General Physics (PHY 2140)

Size: px

Start display at page:

Download "General Physics (PHY 2140)"

Reynard Peters
5 years ago
Views:

General Physics (PHY 140) Lecture 6 lectrodynamics Direct current circuits parallel and series connections Kirchhoff s rules circuits Hours of operation: Monday and Tuesday Wednesday and Thursday

10:00 M to 5:00 PM 10:00 M to 4:00 PM losed Undergraduate students taking PHY100, PHY130, PHY140, PHY170/175 and PHY180/185 will be able to get assistance in this enter with their homework, labwork

1 General Physics (PHY 140) Lecture 6 lectrodynamics Direct current circuits parallel and series connections Kirchhoff s rules circuits Hours of operation: Monday and Tuesday Wednesday and Thursday Friday, Saturday and Sunday Department of Physics and stronomy announces the SpringSummer 007 opening of The Physics esource enter on Monday, May 1 in oom 17 of Physics esearch Building. 10:00 M to 5:00 PM 10:00 M to 4:00 PM losed Undergraduate students taking PHY100, PHY130, PHY140, PHY170/175 and PHY180/185 will be able to get assistance in this enter with their homework, labwork and other issues related to their physics course. The enter will be open: Monday, May 1 to Thursday, ugust, hapter 18 1 Last lecture: 1. urrent and resistance Lightning eview Temperature dependence of resistance Power in electric circuits. D ircuits MF esistors in series ΔQ t = Δ = 1 ( ) o α T To ( Δ ) P= Δ = = Δ = r = ntroduction: elements of electrical circuits branch: branch is a single electrical element or device (resistor, etc.). b b b b b circuit with 5 branches. junction: junction (or node) is a connection point between two or more branches. b b b circuit with 3 nodes. f we start at any point in a circuit (node), proceed through connected electric devices back to the point (node) from which we started, without crossing a node more than one time,, we form a closedpath (or loop)

2 18.1 Sources of MF Steady current (constant in magnitude and direction) ruires a complete circuit path cannot be only resistance cannot be only potential drops in direction of current flow lectromotive Force (MF) provides increase in potential converts some external form of energy into electrical energy Single emf and a single resistor: emf can be thought of as a charge pump = MF ach real battery has some internal resistance B: potential increases by on the source of MF, then decreases by r (because of the internal resistance) Thus, terminal voltage on the battery Δ is Δ = r r B D = = Note: is the same as the terminal voltage when the current is zero (open circuit) 5 6 MF (continued) Now add a load resistance Since it is connected by a conducting wire to the battery terminal voltage is the same as the potential difference across the load resistance Δ = r =, or = r Thus, the current in the circuit is = r r B Power output: = r D Measurements in electrical circuits oltmeters measure Potential Difference (or voltage) across a device by being placed in parallel with the device. mmeters measure current through a device by being placed in series with the device. Note: we ll assume r negligible unless otherwise is stated 7 8

3 Direct urrent ircuits Two Basic Principles: onservation of harge onservation of nergy esistance Networks Ohm s Law: ab = ab a b 18. esistors in series v i 1 1 v 1 By definition, v Thus, would be B 1. Because of the charge conservation, all charges going through the resistor will also go through resistor 1. Thus, currents in 1 and are the same, Δ 1 = =. Because of the energy conservation, total potential drop (between and ) uals to the sum of potential drops between and B and B and, Δ = 1 Δ = 1 = = 1 = esistors in series: notes esistors in series: example nalogous formula is true for any number of resistors, n the electrical circuit below, find voltage across the resistor r 1 in terms of the resistances 1, and potential difference between the battery s terminals. = (series combination) t follows that the uivalent resistance of a series combination of resistors is greater than any of the individual resistors v B v i 1 1 v 1 nergy conservation implies: = 1 = and = with 1 1 = 1,so = Then, ( ) Thus, 1 1 = 1 1 This circuit is known as voltage divider

4 18.3 esistors in parallel By definition, Thus, would be 1 1 = = = = = 1. Since both 1 and are connected to the same battery, potential differences across 1 and are the same, = =. Because of the charge conservation, current, entering the junction, must ual the current leaving this junction, = 1 esistors in parallel: notes nalogous formula is true for any number of resistors, = (parallel combination) t follows that the uivalent resistance of a parallel combination of resistors is always less than any of the individual resistors or 1 = esistors in parallel: example n the electrical circuit below, find current through the resistor 1 in terms of the resistances 1, and total current induced by the battery. 1 1 harge conservation implies: = 1 with 1 =,and = 1 1 Then, 1 =,with = 1 1 Direct current circuits: example Find the currents 1 and and the voltage x in the circuit shown below. 0 7 Ω x 1 4 Ω 1 Ω Strategy: 1. Find current by finding the uivalent resistance of the circuit. Use current divider rule to find the currents 1 and 3. Knowing, find x. Thus, 1 = 1 This circuit is known as current divider

