Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Lecture 7 Circuits Chp. 28 • • • • Cartoon -Kirchoff’s Laws Opening Demo- transmission lines Physlet Topics – Direct Current Circuits – Kirchoff’s Two Rules – Analysis of Circuits Examples – Ammeter and voltmeter – RC circuits • Demos – Ohms Law – Power loss in transmission lines – Resistivity of a pencil – Blowing a fuse • Warm-up problems Transmission line demo Direct Current Circuits 1. The sum of the potential charges 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. b. c. Potential across resistors is negative Sign of E for a battery depends on assumed current flow If you guessed wrong on the sign, your answer will be negative Start in the upper left hand corner. iR1 iR 2 E 2 ir 2 iR 3 E1 ir 1 0 E1 E 2 i R1 R 2 R3 r1 r 2 i E1 E 2 R1 R 2 R3 r1 r 2 Put in numbers. Suppose: R1 R 2 R 3 10 r1 r 2 1 E1 10V E 2 5V 10 5 5 i amp 10 10 10 1 1 32 Suppose: E1 5V E 2 10V amp 5 10 5 i 32 32 We 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. i Eeq. 5(V ) 5 amp Re q. 32() 32 Sign of EMF Battery 1 current flows from - to + in battery +E1 Battery 2 current flows from + to - in battery -E2 In 1 the electrical potential energy increases In 2 the electrical potential energy decreases Example with numbers Quick solution: 3 E 12V 4V 2V 10V i i 1 6 R 16 i i 1 I Eeq. 10 A Re q. 16 Question: What is the current in the circuit? Write down Kirchoff’s loop equation. Loop equation Assume current flow is clockwise. Do the batteries first – Then the current. (12 4 2) i (1 5 5 1 1 3) 0 10 V i 0.625amps 0.625 A 16 Example with numbers (continued) Question: What are the terminal voltages of each battery? 12V: 2V: 4V: V ir 12V 0.625A 1 11.375V V ir 2V 0.625A 1 1.375V V ir 4V 0.625A 1 4.625V Multiloop Circuits Find i, i1, and i2 We now have 3 equations with 3 unknowns. 12 4i1 3(i1 i 2) 0 12 7i1 3i 2 0 multiply by 2 Kirchoff’s Rules 1. in any loop V 2. i 0 i i i in at any junction out Rule 1 – Apply to 2 loops (2 inner loops) a. 12 4i1 3i 0 b. 2i 2 5 4i1 0 Rule 2 a. i i1 i 2 5 4i1 2i 2 0 multiply by 3 24 14i1 6i 2 0 subtract them 15 12i1 6i 2 0 39 26i1 0 39 1. 5 A 26 i 2 0. 5 A i 2.0 A i1 Find the Joule heating in each resistor P=i2R. Is the 5V battery being charged? Method of determinants for solving simultaneous equations i i1 i 2 0 3i 4i1 0 12 0 4i1 2i 2 5 For example solve for i 0 12 5 i 1 3 0 1 1 4 0 4 2 48 20 24 52 2A 1 1 8 12 6 26 4 0 4 2 You try it for i1 and i2. See Appendix in your book on how to use Cramer’s Rule. Another example Find all the currents including directions. Loop 2 i i i2 i1 i Loop 1 i Loop 1 0 8V 4V 4V 3i 2i1 0 8 3i1 3i 2 2i1 0 8 5i1 3i 2 multiply by 2 i = i1+ i2 i2 Loop 2 6i 2 4 2i1 0 6i 2 16 10i1 0 0 12 12i1 0 i1 1A 6i 2 4 2(1A) 0 i 2 1A i 2A Rules for solving multiloop circuits 1. 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. How does a capacitor behave in a circuit with a resistor? Charge capacitor with 9V battery with switch open, then remove battery. Now close the switch. What happens? Discharging a capacitor through a resistor Potential across capacitor = V = Qo C just before you throw switch at time t = 0. V(t) Potential across Resistor = iR Qo Qo i oR i o C RC at t > 0. What is the current I at time t? Q(t) i(t) RC Time constant = RC What is the current? Q Q0e t RC dQ Q0 i e dt RC RC t RC V0 e R t RC Ignore - sign How the charge on a capacitor varies with time as it is being charged Ohmmeter Ammeter Voltmeter Warm up set 7 Warm up set 7 Due 8:00 am Tuesday 1. HRW6 28.TB.05. [119859] In the context of the loop and junctions rules for electrical circuits a junction is: where where where where where a wire is connected to a battery three or more wires are joined a wire is bent a wire is connected to a resistor only two wires are joined 2. HRW6 28.TB.18. [119872] Two wires made of the same material have the same length but different diameter. They are connected in parallel to a battery. The quantity that is NOT the same for the wires is: the electric field the electron drift velocity the current the current density the end-to-end potential difference 3. HRW6 28.TB.26. [119880] The emf of a battery is equal to its terminal potential difference: only when there is no current in the battery only when a large current is in the battery under all conditions under no conditions only when the battery is being charged