3V + Vcf = 0 Vcf = 10.3V Right side moving CW (cdefc) (2.6)(21) -45 + Vcf =0 9.6V + Vcf = 0 Vcf = -9.6V + ΔV moving from ‘c’ to ‘f’, - ΔV moving from ‘f’ to ‘c’ therefore, ‘f’ is higher potential than ‘c’ Diff is due to rounded current Example: Ideal batteries ξ1 = 11.5 V and ξ2 = 4.00V, and the resistances are each 3.2Ω. a) What is the size and direction of current i1? (Take upward to be positive.) ξ1 i1 i2 ξ2 a) Focus on left hand loop ΣV around loop = 0, therefore V across resistor must be 11.5V. Moving CW around loop yields… i1 11.5V i1 = -11.5V/3.2Ω = -3.59A Negative since following direction of ξ1 i2 4.0V b) What is the size and direction of current i2? (Take upward to be positive.) Focus loop on center rectangle Move around loop, CW starting at battery. i2 3.59A ½ i2 i2 4.0V 4V + 3.59(3.2) – i2 (3.2) – (i2/2)(3.2) – i2 (3.2) = 0 i2 =1.94A, down c) At what rate is energy being transferred at the 4.0V battery AND is the battery supplying or absorbing energy? P = I2V2 = (1.94A)(4V) = 7.76W Supplying since current is moving from – to + through the battery using conventional current as default. We expect current to move from + plate to the circuit. 4.0V Kirchoff worksheet Get out RC demo board RC CIRCUITS Often RC circuits are used to control timing. Some examples include windshield wipers, strobe lights, and flashbulbs in a camera, some pacemakers. Initially, at t=0, at the instant a switch closes there is a potential difference of zero across an uncharged capacitor (it acts like a wire…short circuit). As time passes, the capacitor reaches a maximum charge where I = 0 in capacitor (it acts like an open circuit). Resistance is infinite along line of R and C. 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 ξ 1) At instant switch is connected to ‘a’, C starts to charge. I is a maximum in circuit at t = 0, given by ξ/R. 2) As C charges, VC increases as VR decreases, keeping ΣV = 0 around loop (KVL). Deriving V(t) and q(t) for RC circuit using Kirchoff’s Loop Rule. We will move CW (after closing switch) around loop: ξ – iR – q/C = 0 ( at t = 0, q = 0, & i = ε / R ) ξ – (dq/dt)R – q/C = 0 (since i = dq/dt) Rearranging and dividing by R yields dq/dt = -q/RC + ξ / R dq/dt = (-q + ξC ) / RC dq/dt = - 1/RC(q - ξC ) Multiplying last term by C/C and combining yields Factoring out -1/RC dq/dt = - 1/RC(q - ξC ) q C e C dq 1 dt q C RC q 0 dq 1 q C RC Using u –sub (u=q-ξC) for left side yields q C t ln C RC Exponentiating both sides we get ( t ) RC t dt FINALLY! 0 q (t ) C (1 e VC (1 e t t RC RC ) ) Current, i, as function of time q (t ) C (1 e t RC ) q (t ) C Ce t RECALL RC dq t 1 RC 0 (Ce )( ) dt RC dq t RC ( )e dt R i ( )e R t RC d at e ae at dt 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. 1 τ = 1RC. Note: R needs to be in series in some way with C for there to be an effect on the time constant for charging. (explain later) It can be shown that after 1 time constant (RC), VC is 63% of its maximum voltage, Vo. If t = RC, then Vc = Vo (1 – e-(t/RC)) = Vo (1 - e-1) = Vo (1- 0.37) = 0.63Vo 0.63 Vmax 0.37 Imax 1 time constant, RC Discharging After a very long time, the capacitor would be fully charged. If the switch was then moved to ‘b’, ξ i The capacitor would discharge through the resistor as a function of time similar to the previous derivation. Voltage across resistor equals voltage across capacitor at all times in above circuit. Using the loop rule from before where at t = 0, q = Qo ε – iR – q/C = 0 , but ε = 0 (no battery) – (dq/dt)R – q/C = 0 dq q dt RC q t dq 1 Q q RC 0 dt o q t ln Qo RC q t RC e Qo q Qo e t RC dq