Example: In the given circuit: (a) How much power is drawn from the battery? (b) How much current flows through each resistor? And in what direction?

Transcription

1 0.8 Circuits Wired Partially in Series and Partially in Parallel Example: n the given circuit: (a) How much power is drawn from the battery? (b) How much current flows through each resistor? And in what direction? n examining a circuit, you start by trying to find combinations of resistors that are in series and/or in parallel

2 0.8 Circuits Wired Partially in Series and Partially in Parallel 0Ω and 50Ω in series: R S 0Ω + 50Ω 470Ω R P 80Ω and 470Ω in parallel: Ω 80Ω 470Ω R P Ω 0Ω 0Ω and 0Ω in series: R S 0Ω + 0Ω 40Ω (a) P V V R (4V) 40Ω.4W

3 0.8 Circuits Wired Partially in Series and Partially in Parallel V 0 Ω 0 Ω V R 4V 0.0 A 40Ω 0.0 A (left to right) 00 Ω (0 (0.0 A)(0 V 0.0 A V AB V B V A 0 Ω (0 (0.0 A)(0 V 80 Ω 80 Ω 470 Ω VAB V 80 Ω 80 Ω 0.07 A (down) VAB 470 Ω V 470 Ω 0.08 A 0.0 A 0.07 A 0.0 A 0.07 A 0.08 A 0 Ω 50 Ω 0.08 A (down)

4 0.0 Kirchhoff s Rules n many circuits, especially ones involving loops and multiple voltage supplies: The resistors cannot be combined into series and parallel equivalents. For example, in the circuit to the left, the three resistors are clearly not in series, since they do not share the same current Here + by the junction rule.0 V They are also NOT in parallel The presence of the two batteries with different voltage makes it so that the voltage across the three resistors are NOT the same. We need a more general approach to solving battery-resistor networks: Kirchoff s Rules: Based on two basic principles (a) Conservation of charge (b) The potential well-defined and measuable (i.e. single-valued) at any point We also include the basic properties of the circuit elements 4

5 0.0 Kirchhoff s Rules KRCHHOFF S RULES (a) Junction Rule The total current directed into a junction must equal the total current directed out of the junction. 7A A + 5A mportant Point: For each continuous segment of the circuit (between branches), one should label a current with a presumed direction. This DOES NOT need to be the actual direction. After solving the circuit, some of these current will solve to positive values these are flowing in the presumed directions The currents that solve to negative values flow in the direction opposite of the presumed labeling. An analogy can be drawn from the flow of water in channels. The junctions in a circuit is equivalent to junctions in water ways. 5

6 0.0 Kirchhoff s Rules (b) Loop rule: Around any closed circuit loop, the sum of the potential drops equals the sum of the potential rises. The loop rule expresses conservation of energy in terms of the electric potential and states that for a closed circuit loop, the total of all potential rises is the same as the total of all potential drops. S Here we know the equivalent series resistance is R S R +R 6Ω So V/R V / 6Ω A n practice you go around a loop once sequentially Starting at a point on the circuit that is between circuit elements Here we start at S There are no junctions: there is only one current: We label the current counter-clockwise (in the direction of the battery). Traversing the circuit clock-wise: (a) Voltage drop along the current over 5Ω: V R (A)(5 0V (b) Voltage drop along the current over Ω: V R (A)( V (c) Voltage drop from to + side of a battery: V +E V Loop complete! 6 Note V + V + V 0

7 0.0 Kirchhoff s Rules Voltage rise or drop through a resistor or battery. n each case we are traversing from point a to b There are 4 cases: Resistor: along presumed current f we were solving this circuit, we d take the following steps (a) Label the current(s) (b) Eliminate some of the current(s) by enforcing junction rules (c) Determine how many loops we need: N number of remaining unknown, independent currents (d) Traverse through N different (independent) loops and write down the loop rule for each gives N linear equations in the N unknown currents Resistor: against presumed current Voltage source: From to + terminal Voltage source: From + to terminal 7

