Lecture 7 Circuits Ch. 27PowerPoint PresentationDirect Current CircuitsExample (Single Loop Circuit)Slide 5Example with numbersExample with numbers (continued)Multiloop CircuitsMethod of determinants for solving simultaneous equationsSlide 10Another exampleRules for solving multiloop circuits3 bulb questionWhen the bulb #2 is not burnt out:When the bulb #2 is burnt out:How does a capacitor behave in a circuit with a resistor?Discharging a capacitor through a resistorWhat is the current I at time t?What is the current?How the charge on a capacitor varies with time as it is being chargedInstrumentsOhmmeterAmmeterVoltmeterLecture 7 Circuits Ch. 27•Cartoon -Kirchhoff's Laws• Warm-up problems•Topics–Direct Current Circuits–Kirchhoff's Two Rules–Analysis of Circuits Examples–Ammeter and voltmeter–RC circuits•Demos–Three bulbs in a circuit–Power loss in transmission lines–Resistivity of a pencil–Blowing a fuseTransmission line demoDirect Current Circuits1. The sum of the potential drops around a closed loop is zero. This follows from energy conservation and the fact that the electric field is a conservative force.2. The sum of currents into any junction of a closed circuit must equal the sum of currents out of the junction. This follows from charge conservation.Example (Single Loop Circuit)No junction so we don’t need that rule.How do we apply Kirchoff’s rule?Must assume the direction of the current – assume clockwise.Choose a starting point and apply Ohm’s Law as you go around the circuit.a. Potential across resistors is negativeb. Sign of E for a battery depends on assumed current flowc. If you guessed wrong on the sign, your answer will be negativeStart in the upper left hand corner.213212111322210rrRRREEiirEiRirEiRiR++++−==−+−−−−−2132121rrRRREEi++++−=Now let us put in numbers. VEVErrRRR5101102121321==Ω==Ω===Suppose:32511101010510=Ω++++−=ViampVEVE10521==Suppose:32532)105(−=Ω−=ViampWe get a minus sign. It means our assumed direction of current must be reversed.Note that we could have simply added all resistors and get the Req. and added the EMFs to get the Eeq. And simply divided.325)(32)(5R.e.=Ω==VEiqeqampSign of EMFBattery 1 current flows from - to + in battery +E1Battery 2 current flows from + to - in battery -E2In 1 the electrical potential energy increasesIn 2 the electrical potential energy decreasesExample with numbersQuick solution:AEIRVVVVEqeqiiii1610R16102412.e.6131==Ω==+−=∑∑==Question: What is the current in the circuit?AampsViiV625.0625.016100)311551()2412(==Ω==Ω+++++−+−+Write down Kirchoff’s loop equation.Loop equationAssume current flow is clockwise.Do the batteries first – Then the current.Example with numbers (continued)Question: What are the terminal voltages of each battery?V375.111A625.0V12irV =Ω⋅−=−=εV375.11A625.0V2irV =Ω⋅−=−=εV625.41A625.0V4irV =Ω⋅+=−=ε2V:4V:12V:Multiloop CircuitsKirchoff’s Rules1. in any loop2. at any junction0=∑iiV∑∑=outiniiFind i, i1, and i2Rule 1 – Apply to 2 loops (2 inner loops)a.b.Rule 2a.045203412121=+−−=−−iiii21iii +=We now have 3 equations with 3 unknowns.0245037120)(34122121121=−+−=−−=+−−iiiiiii0612150614242121=−+−=−−iiiiAiAiAii0.25.05.1263902639211=====−multiply by 2multiply by 3subtract themFind the Joule heating in each resistor P=i2R.Is the 5V battery being charged?Method of determinants for solving simultaneous equations524012043021121=−+−=+−−=−−iiiiiiiCramer’s Rule says if :333231323222121312111dicibiadicibiadicibia=++=++=++Then,3332221113332221111cbacbacbacbdcbdcbdi =3332221113332221112cbacbacbacdacdacdai =3332221113332221113cbacbacbadbadbadbai =i =0 −1 −1−12 −4 05 +4 −21 −1 −1−3 −4 00 +4 −2=0−4 04 −2⎛⎝⎜⎞⎠⎟−10 −12−2 5⎛⎝⎜⎞⎠⎟−1−12 −45 4⎛⎝⎜⎞⎠⎟1−4 04 −2⎛⎝⎜⎞⎠⎟−10 −3−2 0⎛⎝⎜⎞⎠⎟−1−3 −40 4⎛⎝⎜⎞⎠⎟=24 +48 −208 +6 +12=5226=2AYou try it for i1 and i2.See inside of front cover in your book on how to use Cramer’s Rule.For example solve for iMethod of determinants using Cramers Rule and cofactorsAlso use this to remember how to evaluate cross products of two vectors.Another exampleFind all the currents including directions.211211358023380234480iiiiiiiVVV−−=−−−=−−−++=012120010166024611212=+−=−+−=++−iiiii0)1(2462=++− AiLoop 1Loop 2Ai 11=AiAi212==Loop 1Loop 2i iiii1i2i221iii +=024612=++− iiMultiply eqn of loop 1 by 2 and subtract from the eqn of loop 2Rules for solving multiloop circuits1. Replace series resistors or batteries with their equivalent values.2. Choose a direction for i in each loop and label diagram.3. Write the junction rule equation for each junction.4. Apply the loop rule n times for n interior loops.5. Solve the equations for the unknowns. Use Cramer’s Rule if necessary.6. Check your results by evaluating potential differences.3 bulb questionThe circuit above shows three identical light bulbs attached to an ideal battery. If the bulb#2 burns out, which of the following will occur?a) Bulbs 1 and 3 are unaffected. The total light emitted by the circuit decreases.b) Bulbs 1 and 3 get brighter. The total light emitted by the circuit is unchanged.c) Bulbs 1 and 3 get dimmer. The total light emitted by the circuit decreases.d) Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit is unchanged.e) Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit is unchanged.f) Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit decreases.g) Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit decreases.h) Bulb 1 is unaffected, but bulb 3 gets brighter. The total light emitted by the circuit increases.i) None of the above.When the bulb #2 is not burnt out:R232RRReq=+=R3V2RVI231==RIP,Power2=RVI =For Bulb #1For Bulb #2For Bulb #3RV44.R9V4RIP22211===R3V2II12==RV11.R9VRIP22222===R3V2II13==RV11.R9VRIP22233===1I2II12=2II13=R2RRReq=+=R2VI1=RIP,Power2=RVI =For Bulb #1For Bulb #2For Bulb #3RV25.R4VRIP22211===0I2=0RIP222==R2VII13==RV25.R4VRIP22233===1I13II =So, Bulb #1 gets dimmer and bulb #3 gets brighter. And the total power decreases.f) is the answer.Before total power wasRV66.RVRVP2232eq2b===RV50.R2VRVP22eq2a===After total power is When the bulb #2 is burnt out:How does a capacitor behave in a circuit with a