5 Direct current circuits: example Find the currents 1 and and the voltage x in the circuit shown below. 0 7 Ω x 1 4 Ω 1 Ω Then find current by, First find the uivalent resistance seen by the 0 source: 4 Ω(1 Ω) = 7Ω = 10 Ω 1Ω 4Ω 0 0 = = = 10Ω We now find 1 and directly from the current division rule: (4 Ω) 1 = = 0.5, and = 1 = 1.5 1Ω 4Ω Finally, voltage x is ( ) ( ) x = 4Ω = 1.5 4Ω = 6 17 xample: Determine the uivalent resistance of the circuit as shown. Determine the voltage across and current through each resistor. Determine the power dissipated in each resistor Determine the power delivered by the battery =18 1 =4Ω =3Ω 3 =6Ω 18 xample: Determine the uivalent resistance of the circuit as shown. Determine the voltage across and current through each resistor. Determine the power dissipated in each resistor Determine the power delivered by the battery =18 =18 1 =4Ω =3Ω 3 =6Ω =6Ω 1 =4Ω 3 =Ω So, = / =3 P= = 108W 18.4 Kirchhoff s s rules and D currents The procedure for analyzing complex circuits is based on the principles of conservation of charge and energy They are formulated in terms of two Kirchhoff s s rules: 1. The sum of currents entering any junction must ual the sum of the currents leaving that junction (current or junction rule).. The sum of the potential differences across all the elements around any closedcircuit circuit loop must be zero (voltage or loop rule)

6 1. Junction rule s a consuence of the Law of the conservation of charge, we have: The sum of the currents entering a node (junction point) ual to the sum of the currents leaving. d a c b a b = c d Similar to the water flow in a pipe. a, b, c, and d can each be either a positive or negative number.. Loop rule s a consuence of the Law of the conservation of energy, we have: The sum of the potential differences across all the elements around any closed loop must be zero. 1. ssign symbols and directions of currents in the loop f the direction is chosen wrong, the current will come out with the correct magnitude, but a negative sign (it s s ok).. hoose a direction (cw( or ccw) for going around the loop. ecord drops and rises of voltage according to this: f a resistor is traversed in the direction of the current: = f a resistor is traversed in the direction opposite to the current: = f MF is traversed from to : f MF is traversed from to : 1 Simplest Loop ule xample: Loop rule: more complex illustration Single loop, start at point Battery traversed from to So use (a voltage gain) esistor traversed from to So use (a voltage drop) For the loop we have: 0 = B D Loops can be chosen arbitrarily. For example, the circuit below contains a number of closed paths. Three have been selected for discussion. Suppose that for each element, respective current flows from to signs. v 1 v 1 v v 3 v 10 v4 v 5 v 7 v 6 v 8 Path 1 Path Path 3 3 v 11 v 9 4 6

7 Loop rule: illustration Kirchhoff s s ules: : Singleloop loop circuits b v v 5 v 1 v 4 v 6 v 1 v 3 v 10 v 7 a v 8 Using sum of the drops = 0 Blue path, starting at a v 7 v 10 v 9 v 8 = 0 ed path, starting at b v v 5 v 6 v 8 v 9 v 11 v 1 v 1 = 0 xample: For the circuit below find, 1,, 3, 4 and the power supplied by the 10 volt source Ω 3 15 Ω 40 Ω 5 Ω For convenience, we start at point a and sum voltage drops =0 in the direction of the current = 0 (1). We note that: 1 = 0, = 40, 3 = 15, 4 = 5 () "a" v 11 v 9 Yellow path, starting at b v v 5 v 6 v 7 v 10 v 11 v 1 v 1 = We substitute the above into q. 1 to obtain q. 3 below = 0 (3) Solving this uation gives, = Kirchhoff s s ules: : Singleloop loop circuits (cont.) "a" Using this value of in q. gives: 0 Ω 1 = 10 3 = Ω 40 Ω 5 Ω = 0 4 = P 10(supplied) = 10 = 5 W 18.5 circuits When switch is closed, current flows because capacitor is charging harge across capacitor Q 0.63 Q (We use the minus sign in 10 because the current is entering the terminal) n this case, power is being absorbed by the 10 volt supply. 7 s capacitor becomes charged, the current slows because the voltage across the resistor is c and c gradually approaches. Once capacitor is charged the current is zero is called the time constant ( 1 ) q = Q e t 8 7

Discharging the capacitor in circuit xample : charging an unknown capacitor f a capacitor is charged and the switch is closed, then current flows and the voltage on the capacitor gradually decreases.

8 Discharging the capacitor in circuit xample : charging an unknown capacitor f a capacitor is charged and the switch is closed, then current flows and the voltage on the capacitor gradually decreases. harge across capacitor Q 0.37Q series combination of a 1 kωk resistor and an unknown capacitor is connected to a 1 battery. One second after the circuit is completed, the voltage across the capacitor is 10. Determine the t capacitance of the capacitor. This leads to decreasing charge q = Qe t 9 30 series combination of a 1 kωk resistor and an unknown capacitor is connected to a 1 battery. One second after the circuit is completed, the voltage across the capacitor is 10. Determine the capacitance of the capacitor. Given: =1 kω Ε =1 =10 t = 1 sec Find: =? ecall that the charge is building up according to t q = Q( 1 e ) Q = Thus the voltage across the capacitor changes as q Q 1 t t = = e = 1 e ( ) ( ) This is also true for voltage at t = 1s after the switch is closed, t = 1 e t e = 1 t = log 1 t 1s = = = 46.5μF log 1 10 ( 1,000 ) log 1 Ω 1 31 hapter 19: Magnetism Magnetic field pattern for a horseshoe magnet 3 8

Chapter 28. Direct Current Circuits

Chapter 28. Direct Current Circuits Chapter 28 Direct Current Circuits Circuit Analysis Simple electric circuits may contain batteries, resistors, and capacitors in various combinations. For some circuits, analysis may consist of combining