Qo t RC i ( )e dt RC Neg indicates reverse direction for discharge When discharging, C is now the power source, so the time constant is C multiplied by the equivalent resistance of remainder of circuit. Example 2R Both switches are initially open, and the capacitor is uncharged. What is the current through the battery just after switch S1 is closed? a) Ib = 0 ε b) Ib = ε / (3R) c) Ib = ε /(2R) d) Ib = ε / R What is the current through the battery after switch 1 has been closed a long time? C S1 R S2 Because C acts like a short a) Ib = 0 b) Ib = V/(3R) c) Ib = V/(2R) d) Ib = V/R Because C acts like R = Both switches are initially open, and the capacitor is uncharged. What is the current through the battery just after switch S1 & S2 are closed? a) Ib = 0 b) Ib = ε / (3R) 2R ε c) Ib = ε /(2R) d) Ib = ε / R C S1 After a long time what is the current through the battery? a) Ib = 0 After a long time S1 is opened. What is the voltage across R and 2R after 2τ? c) Ib = ε /(2R) b) Ib = ε /(3R) d) Ib = ε /R V2R = 0, VR = 0.13(VC) = 0.13 (ε/3). Capacitor only had 1/3 voltage across it…in parallel with R. R S2 S Example a) Find VR2 & VR1 after S has been closed for 1τ. 12V R2 Initially, each branch gets 12V since they are in parallel so….VR2 = 12V always, unaffected by R1C R1 C Using loop rule on right moving CCW: – VR2 + VC + VR1 = 0 C is neg on left plate -12V + (0.63(12V)) = - VR1 VR1 = 4.4V b) Find total current at this time (1 τ ) if R1 = 10Ω and R2 = 20Ω 12V 4.4V iR1 ( ) 0.44 A 10 S R2 R1 C I R 2 ( ) 0.60 A R Itotal = 1.04A Could also use this eqn to get iR1 i ( )e R t RC 1 (12 / 10)(e ) 0.44 A Example Each circuit below has a 1.0F capacitor charged to 100V. When the switch is closed: a) Which system will be brightest? b) Which lights will stay on longest? 2…each gets full current Time constant of 1 is 4x as long…2RC vs RC/2 c) Which lights consumes more energy assuming we wait until both can’t be seen? E = Pt… Circuit 2 is 4x the power (2V2/R vs V2/2R) but lasts ¼ as long as 1 1MΩ 3MΩ 18V 6MΩ 1µF A capacitor is initially uncharged and then the switch is closed. a) At t = 0, when switch is closed, find the current in each resistor. b) Find these currents after a long time later. 1MΩ 3MΩ 18V 6MΩ 1µF Initially, C acts as a short but it is in series with 3MΩ . REQ = (6-1 + 3-1)-1 + 1 = 3MΩ iT = 18V / 3MΩ = 6uA i1 = 6uA i3 = 4uA i6 = 2uA 1MΩ 3MΩ 18V b) After a long time, C has infinite resistance… 6MΩ 1µF i3Ω = 0 Use KVL on outside loop… 18 - i1 (1MΩ) – i1 (6MΩ) = 0 i1 = 2.57uA After a terribly long time, the switch is opened. 1MΩ 3MΩ 18V 6MΩ 1µF c) Find the voltage drop on the capacitor just after the switch is opened. Circuit will just be right hand loop Find V6Ω first before switch was opened. i6Ω = 2.57uA, so V6Ω = (2.57)(6) = 15.4V Since C and 6MΩ were in parallel, they have same V, therefore, VC = 15.4V. 1MΩ 3MΩ 18V 6MΩ 1µF d) Find the charge on the capacitor 18s after the switch is opened. 9MΩ is in series with C, so τ = RC = 9s qo = CVo = (1uF )(15.4V) = 15.4uC q(t) = qoe-t/RC = 15.4e-18/9 q = 2.1uC e) How much energy is consumed by the 6MΩ during the discharging process? UC = ½ CVo2 UC = 119uJ Since P = I2R, the energy dissipated is proportional to R, therefore E = 6/9 (119uJ) = 79.3uJ QUESTION: For the circuit shown, ξ = 6V, R1= 7Ω, R2 = 3 Ω, R3 = 6Ω, and R4 =12 Ω . After operating for a long time, equilibrium is established. What is the voltage across the capacitor? Use Kirchoff loop rule. Current in left branch will be 6/10, current in right branch will be 6/18 which gives voltage for R1 and R3 . Top loop is 4.2V -2V = 2.2V=VC