Lecture 18 (Feb. 23, 2009): Chapter 27: Circuits (cont d) Analyzing a circuit with resistors in combination <a href="/keyword/illustrative-examples/" >illustrative examples</a> Analyzing multiloop circuits using Kirchhoff s Rules (conservation of charge and energy) <a href="/keyword/illustrative-examples/" >illustrative examples</a> Resistors in series have the same current passing though all resistors; therefore V = V1 + V2 + V3 + = I*R1 + I*R2 + I*R3 + = I*(R1 + R2 + R3 + ) = I*Req Req = R1 + R2 + R3 + Resistors in parallel have the same voltage across them; therefore I = I1 + I2 + I3 + = V/R1 + V/R2 + V/R3 + = V*(1/R1 + 1/R2 + 1/R3 + ) = V/Req (1/Req) = (1/R1 + 1/R2 + 1/R3 + ) Illustrative Example 18.1: Steps in reducing a combination of resistors to a single equivalent resistor and finding the current in each: 1/(6 ) + 1/(3 ) = 1/(2 ) 4 + 2 = 6 18V/6 = 3A 6V/6 = 1A 6V/3 = 2A 3A*4 = 12V 3A*2 = 6V Illustrative example 18.2: Two identical light bulbs in series and in parallel: Req = 4 I = 2A Itotal = I1+I2 I1 Req = 1 Itotal = 8A I1 = I2 = 4A I2 Illustrative Example 18.3: The diagram shows a combination of five identical 5.00 resistors. Determine the equivalent resistance of this combination (a) if a battery is connected between points F and H; (b) if a battery is connected between points F and G. For the answers see the next slide (a) Redraw the circuit if voltage is applied between F and H: F R R H (a) Redraw the circuit if voltage is applied between F and G: F R R G R H R R The 3 resistors on the right combine to: R G R 1 11 1 4 Req R 2 R 2 R 2 R R Req 2 R Combine them with the other 2 resistors: 1 11 3 Req1 R 2 R 2 R Req1 2R 3 1 1 1 8 Req R ( R 2 3 R ) 5 R Req 5R 8 Kirchhoff s Rules: Many practical resistor networks cannot be reduced to simple series-parallel combinations. Here are two examples. Use Kirchhoff s Rules to solve: (1) Junction Rule (valid at any junction): I 0 conservation of charge (2) Loop Rule (valid for any closed loop): V 0 conservation of energy Illustrative Example 18.4: This is a multiloop circuit with three resistors in which the potential across the resistors is NOT the same because there are batteries inserted between points a and b and between b and c. How do we find the currents through each of the resistors? Apply the junction rule to relate currents; Apply the loop rule to relate potentials Assume you know values for the two emf s and the three resistors. Step 1: identify the currents and assign to each of them an arbitrary direction. (If your numerical answer for a current comes out negative, it means that it flows in the opposite direction from the one you picked.) Step 2: Apply the Junction Rule; at junction b it leads to the equation i2 = i1 + i3 (Note that junction d yields the same equation since the same three currents are involved.) Step 3: Apply the loop rule to the three possible loops left (badb), right (bdcb), and big (badcb). The resulting equations are: left loop (ccw from b): E1 i1 R1 i3 R3 0 right loop (ccw from b): i3 R3 i2 R2 E2 0 big loop (ccw from b): E1 i1 R1 i2 R2 E2 0 Note: These three equations are not independent; you can combine any two and obtain the third one. But with the junction equation you now have three equations for the three unknown currents. The rest is algebra. The sign conventions to follow when applying Kirchoff s Loop Rule while going around a circuit loop: + Solve this problem for the following set of values: E1=8.0V, E2=12V; R1=10 , R2=15 , R3=20 i2 i1 i3 1 E1 i1 R1 i3 R3 0 i3 i1 R1 E1 0.50i1 0.40 R3 1 4 E1 i1 R1 i2 R2 E2 0 i2 E1 E2 i1R1 15 2 i1 3 R2 4 15 2 i1 i1 0.50i1 0.40 which yields i1 61.5mA 3 with these expressions for i3 and i2 the junction equation becomes: Now solve for i2 and i3 : i2 307.7mA and i3 369.3mA