8 0.0 Kirchhoff s Rules Example: Using Kirchhoff s Loop Rule, determine the current in the circuit. Again, this is a one loop circuit, no junctions and branches We have single unknown current, which we presume to be counterclock-wise (CCW). This is an educated guess, because the 4V battery should push the current (of fictitious positive charge carriers) more CCW than the 6V battery tries to push it CW. Going through the loop starting at bottom left corner (point S) ( 6.0 V ( V 0 Rearranging : (0 + 8 V 0 (0 8 V 8 V 0 Ω 0.90 A (clockwise) 8

9 0.0 Kirchhoff s Rules Example: Using Kirchhoff s Rules, determine the power consumed by each of the resistors The circuit has branches and junctions We label currents as shown The presumed directions are motivated by the fact that the 5V battery will push more current right to left than the V in the opposite direction Apply Junction Rule: A B Junction A : Junction B : + + N N, + OUT, OUT +.0 V Notice these constraints are the SAME There will generally be one more junction than constraints on the currents (this is because we are dealing with closed circuit loops Now we choose to eliminate one current based on the one constraint: This is an arbitrary choice. We have to find the power consumed each resistor we have to find all three currents. We will eliminate because it goes through the fewest components. f for example you are only asked for the power in one of the resistors then you should eliminate first a different current (leads to faster solution). 9

10 0.0 Kirchhoff s Rules + A - Loop B We now apply Loop Rule There are more loops () than we need () for solving for the two independent currents: One outer loop and two inner loops There are generally more loops than we need. Loop We need to choose a combination of loops that will include EVERY COMPONENT We pick the two inner loops as shown.0 V Loop: traverse in counter - clock - wise direction starting at B + 5 V Loop : traverse in clock - wise direction starting at B + 5 V Rearranging () and () into the standard form, and dropping S units (where then the currents are solved in amps) : () () (.0 (.0 ( + + (.0.0 V 0 () )(.0 0 () (' ) 5 (' ) 0

11 0.0 Kirchhoff s Rules (' ) (' ) This is a x system of linear equations. f the problems asks for only, for example, the power consumed in the resistor on the V battery branch Then one should solve for first. Here we are free to choose: let us eliminate first. There are a few ways to do this: the one that can be applied to larger (say x) systems: Operate on equations: ( ) + ( ) + 4 (' ) + 5 (' ) ( V 9.0 V 5.0 Ω.8 A substituting intoequation (' ) (.0 + (.0.0 V.0 A -.8 A 0.0 A and.8 A 0.0 A.6 A A - Loop Loop.0 V t is useful to double check your results with a loop that was not used: the outerloop, clockwise starting at point B: +.0 V + (.0 ( V + (0.0 A)(.0 (.6 A)( V V. V 0 (correct!) top : P mid : P bot : P.0 Ω.0 Ω.0 Ω ( ) (.0 (.6 A) (.0 5. W ( ) (.0 (.8 A) ( W ( ) (.0 (0.0 A) ( W B

12 0.0 Kirchhoff s Rules (' ) (' ) This is a x system of linear equations. f the problems asks for only, for example, the power consumed in the resistor on the V battery branch Then one should solve for first. Here we are free to choose: let us eliminate first. There are a few ways to do this: the one that can be applied to larger (say x) systems: Operate on equations: ( ) + ( ) + 4 (' ) + 5 (' ) ( V 9.0 V 5.0 Ω.8 A substituting intoequation (' ) (.0 + (.0.0 V.0 A -.8 A 0.0 A and.8 A 0.0 A.6 A A - Loop Loop.0 V t is useful to double check your results with a loop that was not used: the outerloop, clockwise starting at point B: +.0 V + (.0 ( V + (0.0 A)(.0 (.6 A)( V V. V 0 (correct!) top : P mid : P bot : P.0 Ω.0 Ω.0 Ω ( ) (.0 (.6 A) (.0 5. W ( ) (.0 (.8 A) ( W ( ) (.0 (0.0 A) ( W B