Physics 227: Lecture 11 Circuits, KVL, KCL, Meters

Transcription

1 Physics 227: Lecture 11 Circuits, KVL, KCL, Meters Lecture 10 review: EMF ξ is not a voltage V, but OK for now. Physical emf source has V ab = ξ - Ir internal. Power in a circuit element is P = IV. For a resistor with V = IR, P = IV = I2 R = V 2 /R. Light bulbs are rated in Watts, for fixed V over them. Higher power bulbs have lower resistance, do brightness of bulbs in series is opposite brightness of bulbs in parallel.

2 Definitions Node A node or junction: a place where 2 or more wires meet. Often shown in drawings as a solid circle. Elements (resistors, capacitors,...) in parallel: elements between the same two notes, necessarily have the same voltage over them. Elements in series: elements connected with a wire, with no nodes between them, so that they necessarily have the same current through them. E 2 E 1 and E 2 in series E 1 and E 2 in parallel E 1 E 1 E 2

3 Resistor iclicker Which of the following is true about the circuit shown? A. R1 is in parallel with R2. B. R1 is in parallel with R3. C. R2 is in parallel with R3. D. R1 is in series with R2 and/or R3. E. R2 is in series with R3.

4 Resistor iclicker Which of the following is true about the circuit shown? A. R1 is in parallel with R2. B. R1 is in parallel with R3. C. R2 is in parallel with R3. D. R1 is in series with R2 and/or R3. E. R2 is in series with R3. R2 and R3 are in series - they are connect with a wire and have the same current. No other resistors are in series. No resistors are in parallel - no two connect the same two nodes and necessarily have the same voltage.

5 KVL and KCL Two basic ideas for analyzing circuits: No charge buildup in a circuit: If wires meet, the sum of the currents into (or out of) a node is 0. i I i = 0. I 1 I 2 I 3 ``Kirchoff s Current Law If you go around a loop, the voltage is the same. You return to a position with the same potential and potential energy. i V i = 0. ``Kirchoff s Voltage Law V 4 V 3 V 1 V 2

6 Resistors in series KCL: there is the same current I through each of the resistors. V = V 1 + V 2 + V 3 = I(R 1 + R 2 + R 3 ). e.g.: V 1 / V = [ R 1 / (R 1 + R 2 + R 3 ) ]. We can replace several resistors in series with one equivalent resistor: R eq = R 1 + R 2 + R Resistors in series add. This is opposite the case for capacitors.

7 Resistors in parallel KVL: there is the same voltage V across each of the resistors. The total current I is split between the resistors. I = I 1 + I 2 + I 3 = V/R 1 + V/R 2 + V/R 3 = V(1/R 1 + 1/R 2 + 1/R 3 ). e.g.: I 1 / I = [ 1/R 1 / (1/R 1 + 1/R 2 + 1/R 3 ) ] = (R 2 R 3 ) / (R 1 R 2 + R 2 R 3 + R 1 R 3 ) We can replace several resistors in parallel with one equivalent resistor: 1/R eq = 1/R 1 + 1/R 2 + 1/R Resistors in parallel add inversely. This is opposite the case for capacitors.

8 Simple Networks - series / parallel

9 Simple Networks - series / parallel

10 Simple Networks - series / parallel

11 Simple Networks - series / parallel

12 Less Simple Networks - applying KVL/KCL Sometimes they will be solved as n equations in n unknowns, or perhaps we can more ``artfully find easy quantities first. This circuit cannot be simplified using parallel / series, we need to use KVL / KCL. Define the currents going around loops - do not worry about the actual current direction, if you guess wrong you will get negative numbers. Note the current through an element may be the sum of two loop currents. E.g.: I through r2 = I loop2 - I loop3 As shown, only two of I loop1, I loop2, and I loop3 are independent.

13 Cicuitous iclicker A. I is impossible to figure out. What is I? B. I = 0 A. C. I = 0.5 A. D. I = (8/9) A. E. I = (8/3) A.

14 Cicuitous iclicker What is I? Apply KVL going CCW around the circuit from b I-3I-4I-4-7I = I = 0 I = 1/2 A. I is impossible to figure out. B. I = 0 A. C. I = 0.5 A. D. I = (8/9) A. E. I = (8/3) A.

15 Less Simple Networks - applying KVL/KCL What is I? Use KCL at node a: I = 1 A + 2 A = 3 A What is r? Use KVL around loop 1: -(2 A)(3 Ω) + 12 V - (3 A)r = 0 r = 2 Ω What is ξ? Use KVL around loop 2: -(2 A)(3 Ω) - ξ + (1 A)(1 Ω) = 0 ξ = -5 V

16 Voltmeters and Ammeters Old electro-mechanical systems largely based on forces between current carrying wires and magnetic fields - a topic for the near future. Modern system all ICs.

17 Voltmeters and Ammeters For a voltmeter, you want a large internal meter resistance. Any current flowing through the meter does not flow through the circuit element, and changes what is happening in the circuit. Rest of circuit If R voltmeter is made too small, most of current flows through it, and there is a greater voltage in the rest of the circuit. Similarly, for ammeter you want R small. R voltmeter R to measure Voltage over

18 Voltmeters and Ammeters Ammeters and Voltmeters have a dial to turn. This allows us to select the ratio of R c to R s, varying the current through the meter, so that we can change the magnitude of voltage / current that gives a full scale reading. E.g.: too much current, needle pegged at full scale, increase R to reduce current. Too little current, reading insensitive due to small needle deflection, decrease R to get more current and more needle deflection.

19 Voltmeters and Ammeters Which is the right way to measure the resistance of a resistor?

20 Voltmeters and Ammeters Use an ohmmeter so you do not have any corrections.

21 Current iclicker A constant current source sends a current I through a resistor R. A second resistor R (<< R) is connected in parallel to the first resistor. What happens to the current I through the first resistor? A. I increases. B. I decreases, but not to 0. C. I decreases to 0. D. I stays the same. E. I cannot figure out what happens.

22 Current iclicker A constant current source sends a current I through a resistor R. A second resistor R (<< R) is connected in parallel to the first resistor. What happens to the current I through the first resistor? R R A. I increases. B. I decreases, but not to 0. C. I decreases to 0. D. I stays the same. E. I cannot figure out what happens. With just R, we have V=I 0 R. Resistors in parallel act as a current divider. The current will split between the two so that V = IR = I R, and I+I = I 0, or V (1/R + 1/R ) = I 0.

23 Thank you. See you monday.